Thermodynamics — HOTS Questions

Steam at 100°C is passed into 20 g of water at 10°C when water acquires a temperature of 80°C, the mass of water present will be :

[Take specific heat of water = 1 cal g^-1 °C^-1 and latent heat of steam = 540 cal g^-1]

1. 24 g
2. 31.5 g
3. 42.5 g
4. 22.5 g

Answer

22.5 g

Reason — Given,

• Mass of cold water, m_w = 20 g at 10°C
• Steam at 100°C is passed into it
• Final temperature of the mixture = 80°C
• Specific heat of water, c = 1 cal g^-1 °C^-1
• Latent heat of steam, L = 540 cal g^-1

Heat taken by the cold water in rising from 10°C to 80°C is

$\text Q_{\text{gained}} = \text m_w\text c\Delta \text T = 20 \times 1 \times (80 - 10) \\[1em] = 1400\ \text{cal}$

Heat given out by the steam : Let m gram of steam be condensed. The steam first condenses into water at 100°C, giving out the latent heat mL, and this water then cools from 100°C to 80°C, giving out mcΔT. Hence

$\text Q_{\text{lost}} = \text{mL} + \text{mc}\Delta\theta = \text m \times 540 + \text m \times 1 \times (100 - 80) \\[1em] = \text m \times 540 + \text m \times 20 = \text m \times 560\ \text{cal}$

By the principle of calorimetry, heat lost = heat gained,

$\text m \times 560 = 1400 \quad \Rightarrow \quad \text m = \dfrac{1400}{560} = 2.5\ \text g$

The condensed steam remains in the vessel as water, so the total mass of water present is

$= 20\ \text g + 2.5\ \text g = 22.5\ \text g$

When a system is taken from state i to state f along the path iaf as shown in figure, it is found that Q = 50 cal and W = 20 cal. Along the path ibf, Q = 36 cal. W along the path ibf is :

1. 16 cal
2. 66 cal
3. 14 cal
4. 6 cal

Answer

6 cal

Reason — Given,

• Along the path iaf : Q₁ = 50 cal and W₁ = 20 cal
• Along the path ibf : Q₂ = 36 cal, W₂ = ?

The internal energy is a state function, so the change in the internal energy in going from the state i to the state f is the same along both the paths,

$\Delta \text U_{iaf} = \Delta \text U_{ibf}$

By the first law of thermodynamics, ΔU = Q − W. Therefore

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2$

Substituting the values,

$50 - 20 = 36 - \text W_2 \\[1em] 30 = 36 - \text W_2$

$\text W_2 = 36 - 30 = 6\ \text{cal}$

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is :

1. P₀V₀
2. 2 P₀V₀
3. $\dfrac{\text P_0 \text V_0}{2}$
4. zero

Answer

zero

Reason —

From the graph, the four states of the gas are A (V₀, P₀), B (2V₀, 3P₀), C (V₀, 3P₀) and D (2V₀, P₀). The two straight lines AB and CD cross each other at the point O.

The work done in a thermodynamic process is equal to the area enclosed between the P-V curve and the volume-axis, and the closed figure here is made up of two triangles which meet at O.

Work done in the process O → D → A : This lower triangle is traced clockwise, so the work done is positive,

$\text W_1 = \dfrac{1}{2} \times \text P_0 \times \text V_0$

Work done in the process O → B → C : This upper triangle is traced anticlockwise, so the work done is negative and equal in magnitude,

$\text W_2 = -\dfrac{1}{2} \times \text P_0 \times \text V_0$

The two triangles are of equal area but are described in opposite senses. Hence the total work done by the system in the cycle is

$\text W = \text W_1 + \text W_2 = \dfrac{1}{2}\text P_0\text V_0 - \dfrac{1}{2}\text P_0\text V_0 \\[1em] = \text{zero}$

The figure shows two paths that may be taken by a gas to go from a state A to a state C.

