Kinetic Theory — Practice & Self Evaluation

The kinetic theory of gases assumes that gas molecules:

1. are stationary
2. are in constant random motion
3. interact strongly with each other
4. lose energy during collisions

Answer

are in constant random motion

Reason — According to the first postulate of the kinetic theory of gases, gas molecules move in continuous, random motion in all directions. They travel in straight lines until they collide with other gas molecules or with the walls of the container. The molecules are therefore never stationary, they do not exert forces on one another except during collisions, and the collisions being perfectly elastic, no kinetic energy is lost in them.

The pressure of a gas is due to:

1. collisions of gas molecules with the walls of the container
2. attraction between gas molecules
3. repulsion between gas molecules
4. volume of the gas

Answer

collisions of gas molecules with the walls of the container

Reason — The molecules of a gas keep on colliding with the walls of the vessel and thus transfer momentum to them. By Newton's law of motion, this rate of change of momentum appears as a steady force acting perpendicular to the walls of the vessel. The magnitude of this force per unit area of the walls is called the pressure of the gas. The attraction or repulsion between the molecules is negligible except during collisions, so it plays no part in producing the pressure.

According to kinetic theory, the average kinetic energy of gas molecules is directly proportional to:

1. volume
2. temperature
3. pressure
4. number of molecules

Answer

temperature

Reason — According to the kinetic theory, the average kinetic energy of a gas molecule at absolute temperature T is

$\overline{\text E} = \dfrac{3}{2}\text{kT}$

where k is the Boltzmann constant. Since $\dfrac{3}{2}$ k is a constant, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas, and it does not depend on the volume, the pressure or the number of molecules.

The root mean square speed of gas molecules is proportional to:

1. the square root of the temperature
2. the square of the temperature
3. the temperature
4. the pressure

Answer

the square root of the temperature

Reason — From the kinetic theory, the root-mean-square speed of the molecules of a gas of molecular weight M at absolute temperature T is

$\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

Since 3R/M is a constant for a given gas, the root-mean-square speed of the molecules is directly proportional to the square-root of the absolute temperature of the gas. This is the kinetic interpretation of temperature.

If the temperature of a gas is doubled, the average kinetic energy of its molecules:

1. becomes half
2. remains the same
3. doubles
4. becomes four times

Answer

doubles

Reason — The average kinetic energy of a gas molecule is $\overline{\text E} = \dfrac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. If the absolute temperature is doubled,

$\dfrac{\overline{\text E}_2}{\overline{\text E}_1} = \dfrac{\text T_2}{\text T_1} = \dfrac{2\text T}{\text T} = 2$

Hence the average kinetic energy of the molecules also becomes twice its original value.

The temperature of an ideal gas is directly proportional to the :

1. potential energy of the gas
2. average kinetic energy of gas molecules
3. number of gas molecules
4. volume of the gas

Answer

average kinetic energy of gas molecules

Reason — According to the kinetic theory, the average kinetic energy of a molecule of an ideal gas is $\overline{\text E} = \dfrac{3}{2}\text{kT}$, which gives

$\text T = \dfrac{2\overline{\text E}}{3\text k}$

Thus the absolute temperature of a gas is directly proportional to the average kinetic energy of its molecules. This is called the kinetic interpretation of temperature. In an ideal gas there are no intermolecular forces, so there is no internal potential energy.

At absolute zero temperature, the molecules of an ideal gas:

1. have zero potential energy
2. have maximum kinetic energy
3. have zero kinetic energy
4. move at the speed of light

Answer

have zero kinetic energy

Reason — Since $\overline{\text E} = \dfrac{3}{2}\text{kT}$, the average kinetic energy of the molecules becomes zero when T = 0. Also $\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$ becomes zero at T = 0. Hence absolute zero is that temperature at which the motion of all the molecules of a gas stops. As the kinetic energy can never be negative, no temperature is possible below the absolute zero.

Boyle's law states that at constant temperature, the volume of a given mass of gas is:

1. directly proportional to pressure
2. inversely proportional to pressure
3. directly proportional to temperature
4. inversely proportional to temperature

Answer

inversely proportional to pressure

Reason — Boyle's law states that for a fixed amount of gas at constant temperature, the volume V of a gas is inversely proportional to its pressure P,

$\text P \propto \dfrac{1}{\text V} \quad \text{or} \quad \text P_1 \text V_1 = \text P_2 \text V_2$

Thus, at a constant temperature, if the pressure of a gas increases its volume decreases, and vice-versa.

Charles' law states that at constant pressure, the volume of a gas is:

1. inversely proportional to temperature
2. directly proportional to temperature
3. inversely proportional to pressure
4. independent of temperature

Answer

directly proportional to temperature

Reason — Charles' law states that the volume of a given mass of gas at constant pressure is directly proportional to its absolute temperature,

$\text V \propto \text T \quad \text{or} \quad \dfrac{\text V}{\text T} = \text{a constant}$

Thus, at constant pressure, a gas expands on heating and contracts on cooling.

The pressure law states that at constant volume, the pressure of a gas is:

1. inversely proportional to temperature
2. directly proportional to temperature
3. inversely proportional to volume
4. independent of temperature

Answer

directly proportional to temperature

Reason — The pressure law (Gay-Lussac's law) states that for a fixed amount of gas at constant volume, the pressure P of the gas is directly proportional to its absolute temperature T,

$\text P \propto \text T \quad \text{or} \quad \dfrac{\text P}{\text T} = \text{a constant}$

Thus, when the volume is kept constant, heating a gas raises its pressure in the same ratio as the absolute temperature.

The number of degrees of freedom for a monoatomic gas is:

1. 1
2. 3
3. 5
4. 7

Answer

3

Reason — The molecule of a monoatomic gas, such as helium or argon, consists of a single atom. Its translational motion can take place in any direction in space and can therefore be resolved along the three coordinate axes, giving three independent motions. Hence a monoatomic gas has three degrees of freedom, all translational. A monoatomic molecule can rotate also, but its moment of inertia is so small that the kinetic energy of rotation is insignificant.

A diatomic gas molecule has:

1. 3 degrees of freedom
2. 5 degrees of freedom
3. 6 degrees of freedom
4. 7 degrees of freedom

Answer

5 degrees of freedom

Reason — The molecule of a diatomic gas, such as H₂ or O₂, is made up of two atoms joined rigidly through a bond. It can not only move bodily, giving three translational degrees of freedom, but can also rotate about any of the three coordinate axes. However, its moment of inertia about the axis joining the two atoms is negligible, so it can have only two rotational motions. Hence a diatomic molecule has five degrees of freedom, three translational and two rotational. The vibrational motion does not usually occur at ordinary temperatures.

For a polyatomic gas, the degrees of freedom are:

1. 3
2. 5
3. 6
4. greater than 5

Answer

greater than 5

Reason — A polyatomic molecule, such as H₂O, H₂S or CO₂, can rotate about any of the three coordinate axes. Hence a polyatomic gas has six degrees of freedom, three translational and three rotational, which is greater than 5. If the atoms of the molecule vibrate with respect to each other, it would possess vibrational degrees of freedom also.

The law of equipartition of energy states that each degree of freedom contributes how much energy to the total energy of a system?

1. $\dfrac{1}{2}$ kT

2. kT

3. $\dfrac{3}{2}$ kT

4. 2kT

Answer

$\dfrac{1}{2}$ kT

Reason — According to the law of equipartition of energy, in a classical system of particles which is in equilibrium at absolute temperature T, the average internal (kinetic) energy per particle associated with each degree of freedom is $\dfrac{1}{2}$ kT, where k is Boltzmann's constant. If the particle has f degrees of freedom, its average kinetic energy would be $\dfrac{1}{2}$ f kT.

For a diatomic gas, the average energy per molecule is:

1. $\dfrac{3}{2}$ kT

2. $\dfrac{1}{2}$ kT

3. kT

4. $\dfrac{7}{2}$ kT

Answer

$\dfrac{5}{2}$ kT

Reason — By the law of equipartition of energy, the average energy per molecule is $\dfrac{1}{2}$ f kT, where f is the number of degrees of freedom. A diatomic gas molecule has f = 5, three translational and two rotational degrees of freedom. Hence the average energy per molecule of a diatomic gas is

$\overline{\text E} = \dfrac{1}{2}\text{fkT} = \dfrac{5}{2}\text{kT}$

Note: The value $\dfrac{5}{2}$ kT obtained above is not offered among the four options. The printed answer key marks option 2, that is, $\dfrac{1}{2}$ kT, which is the energy associated with one degree of freedom of a molecule and not the average energy of a diatomic molecule. The textbook's own explanation for this question also states the value as $\dfrac{5}{2}$ kT, so the options appear to be misprinted.*

Which of the following is an assumption of the kinetic theory of gases?

1. Gas molecules are strongly attracted to each other
2. Collisions between gas molecules are elastic
3. Gas molecules have high potential energy
4. All molecules have the same velocity

Answer

Collisions between gas molecules are elastic

Reason — One of the postulates of the kinetic theory is that the collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic, so that no kinetic energy is lost in the collisions. The other statements contradict the postulates, since gas molecules do not exert attractive or repulsive forces on each other except during collisions, they have negligible potential energy, and their speeds vary over a wide range.

The mean free path of gas molecules is defined as:

1. the distance between two gas molecules
2. the average distance a molecule travels before colliding with another molecule
3. the distance between two walls of the container
4. the total distance travelled by a molecule

Answer

the average distance a molecule travels before colliding with another molecule

Reason — The molecules of a gas move in straight lines with constant speeds between two successive collisions, so a particular molecule is found to follow short zig-zag paths of different lengths. These are called the 'free paths' of the molecule. The mean free path is the average distance travelled by a molecule between two successive collisions with other molecules.

The mean free path of a gas molecule is inversely proportional to:

1. the volume of the gas
2. the number density of the gas
3. the temperature of the gas
4. the molecular mass of the gas

Answer

the number density of the gas

Reason — The mean free path of a molecule is

$\lambda = \dfrac{1}{\sqrt2\pi \text n \text d^2}$

where n is the number of molecules per unit volume, that is, the number density, and d is the diameter of a molecule. Hence the mean free path is inversely proportional to the number density of the gas, which is why the mean free path varies inversely as the density of the gas.

According to the kinetic theory, the pressure exerted by a gas is due to:

1. molecular collisions with each other
2. molecular collisions with the walls of the container
3. the gravitational force between molecules
4. the temperature of the gas

Answer

molecular collisions with the walls of the container

Reason — The pressure exerted by a gas on the walls of its container is due to the collisions of the gas molecules with the walls. Each collision imparts a small force on the wall, and the cumulative effect of a very large number of such collisions results in the measurable pressure. The collisions of the molecules with one another do not contribute to the pressure on the walls.

The kinetic energy of one mole of an ideal gas at temperature T is given by:

1. $\dfrac{3}{2}$ RT

2. RT

3. $\dfrac{1}{2}$ RT

4. 2RT

Answer

$\dfrac{3}{2}$ RT

Reason — For 1 mole of a gas of molecular weight M, the kinetic theory gives $\text{PV} = \dfrac{1}{3}\text M\overline{\text v^2}$, and the ideal gas equation gives PV = RT. Hence the kinetic energy of 1 mole of the gas is

$\dfrac{1}{2}\text M\overline{\text v^2} = \dfrac{3}{2}\text{RT}$

Thus the kinetic energy of one mole of an ideal gas at absolute temperature T is $\dfrac{3}{2}$ RT.

If the pressure of a gas is doubled while keeping the temperature constant, its volume will:

1. remain unchanged
2. become half
3. become double
4. become one-fourth

Answer

become half

Reason — By Boyle's law, at constant temperature P₁V₁ = P₂V₂. Putting P₂ = 2P₁,

$\text V_2 = \dfrac{\text P_1 \text V_1}{\text P_2} = \dfrac{\text P_1 \text V_1}{2\text P_1} = \dfrac{\text V_1}{2}$

Hence, on doubling the pressure at constant temperature, the volume of the gas becomes half.

If the temperature of a gas is doubled at constant pressure, the volume will:

1. double
2. become half
3. remain unchanged
4. become four times

Answer

double

Reason — By Charles' law, at constant pressure $\dfrac{\text V_1}{\text T_1} = \dfrac{\text V_2}{\text T_2}$. Putting T₂ = 2T₁,

$\text V_2 = \text V_1 \times \dfrac{\text T_2}{\text T_1} = \text V_1 \times \dfrac{2\text T_1}{\text T_1} = 2\text V_1$

Hence, on doubling the absolute temperature at constant pressure, the volume of the gas also doubles.

If the temperature of a gas is doubled at constant volume, its pressure will:

1. double
2. remain the same
3. become half
4. become one-fourth

Answer

double

Reason — By the pressure law, at constant volume $\dfrac{\text P_1}{\text T_1} = \dfrac{\text P_2}{\text T_2}$. Putting T₂ = 2T₁,

$\text P_2 = \text P_1 \times \dfrac{\text T_2}{\text T_1} = 2\text P_1$

Hence, on doubling the absolute temperature at constant volume, the pressure of the gas doubles.

The internal energy of a monoatomic gas depends on:

1. the temperature only
2. the pressure only
3. the volume only
4. both pressure and volume

Answer

the temperature only

Reason — In an ideal gas there are no intermolecular forces, so there is no internal potential energy and the total internal energy of the gas is entirely the kinetic energy of its molecules. For μ moles of a gas having f degrees of freedom,

$\text U = \dfrac{1}{2}\mu \text{fRT}$

For a monoatomic gas f = 3, so $\text U = \dfrac{3}{2}\mu \text{RT}$. Thus, for a given amount of the gas, the internal energy depends only on the temperature and not on the pressure or the volume.

In an isothermal process, the temperature of the system:

1. increases
2. decreases
3. remains constant
4. fluctuates

Answer

remains constant

Reason — An isothermal process is one which is carried out at a constant temperature, and Boyle's law applies to such a process. For an ideal gas the internal energy depends only on the temperature, so in an isothermal process the internal energy also remains unchanged, and changes in pressure do not affect the energy of the gas directly.

The average velocity of gas molecules is proportional to:

1. $\sqrt{\text T}$
2. T
3. $1/\sqrt{\text T}$
4. 1/T

Answer

$\sqrt{\text T}$

Reason — The average speed of the molecules of a gas, like the root-mean-square speed, is proportional to the square-root of the absolute temperature of the gas,

$\text v \propto \sqrt{\text T}$

This follows from the kinetic theory, since $\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$ and the average speed differs from the rms speed only by a numerical factor.

The pressure of an ideal gas is proportional to:

1. volume
2. number of molecules
3. kinetic energy of molecules
4. square of the temperature

Answer

kinetic energy of molecules

Reason — From the kinetic theory, the pressure of a gas is

$\text P = \dfrac{1}{3}\rho\overline{\text v^2} = \dfrac{2}{3}\left(\dfrac{1}{2}\rho\overline{\text v^2}\right) = \dfrac{2}{3}\text E$

where $\text E \left(= \dfrac{1}{2}\rho\overline{\text v^2}\right)$ is the translational kinetic energy of the gas per unit volume. Thus the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume, that is, the pressure is proportional to the kinetic energy of the molecules.

A gas behaves most ideally at:

1. high pressure and low temperature
2. low pressure and high temperature
3. high pressure and high temperature
4. low pressure and low temperature

Answer

low pressure and high temperature

Reason — At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with it. At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible. Since these are exactly the two properties assumed for a perfect gas, the behaviour of a real gas is most nearly ideal at low pressure and high temperature.

If the pressure of a gas increases while keeping the temperature constant, the mean free path of the molecules will:

1. increase
2. decrease
3. remain the same
4. become zero

Answer

decrease

Reason — The mean free path of the molecules of a gas is

$\lambda = \dfrac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$

so at a constant temperature $\lambda \propto \dfrac{1}{\text P}$. On increasing the pressure the molecules are packed more closely together, so a molecule collides with another after covering a shorter distance and the mean free path decreases.

The pressure exerted by an ideal gas on the walls of a container is given by:

1. PV = nRT
2. P = $\dfrac{1}{3}$ ρv_rms²
3. PV = nRT / V
4. P = $\dfrac{\text F}{\text A}$

Answer

P = $\dfrac{1}{3}$ ρv_rms²

Reason — From the kinetic theory, the pressure exerted by an ideal gas on the walls of the container is

$\text P = \dfrac{1}{3}\dfrac{\text{mn}}{\text V}\overline{\text v^2}$

Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas. Hence

$\text P = \dfrac{1}{3}\rho \text v_{rms}^2$

The relation PV = nRT is the equation of state of the gas, and P = F/A is only the general definition of pressure; neither is derived from the kinetic theory.

If the temperature of an ideal gas is doubled, and the number of molecules is halved, the pressure of the gas (assuming constant volume) will:

1. double
2. remain the same
3. quadruple
4. halve

Answer

remain the same

Reason — For n molecules of a gas, PV = nkT, so at constant volume

$\text P = \dfrac{\text{nkT}}{\text V} \quad \Rightarrow \quad \text P \propto \text{nT}$

Here the temperature is doubled and the number of molecules is halved, so

$\dfrac{\text P_2}{\text P_1} = \dfrac{\left(\dfrac{\text n}{2}\right)(2\text T)}{\text{nT}} = 1$

The two changes cancel each other, and hence the pressure of the gas remains the same.

A gas is compressed to half its initial volume while its temperature is increased to double the original value. The pressure of the gas will:

1. remain the same
2. double
3. quadruple
4. halve

Answer

quadruple

Reason — From the ideal gas equation PV = μRT, for a given mass of gas

$\dfrac{\text P_1 \text V_1}{\text T_1} = \dfrac{\text P_2 \text V_2}{\text T_2}$

Here $\text V_2 = \dfrac{\text V_1}{2}$ and T₂ = 2T₁. Therefore

$\text P_2 = \text P_1 \times \dfrac{\text V_1}{\text V_2} \times \dfrac{\text T_2}{\text T_1} = \text P_1 \times 2 \times 2 = 4\text P_1$

Hence the pressure of the gas becomes four times, that is, it quadruples.

For a gas kept at constant temperature, the ratio of the final volume to the initial volume is 3:1 when the pressure is decreased by:

1. 1/3
2. 2/3
3. 3 times
4. 1/2

Answer

2/3

Reason — By Boyle's law, at constant temperature P₁V₁ = P₂V₂, that is,

$\dfrac{\text P_2}{\text P_1} = \dfrac{\text V_1}{\text V_2}$

Given that $\dfrac{\text V_2}{\text V_1} = \dfrac{3}{1}$,

$\dfrac{\text P_2}{\text P_1} = \dfrac{1}{3} \quad \Rightarrow \quad \text P_2 = \dfrac{\text P_1}{3}$

The fall in the pressure is therefore

$\text P_1 - \text P_2 = \text P_1 - \dfrac{\text P_1}{3} = \dfrac{2}{3}\text P_1$

Hence, the pressure has been decreased by $\dfrac{2}{3}$ of its initial value, the final pressure being one-third of the initial pressure.

Note: The printed answer key marks option 3, "3 times". Since the pressure falls to one-third of its initial value, the decrease is $\dfrac{2}{3}$ of the initial pressure.

A gas with 3 translational and 2 rotational degrees of freedom is heated. According to the law of equipartition of energy, the fraction of total energy associated with rotational motion is:

1. 1/5
2. 2/5
3. 3/5
4. 1/3

Answer

2/5

Reason — By the law of equipartition of energy, each degree of freedom contributes an equal amount $\dfrac{1}{2}$ kT to the energy. The gas has 3 translational and 2 rotational degrees of freedom, so the total number of degrees of freedom is f = 5, of which 2 are rotational.

$\text{Fraction} = \dfrac{\text{rotational degrees of freedom}}{\text{total degrees of freedom}} = \dfrac{2}{5}$

Hence the fraction of the total energy associated with rotational motion is 2/5.

If the mean free path of a gas molecule is decreased by a factor of 4, which of the following could be a possible explanation?

1. The volume of the gas was decreased by a factor of 4 at constant temperature
2. The number of molecules was decreased by a factor of 2
3. The pressure was decreased by a factor of 4
4. The temperature was quadrupled

Answer

The volume of the gas was decreased by a factor of 4 at constant temperature

Reason — The mean free path is inversely proportional to the number density of the gas,

$\lambda = \dfrac{1}{\sqrt2\pi \text n \text d^2}, \qquad \text n = \dfrac{\text N}{\text V}$

If the volume is decreased by a factor of 4 at constant temperature, the same number of molecules N is confined in one-fourth of the volume, so the number density n becomes 4 times and the mean free path becomes one-fourth of its earlier value. Decreasing the number of molecules or the pressure would increase the mean free path, and raising the temperature at constant pressure would also increase it.

Assertion (A): The pressure of a gas decreases if the volume is increased at constant temperature.

Reason (R): According to Boyle's law, the pressure of a gas is inversely proportional to its volume at constant temperature.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At a constant temperature, when the volume of a given mass of gas is increased, the molecules have to travel a larger distance before striking the walls and the number of collisions per unit area per second falls. Hence the pressure of the gas decreases.

Reason (R) is also correct: Boyle's law states that for a fixed amount of gas at constant temperature, the pressure of a gas is inversely proportional to its volume, $\text P \propto \dfrac{1}{\text V}$, or P₁V₁ = P₂V₂.

Since P is inversely proportional to V, an increase in V must be accompanied by a decrease in P. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Gas molecules move with different speeds in random directions.