In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be :

1. 500 J
2. 460 J
3. 300 J
4. 380 J

Answer

460 J

Reason — Given,

• Heat added in the process AB, Q_AB = 400 J
• Heat added in the process BC, Q_BC = 100 J

From the graph, the states are A (2 × 10^-3 m³, 2 × 10⁴ Pa) and C (4 × 10^-3 m³, 6 × 10⁴ Pa), and the path AC is the straight line joining them.

Since the internal energy is a state function, the change in internal energy is the same along the path ABC and along the path AC.

For the path ABC : The work done is zero along AB (isochoric) and along BC the pressure is constant, but the total heat given is

$\text Q_{\text{ABC}} = 400 + 100 = 500\ \text J$

For the closed cycle A → B → C → A : Applying the first law to the complete cycle, in which ΔU = 0, the net heat absorbed equals the net work done, which is the area of the triangle ABC,

$400\ \text J + 100\ \text J + \text Q_{\text C \to \text A} = \dfrac{1}{2} \times (4 \times 10^{4}) \times (2 \times 10^{-3})$

$500 + \text Q_{\text C \to \text A} = 40 \\[1em] \text Q_{\text C \to \text A} = -460\ \text J$

Since the heat absorbed in going from C to A is − 460 J, the heat absorbed by the system in the reverse process, that is, in the process A → C, is

$\text Q_{\text A \to \text C} = 460\ \text J$

The given P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is :

1. P₀V₀
2. $\left(\dfrac{13}{2}\right)$ P₀V₀
3. $\left(\dfrac{11}{2}\right)$ P₀V₀
4. 4 P₀V₀

Answer

$\left(\dfrac{13}{2}\right)$ P₀V₀

Reason —

From the graph the cycle consists of two isochoric and two isobaric processes, the states being (V₀, P₀) → (V₀, 2P₀) → (2V₀, 2P₀) → (2V₀, P₀) → (V₀, P₀). The gas is monoatomic, so

$\text C_v = \dfrac{3}{2}\text R \quad \text{and} \quad \text C_p = \dfrac{5}{2}\text R$

Heat is extracted from the source only in those parts of the cycle in which the heat is absorbed by the gas, that is, when the pressure grows at constant volume and when the volume grows at constant pressure.

Heat absorbed when the pressure grows from P₀ to 2P₀ at the constant volume V₀ :

$\text Q_1 = \text n\text C_v\Delta \text T = \dfrac{3}{2}\text n\text R(\text T_2 - \text T_1) = \dfrac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1)$

$= \dfrac{3}{2}(2\text P_0\text V_0 - \text P_0\text V_0) = \dfrac{3}{2}\text P_0\text V_0 \qquad \text{...(i)}$

Heat absorbed when the volume grows from V₀ to 2V₀ at the constant pressure 2P₀ :

$\text Q_2 = \text n\text C_p\Delta \text T = \dfrac{5}{2}(\text P_2\text V_2 - \text P_1\text V_1)$

$= \dfrac{5}{2}(2\text P_0 \times 2\text V_0 - 2\text P_0 \times \text V_0) = \dfrac{5}{2}(2\text P_0\text V_0) = 5\text P_0\text V_0 \qquad \text{...(ii)}$

In the remaining two processes the pressure and the volume decrease, so heat is rejected by the gas.

Adding (i) and (ii), the total heat extracted from the source in a single cycle is

$\text Q = \text Q_1 + \text Q_2 = \dfrac{3}{2}\text P_0\text V_0 + 5\text P_0\text V_0 \\[1em] = \left(\dfrac{13}{2}\right)\text P_0\text V_0$

100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K) :

1. 2.1 kJ
2. 4.2 kJ
3. 8.4 kJ
4. 84 kJ

Answer

8.4 kJ

Reason — Given,

• Mass of water, m = 100 g = 100 × 10^-3 kg
• Initial temperature = 30°C, final temperature = 50°C, so ΔT = 20°C = 20 K
• Specific heat of water, c = 4184 J kg^-1 K^-1
• The slight expansion of the water is to be ignored

Since the expansion of the water is ignored, the change in volume is zero and hence no external work is done, W = 0. By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = \text Q - 0 = \text Q$