Reason (R): Gas molecules have the same kinetic energy at a given temperature.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: According to the postulates of the kinetic theory, gas molecules move in continuous random motion in all directions, and their speeds vary over a wide range. This is why the root-mean-square speed has to be defined as an average over all the molecules.

Reason (R) is false: The molecules of a gas do not all have the same kinetic energy at a given temperature. Their speeds, and hence their kinetic energies, are distributed over a wide range, and the molecules keep exchanging energy among themselves at every collision. What the kinetic theory fixes is only the average kinetic energy per molecule, $\overline{\text E} = \dfrac{3}{2}\text{kT}$, which depends on the temperature alone.

The Reason would have been correct had it stated that the molecules have the same average kinetic energy at a given temperature.

Therefore, assertion is true but reason is false.

Note: The printed answer key marks option 2, treating the Reason as true. As worded, the Reason asserts that individual molecules have equal kinetic energies, which contradicts the distribution of molecular speeds that the Assertion itself describes.*

Assertion (A): Gases expand to fill the entire volume of their container.

Reason (R): Gas molecules have negligible intermolecular forces.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A gas has neither a definite volume nor a definite shape. It covers the entire volume of the containing vessel, which is why, when the cork of an ammonia bottle is opened, the smell of ammonia spreads in the whole room.

Reason (R) is also correct: According to the postulates of the kinetic theory, gas molecules do not exert attractive or repulsive forces on each other, except during collisions.

Since the molecules hardly attract one another and possess large kinetic energy, nothing holds them together and they continue to move throughout the entire available space. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): An increase in the temperature of a gas increases the root mean square speed of its molecules.

Reason (R): The root mean square speed of gas molecules is directly proportional to the square root of the temperature.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Raising the temperature of a gas increases the average kinetic energy of its molecules, and hence the speed of the molecules increases.

Reason (R) is also correct: From the kinetic theory,

$\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

that is, the root-mean-square speed is directly proportional to the square-root of the absolute temperature.

Since v_rms varies as $\sqrt{\text T}$, a rise in temperature necessarily raises the root-mean-square speed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The kinetic theory of gases assumes that collisions between gas molecules are perfectly elastic.

Reason (R): In elastic collisions, the total kinetic energy of the system is conserved.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: One of the postulates of the kinetic theory is that the collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic.

Reason (R) is also correct: In a perfectly elastic collision the total kinetic energy of the colliding bodies is conserved, so no kinetic energy is lost in the collision.

It is precisely because the collisions involve no loss of kinetic energy that the total kinetic energy of the gas remains constant as long as the temperature is constant, though energy may be transferred between the molecules. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The average kinetic energy of gas molecules is directly proportional to the absolute temperature.

Reason (R): Kinetic energy of gas molecules depends on both temperature and pressure.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: According to the kinetic theory, the average kinetic energy of a gas molecule is $\overline{\text E} = \dfrac{3}{2}\text{kT}$, so it is directly proportional to the absolute temperature of the gas.

Reason (R) is false: The relation $\overline{\text E} = \dfrac{3}{2}\text{kT}$ contains only the temperature. The average kinetic energy of the molecules depends on the temperature alone and not on the pressure. In fact, different gases at the same temperature have the same average kinetic energy per molecule whatever their pressure may be.

Therefore, assertion is true but reason is false.

Assertion (A): At absolute zero temperature, the kinetic energy of gas molecules becomes zero.

Reason (R): Temperature is a measure of the average kinetic energy of gas molecules.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since $\overline{\text E} = \dfrac{3}{2}\text{kT}$, putting T = 0 gives $\overline{\text E}$ = 0. Absolute zero is that temperature at which the motion of all the molecules of the gas stops, so their kinetic energy becomes zero.

Reason (R) is also correct: Temperature is a measure of the average kinetic energy of the molecules of a gas, since $\text T = \dfrac{2\overline{\text E}}{3\text k}$. This is the kinetic interpretation of temperature.

As the temperature is itself the measure of the average kinetic energy, the vanishing of the temperature means the vanishing of the kinetic energy. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The pressure of an ideal gas increases with an increase in temperature, provided the volume is constant.

Reason (R): The pressure of a gas is proportional to the average kinetic energy of the molecules.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By the pressure law, at constant volume $\text P \propto \text T$, so the pressure of an ideal gas increases with a rise in temperature when the volume is held constant.

Reason (R) is also correct: From the kinetic theory $\text P = \dfrac{2}{3}\text E$, where E is the translational kinetic energy of the gas per unit volume. Hence the pressure of a gas is proportional to the average kinetic energy of its molecules.

A rise in temperature increases the average kinetic energy of the molecules, so they strike the walls more frequently and with greater momentum, and the pressure rises. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): For a given amount of gas, the pressure decreases if the volume increases at constant temperature.

Reason (R): The number of collisions per unit area decreases as the volume increases.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Boyle's law, for a given amount of gas at constant temperature $\text P \propto \dfrac{1}{\text V}$, so the pressure decreases when the volume increases.

Reason (R) is also correct: When the volume of the vessel is increased at constant temperature, the molecules have to cover a greater distance between two successive collisions with the walls, so the number of collisions per unit area of the walls per second decreases.

Since the pressure of a gas arises from the momentum transferred to the walls by these collisions, fewer collisions per unit area mean a smaller pressure. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Boyle's law is valid for real gases under all conditions of temperature and pressure.

Reason (R): The behaviour of real gases deviates from ideal gases at high pressures and low temperatures.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Boyle's law is strictly true only for a perfect gas. Real gases obey it only approximately, and they deviate from it appreciably under extreme conditions, so it is not valid for real gases under all conditions of temperature and pressure.

Reason (R) is correct: At high pressures the actual volume of the molecules is no longer negligible, and at low temperatures the intermolecular forces are no longer negligible. Hence the behaviour of real gases deviates from ideal gas behaviour at high pressures and low temperatures.

Therefore, assertion is false but reason is true.

Assertion (A): The volume of a gas increases when its temperature increases at constant pressure.

Reason (R): Charles' law states that volume is directly proportional to absolute temperature at constant pressure.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a gas is heated at constant pressure it expands, so its volume increases with a rise in temperature.

Reason (R) is also correct: Charles' law states that the volume of a given mass of gas at constant pressure is directly proportional to its absolute temperature, $\text V \propto \text T$, or $\dfrac{\text V}{\text T}$ = a constant.

Since the volume is directly proportional to the absolute temperature, a rise in temperature must be accompanied by an increase in volume. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): If the temperature of a gas is doubled, its volume also doubles, provided the pressure remains constant.

Reason (R): According to Charles' law, the volume of a gas is directly proportional to its absolute temperature.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: From Charles' law, at constant pressure $\dfrac{\text V_1}{\text T_1} = \dfrac{\text V_2}{\text T_2}$. Putting T₂ = 2T₁ gives V₂ = 2V₁, so the volume also doubles.

Reason (R) is also correct: Charles' law states that the volume of a gas at constant pressure is directly proportional to its absolute temperature.

The doubling of the volume follows directly from this proportionality, provided the temperature used is the absolute temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The pressure of a gas increases with an increase in temperature if the volume is held constant.

Reason (R): An increase in temperature increases the kinetic energy of gas molecules, leading to more frequent and forceful collisions with the container walls.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By the pressure law, at constant volume $\text P \propto \text T$, so heating a gas in a closed vessel of fixed volume raises its pressure.

Reason (R) is also correct: A rise in temperature increases the average kinetic energy of the molecules, and hence their mean speed. This increases both the momentum transferred in each collision with the walls and the number of collisions per second.

Since the pressure of a gas is the force per unit area produced by these collisions, more frequent and more forceful collisions raise the pressure. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The pressure of an ideal gas decreases as the temperature decreases, at constant volume.

Reason (R): The kinetic energy of gas molecules decreases with decreasing temperature, causing fewer collisions with the container walls.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By the pressure law, at constant volume the pressure of an ideal gas is directly proportional to its absolute temperature, so the pressure decreases as the temperature decreases.

Reason (R) is also correct: Since $\overline{\text E} = \dfrac{3}{2}\text{kT}$, a fall in temperature lowers the average kinetic energy of the molecules and hence their mean speed, so they strike the walls less often and with smaller momentum.

The pressure being the result of these collisions, the weaker and less frequent collisions produce a smaller pressure. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A diatomic gas has 5 degrees of freedom at room temperature.

Reason (R): Diatomic gases possess both translational and rotational degrees of freedom but no vibrational degrees of freedom at room temperature.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A diatomic molecule has five degrees of freedom at room temperature, three with respect to translation and two with respect to rotation.

Reason (R) is also correct: A diatomic molecule can move bodily in three independent directions and can rotate about two of the three coordinate axes, its moment of inertia about the axis joining the two atoms being negligible. The molecule can also vibrate along the line joining the two atoms, but the vibrational motion does not usually occur at ordinary temperatures.

Adding the three translational and the two rotational motions gives exactly five degrees of freedom. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A monoatomic gas has only 3 degrees of freedom.

Reason (R): Monoatomic gases can only move in the three spatial dimensions and have no rotational or vibrational degrees of freedom.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A monoatomic gas molecule, such as He or Ar, consists of a single atom and has three degrees of freedom, all translational.

Reason (R) is also correct: The translational motion of a single atom can take place in any direction in space and can be resolved along the three coordinate axes, giving three independent motions. A monoatomic molecule can rotate also, but its moment of inertia is so small that the kinetic energy of rotation is insignificant, and it has no vibrational motion.

Since only the three translational motions are available, the molecule has exactly three degrees of freedom. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The internal energy of an ideal monoatomic gas depends only on its temperature.

Reason (R): According to the law of equipartition of energy, each degree of freedom contributes $\dfrac{1}{2}$ kT to the energy.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an ideal gas there are no intermolecular forces, so there is no internal potential energy and the total internal energy is entirely the kinetic energy of the molecules. For μ moles of a monoatomic gas $\text U = \dfrac{3}{2}\mu \text{RT}$, which depends only on the temperature.

Reason (R) is also correct: By the law of equipartition of energy, each degree of freedom contributes an average energy $\dfrac{1}{2}$ kT per molecule, so for f degrees of freedom the internal energy of 1 mole is $\text U = \dfrac{1}{2}\text{fRT}$.

As every one of these contributions contains the temperature and nothing else, the internal energy can depend only on the temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In a diatomic gas, rotational degrees of freedom contribute to the internal energy at high temperatures.

Reason (R): The law of equipartition of energy states that each degree of freedom contributes equally to the internal energy.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The rotational degrees of freedom of a diatomic molecule do contribute to the internal energy of the gas, each contributing $\dfrac{1}{2}$ kT per molecule.

Reason (R) is also correct: The law of equipartition of energy states that in a system in equilibrium at absolute temperature T, each degree of freedom contributes an equal average energy $\dfrac{1}{2}$ kT.

However, the two rotational degrees of freedom of a diatomic molecule are already active at ordinary room temperature, and it is the vibrational motion that is excited only at high temperatures. The Reason therefore does not explain the Assertion as it is worded.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion (A): The internal energy of a diatomic gas is greater than that of a monoatomic gas at the same temperature.

Reason (R): Diatomic gases have more degrees of freedom than monoatomic gases.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For μ moles of an ideal gas having f degrees of freedom, $\text U = \dfrac{1}{2}\mu \text{fRT}$. For a monoatomic gas f = 3 and for a diatomic gas f = 5, so at the same temperature $\text U_{diatomic} = \dfrac{5}{2}\mu \text{RT}$ is greater than $\text U_{monoatomic} = \dfrac{3}{2}\mu \text{RT}$.

Reason (R) is also correct: A monoatomic molecule has three degrees of freedom, all translational, while a diatomic molecule has five, three translational and two rotational.

Since each degree of freedom contributes an equal share of energy, the gas with the greater number of degrees of freedom has the greater internal energy at the same temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Vibrational modes do not contribute to the internal energy of diatomic gases at room temperature.

Reason (R): Vibrational energy levels in diatomic gases are typically not excited at low temperatures.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If assertion is false but reason is true.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A diatomic molecule can vibrate along the line joining the two atoms, but this vibrational motion does not usually occur at ordinary temperatures. Hence the vibrational modes make no contribution to the internal energy of a diatomic gas at room temperature, and its energy corresponds to only five degrees of freedom.

Reason (R) is also correct: The vibrational energy levels of a diatomic molecule are not excited at low temperatures, and become excited only when the temperature is raised sufficiently.

Since the vibrational motion is not excited at room temperature, it can contribute nothing to the internal energy there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

An ideal gas is at a temperature of 127°C. It is heated at constant pressure until its volume becomes 1.5 times. What will be the new temperature of the gas?

Answer

Given,

• Initial temperature, T₁ = 127°C = 127 + 273 = 400 K
• Final volume, V₂ = 1.5 V₁
• The pressure remains constant

By Charles' law, at constant pressure $\dfrac{\text V}{\text T}$ = a constant, so

$\dfrac{\text V_1}{\text T_1} = \dfrac{\text V_2}{\text T_2} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \dfrac{\text V_2}{\text V_1}$

$\text T_2 = 400 \times \dfrac{1.5\text V_1}{\text V_1} = 600\ \text K$

$\text t_2 = 600 - 273 = 327^\circ\text C$

Hence, the new temperature of the gas is 327°C.

The pressure of a gas at -173°C temperature is 1 atmosphere. Keeping the volume constant, to what temperature should the gas be heated so that its pressure becomes 2 atmospheres?

Answer

Given,

• Initial temperature, T₁ = −173°C = −173 + 273 = 100 K
• Initial pressure, P₁ = 1 atmosphere
• Final pressure, P₂ = 2 atmospheres
• The volume remains constant

By the pressure law, at constant volume $\dfrac{\text P}{\text T}$ = a constant, so

$\dfrac{\text P_1}{\text T_1} = \dfrac{\text P_2}{\text T_2} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \dfrac{\text P_2}{\text P_1}$

$\text T_2 = 100 \times \dfrac{2}{1} = 200\ \text K$

$\text t_2 = 200 - 273 = -73^\circ\text C$

Hence, the gas should be heated to −73°C.

The ratio of the vapour densities of two gases at the same temperature is 8 : 9. Compare the root-mean-square speeds of their molecules.

Answer

Given,

• Ratio of the vapour densities, ρ₁ : ρ₂ = 8 : 9
• Both the gases are at the same temperature

From the kinetic theory, $\text P = \dfrac{1}{3}\rho\overline{\text v^2}$, so at the same temperature and pressure

$\dfrac{\text v_{1rms}}{\text v_{2rms}} = \sqrt{\dfrac{\text M_2}{\text M_1}} = \sqrt{\dfrac{\rho_2}{\rho_1}}$

Substituting the values,

$\dfrac{\text v_{1rms}}{\text v_{2rms}} = \sqrt{\dfrac{9}{8}} = \dfrac{3}{2\sqrt2}$

Hence, the root-mean-square speeds of the molecules of the two gases are in the ratio $3 : 2\sqrt2$.

At a constant temperature, what is the relation between the pressure P and the density d for an ideal gas?

Answer

From the kinetic theory, the pressure of a gas is

$\text P = \dfrac{1}{3}\text d\overline{\text v^2}$

where d is the density of the gas. At a constant temperature the mean-square speed $\overline{\text v^2}$ of the molecules remains constant, so

$\dfrac{\text P}{\text d} = \dfrac{1}{3}\overline{\text v^2} = \text{a constant}$

Hence, at a constant temperature the pressure of an ideal gas is directly proportional to its density, that is, $\dfrac{\text P}{\text d}$ = constant.

A gas, enclosed in a vessel, has pressure P, volume V and absolute temperature T. Write down the formula for the number of molecules 'n' of the gas.

Answer

For a gas containing n molecules, the gas equation in terms of Boltzmann's constant is

$\text{PV} = \text{nkT}$

Therefore the number of molecules of the gas is

$\text n = \dfrac{\text{PV}}{\text{kT}}$

where P is the pressure, V the volume, T the absolute temperature and k the Boltzmann constant.

What is the relation between absolute temperature T of a gas and average kinetic energy of its one molecule $\overline{\text E}$? Explain, writting necessary formula.

Answer

According to the kinetic theory, the average kinetic energy of one molecule of a gas at absolute temperature T is

$\overline{\text E} = \dfrac{3}{2}\text{kT}$

where k is Boltzmann's constant.

Since $\dfrac{3}{2}$ k is a constant, $\overline{\text E} \propto \text T$.

Hence, in an ideal gas the average kinetic energy of one molecule is directly proportional to the absolute temperature of the gas. This relation contains no term for the mass of the molecule, which means that different gases at the same temperature have the same average kinetic energy per molecule.

Although the root-mean-square speed of gas molecules is of the order of the speed of sound in that gas, yet on opening a bottle of ammonia in one corner of a room its smell takes time in reaching the other corner. Explain, why?

Answer

The molecules of ammonia do not travel in straight lines across the room. As they move, they continuously collide with the molecules of air and with one another, and after each collision the magnitudes and the directions of their speeds change.

Because of these frequent collisions, an ammonia molecule follows a short zig-zag path, its mean free path being very small. Hence, although the root-mean-square speed of the molecules is very large, they are not able to advance speedily in one particular direction.

Hence, the smell of ammonia takes time in reaching the other corner of the room.

What is the relation between absolute temperature T of a gas and average kinetic energy E of a gas molecule? Explain by writing the necessary formula.

Answer

According to the kinetic theory, the average kinetic energy of a molecule of a gas at absolute temperature T is

$\text E = \dfrac{3}{2}\text{kT}$

where k is Boltzmann's constant, k = R/N.

Since $\dfrac{3}{2}$ k is a constant,

$\text E \propto \text T$

Hence, the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. This is known as the kinetic interpretation of temperature.

How much will be the kinetic energy of a gas at the absolute zero? Explain with reason.

Answer

The kinetic energy of a gas at the absolute zero is zero.

The average kinetic energy of a molecule of a gas is $\text E = \dfrac{3}{2}\text{kT}$, that is, E ∝ T. Putting T = 0 gives E = 0.

The same result follows from $\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$, which becomes zero at T = 0. Hence absolute zero is that temperature at which the motion of all the molecules of the gas stops.

A box contains equal number of molecules of hydrogen and oxygen. If there is a fine hole in the box, then which gas will leak rapidly? Why?

Answer

Hydrogen will leak out more rapidly.

The rate at which a gas escapes through a fine hole is proportional to the root-mean-square speed of its molecules, and

$\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \dfrac{1}{\sqrt{\text M}}$

Both the gases are in the same box at the same temperature. The molecular weight of hydrogen (2) is much less than that of oxygen (32), so the hydrogen molecules have the larger root-mean-square speed and reach the hole more often.

Write down the formula for the total internal energy of monoatomic and diatomic gases in terms of absolute temperature and universal gas constant.

Answer

The molar internal energy of an ideal gas having f degrees of freedom is $\text U_m = \dfrac{1}{2}\text{fRT}$, where R is the universal gas constant and T the absolute temperature.

Monoatomic gas : f = 3, so

$\text U_m = \dfrac{3}{2}\text{RT}\ \text{J mol}^{-1}$

Diatomic gas : f = 5 at ordinary temperatures, so

$\text U_m = \dfrac{5}{2}\text{RT}\ \text{J mol}^{-1}$

For μ moles of the gas the total internal energy is $\text U = \mu \text U_m$ measured in joule.

A mosquito is flying in a room. What are the degrees of freedom of its flight?

Answer

The degrees of freedom of the flight of the mosquito are 3.

The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo. A mosquito flying in a room moves bodily and its translational motion can take place in any direction in space, so it can be resolved along the three coordinate axes. Hence it has three independent motions, all translational.

Do the number of degrees of freedom of a gas molecule change on raising temperature of the gas?

Answer

Yes, the number of degrees of freedom increases on raising the temperature of the gas.

At ordinary temperatures the vibrational motion of the atoms within a molecule does not occur, so a diatomic molecule has only five degrees of freedom, three translational and two rotational.

On raising the temperature of the gas the vibrational motion of the atoms of the molecule is also promoted, and the molecule then possesses vibrational degrees of freedom in addition. Hence the number of degrees of freedom increases.

What is the relation between the pressure and kinetic energy per unit volume of a gas?

Answer

From the kinetic theory, the pressure of a gas is

$\text P = \dfrac{1}{3}\rho\overline{\text v^2} = \dfrac{2}{3}\left(\dfrac{1}{2}\rho\overline{\text v^2}\right)$

$\text P = \dfrac{2}{3}\text E$

where $\text E \left(= \dfrac{1}{2}\rho\overline{\text v^2}\right)$ is the translational kinetic energy of the gas per unit volume.

Hence, the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume.

Write the formula for the pressure of an ideal gas in terms of mass of molecules, their number and speed on the basis of kinetic theory.

Answer

On the basis of the kinetic theory, the pressure of an ideal gas is

$\text P = \dfrac{1}{3}\dfrac{\text{mn}}{\text V}\text v_{rms}^2$

where

• m = mass of one molecule of the gas
• n = total number of molecules of the gas
• V = volume of the gas
• v_rms = root-mean-square speed of the molecules

Write down the SI unit of gas constant.

Answer

From the gas equation PV = μRT, we have $\text R = \dfrac{\text{PV}}{\mu \text T}$. Therefore

$\text{unit of R} = \dfrac{\text{unit of pressure} \times \text{unit of volume}}{\text{mole} \times \text{unit of absolute temperature}} \\[1em] = \dfrac{\text{newton metre}^{-2} \times \text{metre}^3}{\text{mole K}} = \dfrac{\text{joule}}{\text{mole K}}$

Hence, the SI unit of the gas constant is J/(mol-K) or J mol^-1 K^-1.