The heat taken by the water is

$\text Q = \text{mc}\Delta \text T = (100 \times 10^{-3}) \times 4184 \times (50 - 30)$

$= 0.1 \times 4184 \times 20 \\[1em] = 8368\ \text J = 8.368\ \text{kJ} \approx 8.4\ \text{kJ}$

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

1. $\dfrac{(γ - 1)}{2\text R} \text M v^2$ K

2. $\dfrac{(γ - 1)}{2 (γ + 1)\text R} \text M v^2$ K

3. $\dfrac{(γ - 1)}{2 γ \text R} \text M v^2$ K

4. $\dfrac{γ \text M v^2}{2\text R}$ K

Answer

$\dfrac{(γ - 1)}{2\text R} \text M v^2$ K

Reason — Given,

• Molecular mass of the gas = M
• Ratio of specific heats = γ
• Speed of the vessel = v, and the vessel is suddenly brought to rest
• The vessel is thermally insulated, so no heat is lost to the surroundings

Let the vessel contain μ mole of the gas. The mass of the gas is then μM, and the kinetic energy of the gas on account of the motion of the vessel is

$\text K = \dfrac{1}{2}(\mu \text M)v^2$

When the vessel is suddenly brought to rest, this ordered kinetic energy of the gas is converted into the internal energy of the gas, and hence the temperature of the gas rises. Since no heat is lost, the whole of it appears as the increase in internal energy.

If ΔT be the rise in temperature of μ mole of the gas, then

$\dfrac{1}{2}(\mu \text M)v^2 = \mu\text C_v\Delta \text T$

For an ideal gas the molar specific heat at constant volume is

$\text C_v = \dfrac{\text R}{\gamma - 1}$

Substituting this value,

$\dfrac{1}{2}\mu \text M v^2 = \mu \times \dfrac{\text R}{\gamma - 1} \times \Delta \text T$

Cancelling μ and solving for ΔT,

$\Delta \text T = \dfrac{\text M v^2(\gamma - 1)}{2\text R}\ \text K$

Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)

1. 15.4%
2. 9.1%
3. 10.5%
4. 12.5%

Answer

15.4%

Reason —

From the graph the cycle ABCDA consists of two isochoric and two isobaric lines, the states being A (V₀, P₀), B (V₀, 2P₀), C (2V₀, 2P₀) and D (2V₀, P₀). Helium is a monoatomic gas, so

$\text C_v = \dfrac{3}{2}\text R \quad \text{and} \quad \text C_p = \dfrac{5}{2}\text R$

Work done in the complete cycle : No work is done along AB and CD, since the volume does not change there. Therefore

$\text W = 0 + 2\text P_0(2\text V_0 - \text V_0) + 0 - \text P_0(2\text V_0 - \text V_0)$

$= 2\text P_0\text V_0 - \text P_0\text V_0 = \text P_0\text V_0$

Heat given to the gas from A → B (isochoric) :

$\text Q_1 = \text n\text C_v\Delta \text T = \dfrac{3}{2}\text n\text R\Delta \text T$

But for this path n R ΔT = V₀ ΔP = V₀(2P₀ − P₀) = P₀V₀, so

$\text Q_1 = \dfrac{3}{2}\text P_0\text V_0$

Heat given to the gas from B → C (isobaric) :

$\text Q_2 = \text n\text C_p\Delta \text T = \dfrac{5}{2}\text n\text R\Delta \text T$

But for this path n R ΔT = 2P₀ ΔV = 2P₀V₀, so

$\text Q_2 = \dfrac{5}{2} \times 2\text P_0\text V_0 = 5\text P_0\text V_0$

In the processes C → D and D → A heat is rejected by the gas. Hence the efficiency of the cycle is

$\eta = \dfrac{\text{work done by the gas}}{\text{heat absorbed by the gas}} \times 100 = \dfrac{\text P_0\text V_0}{\text Q_1 + \text Q_2} \times 100$

$= \dfrac{\text P_0\text V_0}{\dfrac{3}{2}\text P_0\text V_0 + 5\text P_0\text V_0} \times 100 = \dfrac{\text P_0\text V_0}{\dfrac{13}{2}\text P_0\text V_0} \times 100$

$= \dfrac{2}{13} \times 100 = 15.4$

In a process carried on an ideal gas, W = 0 and Q < 0 . Then, of the gas :

1. the temperature will fall
2. the volume will increase
3. the pressure will remain constant
4. the temperature will rise

Answer

the temperature will fall

Reason — Given, for the process W = 0 and Q < 0.