What is meant by Boltzmann constant?

Answer

The Boltzmann constant k is the gas constant per molecule. It is the ratio of the universal gas constant R to Avogadro's number N,

$\text k = \dfrac{\text R}{\text N} = \dfrac{8.31\ \text{J mol}^{-1}\text K^{-1}}{6.02 \times 10^{23}\ \text{mol}^{-1}} = 1.38 \times 10^{-23}\ \text{J K}^{-1}$

With this constant the gas equation for n molecules of a gas takes the form PV = nkT, and the average kinetic energy of one molecule is $\dfrac{3}{2}$ kT.

Write the unit of Boltzmann constant.

Answer

The unit of the Boltzmann constant is J/K (joule per kelvin).

Since k = R/N, its unit is the unit of R, that is J mol^-1 K^-1, divided by Avogadro's number, which is a pure number per mole. Hence the unit of k is J K^-1.

Write the relation between gas constant (R) and Boltzmann constant (k).

Answer

The Boltzmann constant is the gas constant per molecule, so

$\text k = \dfrac{\text R}{\text N}$

where R is the universal gas constant and N is Avogadro's number, that is, the number of molecules in 1 mole of a gas.

On the basis of kinetic theory, write the formula for pressure of an ideal gas.

Answer

On the basis of the kinetic theory, the pressure of an ideal gas is

$\text P = \dfrac{1}{3}\rho\overline{\text v^2}$

where ρ is the density of the gas and $\overline{\text v^2}$ is the mean-square speed of its molecules.

What is the volume of one mole gas at normal temperature and pressure?

Answer

The volume of one mole of a gas at normal temperature and pressure is 22.4 L, that is, 22.4 × 10^-3 m³.

This quantity of the gas contains 6.02 × 10²³ molecules, which is Avogadro's number.

What is the value of the ratio of the two specific heats (C_p/C_v) of a monoatomic gas?

Answer

A molecule of a monoatomic gas has f = 3 degrees of freedom. The ratio of the two specific heats is

$\gamma = \dfrac{\text C_p}{\text C_v} = 1 + \dfrac{2}{\text f} = 1 + \dfrac{2}{3} = \dfrac{5}{3}$

Hence, for a monoatomic gas $\dfrac{\text C_p}{\text C_v} = \dfrac{5}{3} = 1.67$.

At which temperature will the mean kinetic energy of molecules of a gas be 1/3 of its kinetic energy at 27°C.

Answer

Given,

• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Final mean kinetic energy, $\overline{\text E}_2 = \dfrac{1}{3}\overline{\text E}_1$

The mean kinetic energy of a molecule is $\overline{\text E} = \dfrac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. Therefore

$\dfrac{\overline{\text E}_2}{\overline{\text E}_1} = \dfrac{\text T_2}{\text T_1} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \dfrac{\overline{\text E}_2}{\overline{\text E}_1}$

$\text T_2 = 300 \times \dfrac{1}{3} = 100\ \text K$

Hence, at 100 K the mean kinetic energy of the molecules will be one-third of its value at 27°C.

State the relation between root-mean-square speed of the molecules of a gas and the speed of sound in that gas.

Answer

The speed of sound in a gas is

$\text v = \sqrt{\dfrac{\gamma \text P}{\rho}}$

and the root-mean-square speed of the molecules of the gas is

$\text v_{rms} = \sqrt{\dfrac{3\text P}{\rho}}$

Dividing the first by the second,

$\dfrac{\text v}{\text v_{rms}} = \sqrt{\dfrac{\gamma}{3}} \quad \Rightarrow \quad \text v = \sqrt{\dfrac{\gamma}{3}}\ \text v_{rms}$

Hence, the speed of sound in a gas is $\sqrt{\dfrac{\gamma}{3}}$ times the root-mean-square speed of its molecules, where γ is the ratio of the two specific heats of the gas.

The volume of a vessel A is twice the volume of another vessel B, and both of them are filled with the same gas. If the gas in A is at twice the temperature and twice the pressure in comparison to the gas in B, what is the ratio of gas molecules in A and B?

Answer

Given,

• V_A = 2V_B
• T_A = 2T_B
• P_A = 2P_B

For n molecules of a gas, PV = nkT, so

$\text n = \dfrac{\text{PV}}{\text{kT}}$

Taking the ratio for the two vessels,

$\dfrac{\text n_{\text A}}{\text n_{\text B}} = \left(\dfrac{\text P_{\text A}}{\text P_{\text B}}\right)\left(\dfrac{\text V_{\text A}}{\text V_{\text B}}\right)\left(\dfrac{\text T_{\text B}}{\text T_{\text A}}\right)$

Substituting the values,

$\dfrac{\text n_{\text A}}{\text n_{\text B}} = 2 \times 2 \times \dfrac{1}{2} = \dfrac{2}{1}$

Hence, the ratio of the number of gas molecules in A and B is 2 : 1.

The temperature of an ideal gas is increased from 150 K to 600 K. If the root-mean-square speed of molecules at 150 K be v, then find its value at 600 K.

Answer

Given,

• T₁ = 150 K, v_{1 rms} = v
• T₂ = 600 K

From the kinetic theory, $\text v_{rms} \propto \sqrt{\text T}$. Therefore

$\dfrac{\text v_{2rms}}{\text v_{1rms}} = \sqrt{\dfrac{\text T_2}{\text T_1}} = \sqrt{\dfrac{600}{150}} = \sqrt4 = 2$

$\text v_{2rms} = 2\text v$

Hence, the root-mean-square speed of the molecules at 600 K is 2v.

What will be the ratio of the rms speeds of hydrogen molecules to that of oxygen molecules at same temperature?

Answer

Given,

• Molecular weight of hydrogen, M_H = 2
• Molecular weight of oxygen, M_O = 32
• Both the gases are at the same temperature

The ratio of the rms speeds of the molecules of two different gases at the same temperature is inversely proportional to the square-root of their molecular weights,

$\dfrac{(\text v_{rms})_{\text H}}{(\text v_{rms})_{\text O}} = \sqrt{\dfrac{\text M_{\text O}}{\text M_{\text H}}} \\[1em] = \sqrt{\dfrac{32}{2}} = \sqrt{16} = \dfrac{4}{1}$

Hence, the ratio of the rms speeds of hydrogen molecules to that of oxygen molecules is 4 : 1.

If the temperature of oxygen gas be raised from 0°C to 273°C, in what ratio will the mean kinetic energy of its molecules be increased?

Answer

Given,

• Initial temperature, T₁ = 0°C = 273 K
• Final temperature, T₂ = 273°C = 273 + 273 = 546 K

The mean kinetic energy of a molecule is $\overline{\text E} = \dfrac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. Therefore

$\dfrac{\overline{\text E}_2}{\overline{\text E}_1} = \dfrac{\text T_2}{\text T_1} = \dfrac{546}{273} = \dfrac{2}{1}$

Hence, the mean kinetic energy of the molecules will be increased in the ratio 2 : 1, that is, it will become twice.

There are n molecules of gas in a box. If the number of molecules is increased to 2n, what will be the effect on the pressure of the gas? On the total kinetic energy of the gas? On the root-mean-square speed of the molecules?

Answer

Given,

• The number of molecules in the box is increased from n to 2n, the volume and the temperature remaining unchanged

(i) Pressure of the gas : For n molecules, PV = nkT, so at constant V and T,

$\text P \propto \text n$

Hence the pressure of the gas will be doubled.

(ii) Total kinetic energy of the gas : The average kinetic energy of one molecule is $\dfrac{3}{2}$ kT, which depends only on the temperature. The total kinetic energy of the gas is $\text n \times \dfrac{3}{2}\text{kT}$, which is proportional to n.

Hence the total kinetic energy of the gas will also be doubled.

(iii) Root-mean-square speed : Since $\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$, the rms speed depends only upon the temperature of the gas and its molecular weight, and not on the number of molecules.

Hence the root-mean-square speed of the molecules will remain the same.

From the ideal gas equation PV = μ RT, find the unit of R.

Answer

From the ideal gas equation PV = μRT, we have

$\text R = \dfrac{\text{PV}}{\mu \text T}$

Therefore the unit of R is

$\text{unit of R} = \dfrac{\text{unit of pressure} \times \text{unit of volume}}{\text{mole} \times \text{unit of absolute temperature}}$

$= \dfrac{\text{newton metre}^{-2} \times \text{metre}^3}{\text{mole K}} = \dfrac{\text{newton metre}}{\text{mole K}}$

$= \dfrac{\text{joule}}{\text{mole K}}$

Hence, the unit of R is J mol^-1 K^-1, that is, J/(mol-K).

Since a difference of 1 K is equal to a difference of 1°C, the unit of R can also be written as J mol^-1 °C^-1.

Show that in ideal gas equation, the value of universal gas constant R is 8.31 J/(mol-K).

Answer

Given,

• Number of moles, μ = 1 mol
• Normal pressure, P = 1.013 × 10⁵ N m^-2
• Normal temperature, T = 0°C = 273 K
• Volume of 1 mole of a gas at N.T.P., V = 22.4 L = 22.4 × 10^-3 m³

According to Avogadro's hypothesis, at normal temperature and normal pressure the volume of 1 mole of any gas is 22.4 litre.

From the ideal gas equation PV = μRT,

$\text R = \dfrac{\text{PV}}{\mu \text T}$

Substituting the values,

$\text R = \dfrac{(1.013 \times 10^5\ \text N\text m^{-2}) \times (22.4 \times 10^{-3}\ \text m^3)}{1\ \text{mol} \times 273\ \text K} \\[1em] = \dfrac{2269.12}{273}\ \text{N m mol}^{-1}\text K^{-1}$

$\text R = 8.31\ \text{J mol}^{-1}\text K^{-1}$

Hence, the value of the universal gas constant R is 8.31 J/(mol-K).

Assuming the relation $\text P = \dfrac{1}{3}ρ\overline{\text v^2}$ of kinetic theory, prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Answer

From the kinetic theory, the pressure of a gas is

$\text P = \dfrac{1}{3}\rho\overline{\text v^2}$

For 1 gram-molecule of the gas, if M is the molecular weight and V the volume, then the density is $\rho = \dfrac{\text M}{\text V}$. Hence

$\text P = \dfrac{1}{3}\dfrac{\text M}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \overline{\text v^2} = \dfrac{3\text{PV}}{\text M}$

The average kinetic energy of 1 gram-molecule of the gas is

$\dfrac{1}{2}\text M\overline{\text v^2} = \dfrac{1}{2}\text M\left(\dfrac{3\text P}{\rho}\right) = \dfrac{1}{2}\text M\left(\dfrac{3\text{RT}}{\rho \text V}\right) \qquad [\because\ \text{PV} = \text{RT}]$

Since M = ρV,

$\dfrac{1}{2}\text M\overline{\text v^2} = \dfrac{3}{2}\text{RT}$

There are N (Avogadro's number) molecules in 1 gram-molecule of the gas. Hence the average kinetic energy of one molecule is

$\overline{\text E} = \dfrac{(3/2)\text{RT}}{\text N} = \dfrac{3}{2}\left(\dfrac{\text R}{\text N}\right)\text T = \dfrac{3}{2}\text{kT}$

where k (= R/N) is Boltzmann's constant.

Since $\dfrac{3}{2}$ k is a constant,

$\overline{\text E} \propto \text T$

Hence, the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

A mixture of helium and hydrogen gases is filled in a vessel at 30°C. Compare the root-mean-square velocities of the molecules of these gases at this temperature. Atomic weight of helium = 4.

Answer

Given,

• Atomic weight of helium, M_He = 4
• Molecular weight of hydrogen, M_H = 2
• Both the gases are in the same vessel at 30°C

The rms velocity of the molecules of a gas is

$\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$

At the same temperature, the ratio of the rms velocities of the molecules of two different gases is inversely proportional to the square-root of their molecular weights,

$\dfrac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text H}} = \sqrt{\dfrac{\text M_{\text H}}{\text M_{\text{He}}}}$

Substituting the values,

$\dfrac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text H}} = \sqrt{\dfrac{2}{4}} = \dfrac{1}{\sqrt2}$

Hence, the root-mean-square velocities of the molecules of helium and hydrogen are in the ratio $1 : \sqrt2$.

The mass of a molecule of krypton is 2.25 times the mass of a hydrogen molecule. A mixture of equal masses of these gases is enclosed in a vessel. Calculate at any constant temperature the ratio of the root-mean-square velocities of the molecules of krypton and hydrogen gases.

Answer

Given,

• Mass of a krypton molecule, m_Kr = 2.25 m_H
• Equal masses of the two gases are enclosed in the vessel
• The temperature is constant

The rms velocity of the molecules of a gas is

$\text v_{rms} = \sqrt{\dfrac{3\text{kT}}{\text m}} \quad \Rightarrow \quad \text v_{rms} \propto \dfrac{1}{\sqrt{\text m}}$

where m is the mass of one molecule. Hence, at the same temperature,

$\dfrac{(\text v_{rms})_{\text{Kr}}}{(\text v_{rms})_{\text H}} = \sqrt{\dfrac{\text m_{\text H}}{\text m_{\text{Kr}}}}$

Substituting m_Kr = 2.25 m_H,

$\dfrac{(\text v_{rms})_{\text{Kr}}}{(\text v_{rms})_{\text H}} = \sqrt{\dfrac{\text m_{\text H}}{2.25\text m_{\text H}}} \\[1em] = \sqrt{\dfrac{1}{2.25}} = \dfrac{1}{1.5} = \dfrac{2}{3}$

Hence, the root-mean-square velocities of the molecules of krypton and hydrogen are in the ratio 2 : 3.

The masses of the two gases taken do not affect the result, since the rms velocity depends only on the temperature and the mass of a molecule.

1 cm³ of hydrogen and 1cm³ of oxygen are at N.T.P. Explain, with reason, which will have a larger number of molecules?

Answer

Both will have an equal number of molecules.

From the kinetic theory, for two gases occupying equal volumes V,

$\text{PV} = \dfrac{1}{3}\text m_1 \text n_1 \overline{\text v_1^2} \quad \text{and} \quad \text{PV} = \dfrac{1}{3}\text m_2 \text n_2 \overline{\text v_2^2}$

Since the two samples are at the same pressure and occupy the same volume,

$\text m_1 \text n_1 \overline{\text v_1^2} = \text m_2 \text n_2 \overline{\text v_2^2}$

Both the gases are at the same temperature, so their average kinetic energies of translation per molecule are equal,

$\dfrac{1}{2}\text m_1 \overline{\text v_1^2} = \dfrac{1}{2}\text m_2 \overline{\text v_2^2}$

Applying this result in the previous expression,

$\text n_1 = \text n_2$

Hence, 1 cm³ of hydrogen and 1 cm³ of oxygen at N.T.P. contain an equal number of molecules. This is Avogadro's law, which states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.

In the upper part of the atmosphere the kinetic temperature of air is of the order of 1000 K, even then one feels severe cold there. Why?

Answer

The sensation of heat depends upon the quantity of heat absorbed by the skin, that is, upon the heat-density of the surroundings, and not upon the kinetic temperature alone.

As we go up in the atmosphere, the number of air molecules per unit volume goes on decreasing. Hence the quantity of heat per unit volume also goes on decreasing.

The kinetic temperature is the measure of the translational kinetic energy per molecule. In the upper part of the atmosphere the kinetic energy per molecule is quite large, so the kinetic temperature is high, but the molecules are so few that the heat-density is very low.

Hence, in the upper part of the atmosphere one feels severe cold although the kinetic temperature there is of the order of 1000 K.

The graph shows the variation of the product PV with respect to the pressure (P) of given masses of three gases A, B and C. The temperature is kept constant. State with proper arguments which of the three gases is ideal.

Answer

For a fixed mass of an ideal gas at a constant temperature, Boyle's law gives

$\text{PV} = \text{a constant}$

so the graph of PV against P must be a straight line parallel to the P-axis for an ideal gas.

In the given graph, the line for the gas C is parallel to the P-axis, showing that the product PV keeps the same value at all pressures.

For the gas B the product PV increases with pressure, and for the gas A it decreases with pressure. Hence these two gases do not obey Boyle's law at all pressures and are real gases.

Hence, the gas C is ideal, because it obeys Boyle's law at all pressures.

In what respect real gases differ from ideal gas? In which conditions a real gas behaves like an ideal gas?

Answer

Difference between real gases and an ideal gas :

Conditions under which a real gas behaves like an ideal gas :

At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature the behaviour of a real gas is approximately ideal.

When do the real gases obey more correctly the gas equation PV/RT?

Answer

The two chief properties of the molecules of an ideal gas are that their volume is zero and that there is no mutual force between them.

At low pressure the volume of the gas is large, so the volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature the real gases obey the gas equation $\dfrac{\text{PV}}{\text{RT}}$ more correctly, that is, their behaviour is approximately ideal.

How does the mean free path of a gas depend upon temperature and pressure?

Answer

The mean free path of the molecules of a gas is

$\lambda = \dfrac{1}{\sqrt2\pi \text n \text d^2}$

where n is the number of molecules per unit volume and d is the diameter of a molecule. For N molecules of an ideal gas, PV = NkT gives

$\text n = \dfrac{\text N}{\text V} = \dfrac{\text P}{\text{kT}}$

Substituting this value of n,

$\lambda = \dfrac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$

Dependence on temperature : $\lambda \propto \text T$, that is, the mean free path is directly proportional to the absolute temperature of the gas.

Dependence on pressure : $\lambda \propto \dfrac{1}{\text P}$, that is, the mean free path is inversely proportional to the pressure of the gas.

What is Avogadro's number? How does it have the same value for all substances?

Answer

Avogadro's number : The Avogadro number is the number of molecules in 1 mole of a substance. It is denoted by N_A and its value is 6.0221 × 10²³.

Thus 1 mole of a substance may also be defined as the mass of the substance containing 6.0221 × 10²³ molecules.

Why it has the same value for all substances : Let 1 mole of a substance A have an absolute mass M_A in gram and let a molecule of A have an absolute mass m_A. A ₆C¹² atom has an absolute mass m_C and 1 mole of ₆C¹² has a mass of exactly 12 g. Since the mass of 1 mole is proportional to the mass of its molecule,

$\dfrac{\text M_{\text A}}{12} = \dfrac{\text m_{\text A}}{\text m_{\text C}} \quad \Rightarrow \quad \dfrac{\text M_{\text A}}{\text m_{\text A}} = \dfrac{12}{\text m_{\text C}}$

The ratio $\dfrac{\text M_{\text A}}{\text m_{\text A}}$ is the number of molecules in 1 mole of the substance A, which is the Avogadro number N_A. The right-hand side contains only the mass of a carbon-12 atom and does not depend on the substance A at all.

Hence, the Avogadro number is the same for all substances. It is essentially a counting unit fixed by convention, just as a dozen is.

What do you mean by degrees of freedom of a gas?

Answer

The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

For a gas molecule these independent motions may be translational, rotational or vibrational.

Monoatomic gas : The molecule consists of a single atom, and its translational motion can be resolved along the three coordinate axes. Hence it has three degrees of freedom, all translational.

Diatomic gas : The molecule is made up of two atoms joined rigidly. It can move bodily in three independent directions and can rotate about two of the three coordinate axes, its moment of inertia about the axis joining the two atoms being negligible. Hence it has five degrees of freedom, three translational and two rotational.

Polyatomic gas : The molecule can rotate about any of the three coordinate axes, so it has six degrees of freedom, three translational and three rotational.

Kinetic Theory of Gases and Laws

The kinetic theory of gases provides a molecular-level understanding of gas behaviour. According to this theory, gas molecules are in constant random motion, colliding elastically with each other and the walls of the container. The pressure exerted by a gas is due to these molecular collisions with the container walls. Boyle's law, which is derived from kinetic theory, states that the pressure of a gas is inversely proportional to its volume at constant temperature. Charles' law explains that the volume of a gas is directly proportional to its temperature at constant pressure. Gay-Lussac's law, on the other hand, relates the pressure of a gas to its temperature when the volume is held constant. The average kinetic energy of gas molecules is directly proportional to the absolute temperature, giving a kinetic interpretation of temperature. This theory is further extended by the law of equipartition of energy, which states that energy is distributed equally among all available degrees of freedom.

(i) What is the primary cause of gas pressure according to the kinetic theory?

1. The volume of the gas
2. The mass of the gas molecules
3. The collisions of gas molecules with the walls of the container
4. The energy from the external environment.

(ii) According to Boyle's law, how does the pressure of a gas change if its volume is doubled at constant temperature?

1. It remains the same
2. It doubles
3. It decreases by half
4. It quadruples

(iii) How does the volume of a gas change according to Charles' law when its temperature is increased at constant pressure?

1. It increases
2. It decreases
3. It remains the same
4. It depends on the pressure

(iv) Which law relates the pressure of a gas to its temperature at constant volume?

1. Boyle's law
2. Charles' law
3. Avogadro's law
4. Gay-Lussac's law

(v) What does the law of equipartition of energy state?

1. The energy is concentrated in the translational degrees of freedom only
2. Energy is distributed equally among all available degrees of freedom
3. Energy is proportional to the pressure of the gas
4. The energy of a gas depends only on its mass

Answer

(i) The collisions of gas molecules with the walls of the container

The molecules of a gas keep on colliding with the walls of the vessel and thus transfer momentum to them. A steady force therefore acts perpendicular to the walls, and the magnitude of this force per unit area of the walls is called the pressure of the gas.