By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W$

Since W = 0, we get ΔU = Q. As Q is negative, ΔU is also negative, that is, the internal energy of the gas decreases.

For an ideal gas the change in internal energy is related to the change in temperature by

$\Delta \text U = \mu\text C_v\Delta \text T$

Since ΔU < 0 and both μ and C_v are positive quantities, ΔT must be negative. Hence the temperature of the gas will fall.

The volume cannot increase, because W = P ΔV = 0 means the volume stays constant.

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement.

1. The change in internal energy in whole cyclic process is 250 R.
2. The change in internal energy in the process CA is 700 R.
3. The change in internal energy in the process AB is − 350 R.
4. The change in internal energy in the process BC is − 500 R

Answer

The change in internal energy in the process BC is − 500 R.

Reason — Given,

• One mole of a diatomic ideal gas, so the degrees of freedom f = 5 and $\text C_v = \dfrac{5}{2}\text R$
• T_A = 400 K, T_B = 800 K, T_C = 600 K
• The process BC is adiabatic

For one mole of a gas the change in internal energy is

$\Delta \text U = \text C_v\Delta \text T = \dfrac{1}{2}f\text R\Delta \text T = \dfrac{5}{2}\text R\Delta \text T$

For the complete cyclic process : The gas returns to its initial state, so

$\Delta \text U = 0$

Hence option 1 is incorrect.

For the process CA :

$\Delta \text U = \dfrac{5}{2}\text R(400 - 600) = \dfrac{5}{2}\text R \times (-200) = -500\ \text R$

Hence option 2 is incorrect.

For the process AB :

$\Delta \text U = \dfrac{5}{2}\text R(800 - 400) = \dfrac{5}{2}\text R \times 400 = 1000\ \text R$

Hence option 3 is incorrect.

For the process BC :

$\Delta \text U = \dfrac{5}{2}\text R(600 - 800) = \dfrac{5}{2}\text R \times (-200) = -500\ \text R$

Hence the correct statement is that the change in internal energy in the process BC is − 500 R.

Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol K) :

1. 62 J
2. 104 J
3. 124 J
4. 208 J

Answer

208 J

Reason — Given,

• Number of moles of helium, μ = 2
• Initial temperature = 30°C, final temperature = 35°C, so ΔT = 5°C = 5 K
• R = 8.31 J mol^-1 K^-1
• The balloon is fully expandable and requires no energy in its expansion

Since the balloon is fully expandable and needs no energy to expand, the gas inside always stays at the constant atmospheric pressure. The heating is therefore an isobaric process, for which the heat required is

$\text Q = \mu\text C_p\Delta \text T$

Helium is a monoatomic gas, for which

$\text C_p = \dfrac{5}{2}\text R$

Substituting the values,

$\text Q = 2 \times \dfrac{5}{2}\text R \times 5 = 25\ \text R$

$= 25 \times 8.31 \\[1em] = 207.75\ \text J \approx 208\ \text J$

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T₁, pressure P₁ and volume V₁ and the spring is in its relaxed state. The gas is then heated very slowly to temperature T₂, pressure P₂ and volume V₂. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement is :

1. If V₂ = 2 V₁ and T₂ = 3 T₁, then the energy stored in the spring is $\dfrac{1}{4}$ P₁V₁.
2. If V₂ = 2 V₁ and T₂ = 3 T₁, then the change in internal energy is 3 P₁V₁.
3. If V₂ = 3 V₁ and T₂ = 4 T₁, then the work done by the gas is $\dfrac{7}{3}$ P₁V₁.
4. If V₂ = 3 V₁ and T₂ = 4 T₁, then the heat supplied to the gas is $\dfrac{17}{6}$ P₁V₁