(ii) It decreases by half

By Boyle's law, at constant temperature P₁V₁ = P₂V₂. Putting V₂ = 2V₁,

$\text P_2 = \dfrac{\text P_1 \text V_1}{2\text V_1} = \dfrac{\text P_1}{2}$

Hence, on doubling the volume the pressure falls to half its original value.

(iii) It increases

Charles' law states that at constant pressure the volume of a gas is directly proportional to its absolute temperature, V ∝ T. Hence, when the temperature is increased at constant pressure, the volume of the gas increases.

(iv) Gay-Lussac's law

Gay-Lussac's law, also called the pressure law, states that for a fixed mass of gas at constant volume the pressure is directly proportional to the absolute temperature, P ∝ T.

(v) Energy is distributed equally among all available degrees of freedom

According to the law of equipartition of energy, in a system in equilibrium at absolute temperature T, the average energy associated with each degree of freedom is $\dfrac{1}{2}$ kT, where k is Boltzmann's constant. The energy is therefore shared equally among all the available degrees of freedom.

Molecular Motion and Mean Free Path

In a gas, molecules are in constant random motion and travel in straight lines between collisions. The average distance a molecule travels before colliding with another molecule is known as the mean free path. The mean free path depends on the temperature, pressure, and size of the gas molecules. As the pressure of a gas increases, the mean free path decreases because the gas molecules are closer together. Similarly, the temperature of the gas affects the speed of the molecules; higher temperatures increase molecular speed and collision frequency. This results in higher pressure if the volume is kept constant. The relationship between the pressure, volume, and temperature of a gas is explained by the ideal gas law PV = nRT.

(i) What is the mean free path of a gas molecule?

1. The distance between two gas molecules
2. The average distance a molecule travels before colliding with another molecule
3. The time taken for a molecule to travel from one side of the container to the other
4. The speed of gas molecules

(ii) What happens to the mean free path of gas molecules if the pressure is increased at constant temperature?

1. It increases
2. It decreases
3. It remains constant
4. It becomes zero

(iii) How does the temperature of a gas affect the speed of gas molecules?

1. Increasing temperature decreases molecular speed
2. Increasing temperature increases molecular speed
3. Temperature does not affect molecular speed
4. Molecular speed decreases as temperature increases

(iv) According to the ideal gas law, what will happen to the pressure of a gas if the volume is kept constant and the temperature is increased?

1. It decreases
2. It remains constant
3. It increases
4. It depends on the molecular size

(v) Which of the following factors affects the mean free path of a gas?

1. The type of container
2. The volume of the gas only
3. The temperature, pressure, and size of gas molecules
4. Only the temperature of the gas

Answer

(i) The average distance a molecule travels before colliding with another molecule

Between two successive collisions a molecule moves in a straight line with a constant speed, so it follows short zig-zag paths of different lengths. The mean of these free paths is called the mean free path.

(ii) It decreases

The mean free path is $\lambda = \dfrac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$, so at constant temperature $\lambda \propto \dfrac{1}{\text P}$. On increasing the pressure the molecules are packed more closely together, so a molecule collides with another after covering a shorter distance.

(iii) Increasing temperature increases molecular speed

The root-mean-square speed of the molecules is $\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}}$, that is, $\text v_{rms} \propto \sqrt{\text T}$. Hence a rise in temperature increases the speed of the molecules and also their collision frequency.

(iv) It increases

From the ideal gas law PV = nRT, at constant volume P ∝ T. Hence, when the temperature is increased and the volume is kept constant, the pressure of the gas increases.

(v) The temperature, pressure, and size of gas molecules

Since $\lambda = \dfrac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$, the mean free path depends on the absolute temperature T, on the pressure P, and on the diameter d of the gas molecules.

Pressure and Degrees of Freedom

The pressure exerted by a gas on the walls of a container is directly related to the kinetic energy of the gas molecules. For an ideal gas, pressure is given by the equation $\text P = \dfrac{1}{3}ρ\text v_{rms}^2$, where ρ is the density and v_rms is the root mean square (rms) speed of the molecules. According to the law of equipartition of energy, each degree of freedom contributes $\dfrac{1}{2}$ kT to the total energy, where T is the temperature and k is the Boltzmann constant. Monoatomic gases have three translational degrees of freedom, while diatomic gases have additional rotational degrees of freedom. At higher temperatures, vibrational degrees of freedom may also be excited in diatomic gases, contributing to the internal energy.

(i) How does the pressure of a gas change if the root mean square (rms) speed of the molecules increases?

1. Pressure decreases
2. Pressure increases
3. Pressure remains constant
4. Pressure is independent of molecular speed

(ii) How many translational degrees of freedom does a monoatomic gas have?

1. 2
2. 3
3. 5
4. 6

(iii) According to the law of equipartition of energy, how much energy is associated with each degree of freedom for a gas at temperature T?

1. kT
2. $\dfrac{1}{2}$ kT
3. 2kT
4. $\dfrac{3}{2}$ kT

(iv) What is the contribution of rotational degrees of freedom to the total energy of a diatomic gas at room temperature?

1. Zero
2. Equal to translational energy
3. Less than translational energy
4. More than translational energy

(v) What additional degrees of freedom become available for diatomic gases at high temperatures?

1. Translational
2. Rotational
3. Vibrational
4. Gravitational

Answer

(i) Pressure increases

From the kinetic theory, $\text P = \dfrac{1}{3}\rho \text v_{rms}^2$, so the pressure is proportional to the square of the rms speed. An increase in the rms speed makes the collisions of the molecules with the walls both more frequent and more forceful, and hence the pressure increases.

(ii) 3

The molecule of a monoatomic gas consists of a single atom whose translational motion can be resolved along the three coordinate axes. Hence it has three degrees of freedom, all translational.

(iii) $\dfrac{1}{2}$ kT

According to the law of equipartition of energy, the average energy associated with each degree of freedom of a particle in a system in equilibrium at absolute temperature T is $\dfrac{1}{2}$ kT, where k is Boltzmann's constant.

(iv) Less than translational energy

A diatomic molecule has three translational and two rotational degrees of freedom at room temperature. By the law of equipartition of energy, the energy in the translational modes is $3 \times \dfrac{1}{2}\text{kT} = \dfrac{3}{2}\text{kT}$, while the energy in the rotational modes is $2 \times \dfrac{1}{2}\text{kT} = \text{kT}$. Since $\text{kT} \lt \dfrac{3}{2}\text{kT}$, the rotational contribution is less than the translational contribution.

Note: The printed answer key marks option 2, "Equal to translational energy". A diatomic molecule has only two rotational degrees of freedom against three translational ones, so by the law of equipartition of energy the rotational contribution is kT against a translational contribution of $\dfrac{3}{2}$ kT. Option 3 is therefore the correct choice.*

(v) Vibrational

At ordinary temperatures the vibrational motion of a diatomic molecule does not occur. At high temperatures this motion is promoted, so vibrational degrees of freedom become available and contribute to the internal energy.

The molecules of a gas are widely separated from each other and there is empty space between them in which they keep on moving rapidly in all possible directions. Gas molecules have large kinetic energy and hence they continue to move throughout the entire available space. The molecules of a gas keep on colliding with the walls of the vessel in which they contained and thus transferring momentum to them. Thus, a steady force acts perpendicular to the walls of the vessel. The magnitude of this force per unit area of the walls is called the pressure of the gas.

(i) Write the formula for the pressure of an ideal gas.

(ii) Explain Boyle's law on the basis of pressure of an ideal gas.

(iii) Two vessels of the same volume are filled with the same gas at the same temperature. If the pressure of the gas in these vessels be in the ratio 1 : 2, then find the ratio of the root-mean-square speeds of the molecules.

Answer

(i) Formula for the pressure of an ideal gas :

From the kinetic theory, the pressure of an ideal gas is

$\text P = \dfrac{1}{3}\dfrac{\text{mn}}{\text V}\text v_{rms}^2$

where

• m = mass of one molecule of the gas
• n = total number of molecules of the gas
• V = volume of the gas
• v_rms = root-mean-square speed of the molecules

(ii) Explanation of Boyle's law :

Boyle's law states that the product of the pressure and the volume of a given mass of gas at constant temperature is constant. From the kinetic theory,

$\text P = \dfrac{1}{3}\dfrac{\text{mn}}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \dfrac{1}{3}\text{mn}\overline{\text v^2}$

Here mn is the mass of the gas, which is constant for a given mass of gas. If the temperature remains constant, the mean-square speed $\overline{\text v^2}$ of the molecules also remains constant, because $\overline{\text v^2} \propto \text T$. Therefore the whole of the right-hand side is constant,

$\text{PV} = \text{constant}$

This is Boyle's law.

(iii) Given,

• Both the vessels are of the same volume and are filled with the same gas at the same temperature
• Ratio of the pressures, P₁ : P₂ = 1 : 2

The root-mean-square speed of the molecules of a gas is

$\text v_{rms} = \sqrt{\dfrac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

so it depends only upon the absolute temperature of the gas and its molecular weight, and not upon the pressure. Since the same gas is at the same temperature in both the vessels,

$\dfrac{(\text v_{rms})_1}{(\text v_{rms})_2} = \dfrac{1}{1}$

Hence, the ratio of the root-mean-square speeds of the molecules in the two vessels is 1 : 1.

What is an ideal gas? Under what conditions of pressure and temperature can a gas be assumed as an ideal gas? Determine the gas constant for one gram-molecule of a gas.

Answer

Ideal gas : In practice, gases do not obey Boyle's law, Charles' law and the law of pressure strictly under all conditions of temperature and pressure. Experiments show that as the pressure is decreased, different gases become more and more similar in behaviour and more and more obedient to the gas laws. So we may imagine a gas whose properties are similar to the properties of a real gas at infinitely low pressure. Such an imaginary gas is called a 'perfect gas' or an 'ideal gas'.

The following properties are imagined in a perfect gas :

(i) It strictly obeys Boyle's law, Charles' law and the law of pressure under all conditions of temperature and pressure.

(ii) Its pressure coefficient and volume coefficient are exactly equal to each other.

(iii) Its molecules are infinitesimally small.

(iv) There is no force of attraction between its molecules. Hence a perfect gas cannot be converted into the liquid or the solid state.

Conditions under which a gas can be assumed to be ideal :

At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature a gas can be assumed to be an ideal gas. In practice, the gases which are difficult to liquefy, such as oxygen, nitrogen, hydrogen and helium, can be considered as perfect.

Gas constant for one gram-molecule : For 1 gram-molecule (1 mole) of a gas the equation of state is

$\text{PV} = \text{RT} \quad \Rightarrow \quad \text R = \dfrac{\text{PV}}{\text T}$

According to Avogadro's hypothesis, at normal temperature (0°C or 273 K) and normal pressure (1 atmosphere or 1.013 × 10⁵ N m^-2) the volume of 1 mole of a gas is 22.4 litre, that is, 22.4 × 10^-3 m³.