Answer

1. If V₂ = 2 V₁ and T₂ = 3 T₁, then the energy stored in the spring is $\dfrac{1}{4}$ P₁V₁.

2. If V₂ = 2 V₁ and T₂ = 3 T₁, then the change in internal energy is 3 P₁V₁.

3. If V₂ = 3 V₁ and T₂ = 4 T₁, then the work done by the gas is $\dfrac{7}{3}$ P₁V₁.

Reason —

The gas is monoatomic, so $\text C_v = \dfrac{3}{2}\text R$ and the total internal energy of the gas is $\text U = \dfrac{3}{2}\text{nRT} = \dfrac{3}{2}\text{PV}$.

Force on the piston : Initially the spring is in its relaxed state, so the initial gas pressure P₁ is balanced by the external pressure alone. When the piston moves out by x, the spring is deformed and exerts a force kx on it. The new force balance on the piston therefore gives

$\text P_2\text A = \text P_1\text A + \text{kx} \qquad \text{or} \qquad \text{kx} = (\text P_2 - \text P_1)\text A$

Since the heating is carried out very slowly, the pressure rises in step with the displacement, so P varies linearly with V between the initial and the final states.

Checking option 1 : For V₂ = 2V₁ and T₂ = 3T₁, the gas equation gives

$\dfrac{\text P_1\text V_1}{\text T_1} = \dfrac{\text P_2\text V_2}{\text T_2} = \dfrac{\text P_2(2\text V_1)}{3\text T_1} \quad \Rightarrow \quad \text P_2 = \dfrac{3\text P_1}{2}$

The energy stored in the spring is

$\dfrac{1}{2}\text{kx}^2 = \dfrac{1}{2}(\text{kx})\text x = \dfrac{1}{2}(\text P_2 - \text P_1)\text A\text x = \dfrac{1}{2}(\text P_2 - \text P_1)(\text V_2 - \text V_1)$

since A x is the change in volume. Substituting the values,

$= \dfrac{1}{2}\left(\dfrac{3}{2}\text P_1 - \text P_1\right)(2\text V_1 - \text V_1) = \dfrac{1}{2} \times \dfrac{1}{2}\text P_1 \times \text V_1 = \dfrac{1}{4}\text P_1\text V_1$

Hence option 1 is correct.

Checking option 2 : With P₂ = $\dfrac{3}{2}$P₁ and V₂ = 2V₁,

$\Delta \text U = \dfrac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1) = \dfrac{3}{2}\left(\dfrac{3}{2}\text P_1 \times 2\text V_1 - \text P_1\text V_1\right)$

$= \dfrac{3}{2}(3\text P_1\text V_1 - \text P_1\text V_1) = \dfrac{3}{2} \times 2\text P_1\text V_1 = 3\text P_1\text V_1$

Hence option 2 is correct.

Checking option 3 : For V₂ = 3V₁ and T₂ = 4T₁,

$\dfrac{\text P_1\text V_1}{\text T_1} = \dfrac{\text P_2(3\text V_1)}{4\text T_1} \quad \Rightarrow \quad \text P_2 = \dfrac{4\text P_1}{3}$

Since the pressure varies linearly with the volume, the work done by the gas is the area under the straight line joining the initial and the final states,

$\text W = \dfrac{1}{2}(\text P_1 + \text P_2)(\text V_2 - \text V_1) = \dfrac{1}{2}\left(\text P_1 + \dfrac{4}{3}\text P_1\right)(3\text V_1 - \text V_1)$

$= \dfrac{1}{2} \times \dfrac{7}{3}\text P_1 \times 2\text V_1 = \dfrac{7}{3}\text P_1\text V_1$

Hence option 3 is correct.