Substituting the values,

$\text R = \dfrac{(1.013 \times 10^5\ \text N\text m^{-2}) \times (22.4 \times 10^{-3}\ \text m^3)}{1\ \text{mol} \times 273\ \text K} \\[1em] = 8.31\ \text{N m mol}^{-1}\text K^{-1}$

$\text R = 8.31\ \text{J mol}^{-1}\text K^{-1}$

Hence, the gas constant for one gram-molecule of a gas is 8.31 J mol^-1 K^-1. As it has the same value for all gases, it is called the universal gas constant.

Establish the equation PV = RT for an ideal gas and obtain the dimensional formula for R.

Answer

Establishment of the equation PV = RT :

The volume, pressure and temperature of a given mass of a gas are inter-related. Suppose that initially the pressure, volume and absolute temperature of a given mass of a gas are P₁, V₁ and T₁, and that they finally change to P₂, V₂ and T₂. This change may be divided into two parts.

(1) Change of pressure at constant temperature : Let the temperature be kept constant at T₁ and the pressure be changed from P₁ to P₂. If the volume changes from V₁ to V', then by Boyle's law

$\text P_2 \text V' = \text P_1 \text V_1 \quad \Rightarrow \quad \text V' = \dfrac{\text P_1 \text V_1}{\text P_2} \qquad \ldots(\text i)$

(2) Change of temperature at constant pressure : Now let the pressure be kept constant at P₂ and the absolute temperature be changed from T₁ to T₂. If the volume changes from V' to V₂, then by Charles' law

$\dfrac{\text V'}{\text T_1} = \dfrac{\text V_2}{\text T_2} \quad \Rightarrow \quad \text V' = \dfrac{\text T_1 \text V_2}{\text T_2} \qquad \ldots(\text{ii})$

From equations (i) and (ii),

$\dfrac{\text P_1 \text V_1}{\text P_2} = \dfrac{\text T_1 \text V_2}{\text T_2} \quad \Rightarrow \quad \dfrac{\text P_1 \text V_1}{\text T_1} = \dfrac{\text P_2 \text V_2}{\text T_2}$

Hence, for a given mass of a gas the value of $\dfrac{\text{PV}}{\text T}$ remains constant. If this constant is taken as R for 1 gram-molecule of the gas, then

$\dfrac{\text{PV}}{\text T} = \text R \quad \Rightarrow \quad \text{PV} = \text{RT}$

Since the gas laws are strictly true only for an ideal gas, this equation is also strictly true only for an ideal gas, and it is therefore called the ideal gas equation. For μ moles of a gas it takes the form PV = μRT.

Dimensional formula for R : From the gas equation,

$\text R = \dfrac{\text{PV}}{\text T}$

The dimensions of pressure are [M L^-1 T^-2], those of volume are [L³] and those of temperature are [θ]. Therefore

$[\text R] = \dfrac{[\text M \text L^{-1}\text T^{-2}][\text L^3]}{[\theta]} = \dfrac{[\text M \text L^2 \text T^{-2}]}{[\theta]}$

$[\text R] = [\text M \text L^2 \text T^{-2}\theta^{-1}]$

Hence, the dimensional formula for R is [M L² T^-2 θ^-1], the mole being dimensionless.

Establish a relation between the pressure and kinetic energy per unit volume of a gas.

Answer

Expression for the pressure of a gas :

Consider an ideal gas contained in a cubical vessel whose walls are perfectly elastic. Let l be the length of each side of the cube, m the mass of each molecule and n the total number of molecules in the gas. The vessel is placed so that its three edges lie along the rectangular coordinate axes.

Let v be the velocity of a molecule and let v_x, v_y and v_z be its components along the X-, Y- and Z-axes, so that

$\text v^2 = \text v_x^2 + \text v_y^2 + \text v_z^2$

Consider the motion of this molecule along the X-axis. When it collides with the face A, the collision being perfectly elastic, its velocity changes from v_x to −v_x. Hence the change in its momentum in one collision is

$\Delta \text p = -\text{mv}_x - (\text{mv}_x) = -2\text{mv}_x$

By the law of conservation of momentum, the momentum transferred to the wall A in one collision is + 2 mv_x.

After rebounding from A the molecule collides with the opposite wall and returns, so it covers a distance 2l between two successive collisions with A. Since its velocity is v_x, it takes $\dfrac{2\text l}{\text v_x}$ second between two collisions with A, that is, it collides with A $\dfrac{\text v_x}{2\text l}$ times per second.

Therefore the momentum transferred to the wall A in 1 second is

$= 2\text{mv}_x \times \dfrac{\text v_x}{2\text l} = \dfrac{\text m}{\text l}\text v_x^2$

According to Newton's law of motion, the rate of change of momentum is equal to the force exerted. Since there are n molecules, the total force exerted on the wall A is

$\dfrac{\text m}{\text l}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right)$

The area of the wall is l², so the pressure on the wall A is

$\text P = \dfrac{\text m}{\text l^3}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right) \\[1em] = \dfrac{\text{mn}}{\text l^3}\left(\dfrac{\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2}{\text n}\right)$

But l³ = V, the volume of the cube, and the bracketed quantity is the average value $\overline{\text v_x^2}$ of v_x² for all the n molecules. Therefore

$\text P = \dfrac{\text{mn}}{\text V}\overline{\text v_x^2} \qquad \ldots(\text i)$

Since v² = v_x² + v_y² + v_z² holds for every molecule, it also holds for the mean values,

$\overline{\text v^2} = \overline{\text v_x^2} + \overline{\text v_y^2} + \overline{\text v_z^2} \qquad \ldots(\text{ii})$

The directions of motion of the molecules are quite at random, that is, all directions are equally possible. Hence

$\overline{\text v_x^2} = \overline{\text v_y^2} = \overline{\text v_z^2}$

From equation (ii),

$\overline{\text v^2} = 3\overline{\text v_x^2} \quad \Rightarrow \quad \overline{\text v_x^2} = \dfrac{1}{3}\overline{\text v^2}$

Substituting this value of $\overline{\text v_x^2}$ in equation (i),

$\text P = \dfrac{1}{3}\dfrac{\text{mn}}{\text V}\overline{\text v^2} \qquad \ldots(\text{iii})$

This is the expression for the pressure of the gas, $\overline{\text v^2}$ being the mean-square velocity of the molecules. Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas, and the above expression can be written as

$\text P = \dfrac{1}{3}\rho\overline{\text v^2}$

Relation between pressure and kinetic energy per unit volume :

The expression obtained above may be written as

$\text P = \dfrac{1}{3}\rho\overline{\text v^2} = \dfrac{2}{3}\left(\dfrac{1}{2}\rho\overline{\text v^2}\right)$

Here $\dfrac{1}{2}\rho\overline{\text v^2}$ is the translational kinetic energy of the gas per unit volume, since ρ is the mass of the gas per unit volume. Denoting it by E,

$\text P = \dfrac{2}{3}\text E$

Hence, the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume.

How is temperature interpreted in terms of the kinetic theory of the gases? Show that the root-mean-square velocity of molecules is proportional to the square-root of absolute temperature of the gas. Also, show that the mean kinetic energy of a molecule is $\dfrac{3}{2}$ kT.

Answer

Kinetic interpretation of temperature :

Suppose 1 mole of a gas is at an absolute temperature T and has a volume V. Let the mass of a molecule of the gas be m and the mean-square speed of its molecules be $\overline{\text v^2}$. There are N (Avogadro's number) molecules in 1 mole. According to the kinetic theory, the pressure of the gas is

$\text P = \dfrac{1}{3}\dfrac{\text{mN}}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \dfrac{1}{3}\text{mN}\overline{\text v^2}$

But mN, the mass of one molecule multiplied by the number of molecules in 1 mole, is the mass M of 1 mole of the gas. Therefore

$\text{PV} = \dfrac{1}{3}\text M\overline{\text v^2}$

The ideal gas equation for 1 mole is PV = RT. Hence

$\text{RT} = \dfrac{1}{3}\text M\overline{\text v^2} \quad \Rightarrow \quad \overline{\text v^2} = \dfrac{3\text{RT}}{\text M} \qquad \ldots(\text i)$

Taking the square-root,

$\text v_{rms} = \sqrt{\overline{\text v^2}} = \sqrt{\dfrac{3\text{RT}}{\text M}} \qquad \ldots(\text{ii})$

Since 3R/M is a constant for a given gas,

$\text v_{rms} \propto \sqrt{\text T}$

Hence, the root-mean-square velocity of the molecules of a gas is directly proportional to the square-root of the absolute temperature of the gas. This is the kinetic interpretation of temperature: the faster the motion of the molecules, the higher is the temperature of the gas.

It is also clear from this relation that if the absolute temperature becomes zero, the motion of the molecules ceases (v_rms = 0). Hence absolute zero is that temperature at which the motion of all the molecules of the gas stops. Since v_rms can never be negative, no temperature is possible below the absolute zero.

Mean kinetic energy of a molecule :

The kinetic energy of 1 mole of the gas is

$\dfrac{1}{2}\text M\overline{\text v^2} = \dfrac{1}{2}\text M\left(\dfrac{3\text{RT}}{\text M}\right) \qquad [\text{from eq. (i)}] \\[1em] = \dfrac{3}{2}\text{RT}$

There are N molecules in 1 mole of the gas. Therefore the average kinetic energy of one molecule is

$\overline{\text E} = \dfrac{(3/2)\text{RT}}{\text N} = \dfrac{3}{2}\left(\dfrac{\text R}{\text N}\right)\text T$

The ratio R/N is a constant which is called Boltzmann's constant and is denoted by k. Hence

$\overline{\text E} = \dfrac{3}{2}\text{kT}$

Hence, the mean kinetic energy of a molecule of a gas is $\dfrac{3}{2}$ kT.

This expression does not contain the mass of the molecule, which means that different gases at the same temperature have the same average kinetic energy per molecule.

Prove that $\dfrac{\text v_{rms}^2}{\text T} = \dfrac{3\text R}{\text M}$, where M is the molecular weight of gas, R is gas constant, T is absolute temperature and v_rms is root-mean-square speed. If the temperature of oxygen changes from 0°C to 273°C, what will be the change in its mean kinetic energy?

Answer

Proof : According to the kinetic theory, the pressure of a gas is

$\text P = \dfrac{1}{3}\dfrac{\text M}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \dfrac{1}{3}\text M\overline{\text v^2}$

where M is the molecular weight of the gas, that is, the mass of 1 mole, and V is the volume of 1 mole.

The ideal gas equation for 1 mole of the gas is

$\text{PV} = \text{RT}$

Comparing the two expressions,

$\dfrac{1}{3}\text M\overline{\text v^2} = \text{RT}$

$\overline{\text v^2} = \dfrac{3\text{RT}}{\text M}$