Checking option 4 : For the same case the change in internal energy is

$\Delta \text U = \dfrac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1) = \dfrac{3}{2}\left(\dfrac{4}{3}\text P_1 \times 3\text V_1 - \text P_1\text V_1\right)$

$= \dfrac{3}{2}(4\text P_1\text V_1 - \text P_1\text V_1) = \dfrac{9}{2}\text P_1\text V_1$

By the first law of thermodynamics, the heat supplied is

$\Delta \text Q = \Delta \text U + \text W = \dfrac{9}{2}\text P_1\text V_1 + \dfrac{7}{3}\text P_1\text V_1 = \dfrac{27 + 14}{6}\text P_1\text V_1 = \dfrac{41}{6}\text P_1\text V_1$

This is not $\dfrac{17}{6}\text P_1\text V_1$, so option 4 is incorrect.

Hence the correct statements are 1, 2 and 3.

Note: The printed answer key of the textbook gives option (b) alone, its hint having taken the spring force as P₂A = kx. Since the spring is relaxed in the initial state, the spring force accounts only for the rise in pressure, so kx = (P₂ − P₁)A. With this relation options 1, 2 and 3 are all correct, and the heat supplied in option 4 works out to $\dfrac{41}{6}$ P₁V₁.

The work of 146 kJ is performed in order to compress on kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is ( R = 8.3 J mol^-1 K^-1):

1. monoatomic
2. diatomic
3. triatomic
4. a mixture of monoatomic and diatomic

Answer

diatomic

Reason — Given,

• Work done on the gas, W = − 146 kJ = − 146 × 10³ J
• Number of moles, μ = 1 kilo mole = 1 × 10³ mol
• Rise in temperature, ΔT = 7°C = 7 K
• R = 8.3 J mol^-1 K^-1

According to the first law of thermodynamics,

$\text Q = \Delta \text U + \text W$

For an adiabatic process Q = 0, so

$\Delta \text U = -\text W$

But the change in internal energy is also ΔU = μ C_v ΔT. Therefore

$\text C_v = \dfrac{-\text W}{\mu\Delta \text T} = \dfrac{-(-146 \times 10^{3})}{(1 \times 10^{3}) \times 7}$

$= \dfrac{146 \times 10^{3}}{7 \times 10^{3}} = 20.8\ \text{J mol}^{-1}\ \text K^{-1}$

For a diatomic gas the molar specific heat at constant volume is

$\text C_v = \dfrac{5}{2}\text R = \dfrac{5}{2} \times 8.3 = 20.75\ \text{J mol}^{-1}\ \text K^{-1}$

Since the calculated value agrees with this, the gas is diatomic.

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T₀, while Box B contains one mole of helium at temperature $\dfrac{7}{3}$ T₀. The boxes are then put in thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore heat capacity of boxes). Then, the final temperature of the gases, T_f in terms of T₀ is :

1. T_f = $\dfrac{5}{2}$ T₀
2. T_f = $\dfrac{3}{7}$ T₀
3. T_f = $\dfrac{7}{3}$ T₀
4. T_f = $\dfrac{3}{2}$ T₀

Answer

T_f = $\dfrac{3}{2}$ T₀

Reason — Given,

• Box A : 1 mole of nitrogen (diatomic) at temperature T₀, so $\text C_{v_1} = \dfrac{5}{2}\text R$
• Box B : 1 mole of helium (monoatomic) at temperature $\dfrac{7}{3}$T₀, so $\text C_{v_2} = \dfrac{3}{2}\text R$
• The heat capacity of the boxes is to be ignored

The boxes are rigid, so the volume of each gas remains constant and no work is done. The heat lost by the hotter gas is therefore gained by the cooler gas, and the total change in internal energy is zero,

$\mu_1\text C_{v_1}\Delta \text T_1 + \mu_2\text C_{v_2}\Delta \text T_2 = 0$

Substituting the values,

$1 \times \left(\dfrac{5}{2}\text R\right) \times (\text T_f - \text T_0) + 1 \times \left(\dfrac{3}{2}\text R\right) \times \left(\text T_f - \dfrac{7}{3}\text T_0\right) = 0$

Multiplying throughout by $\dfrac{2}{\text R}$,

$5(\text T_f - \text T_0) + 3\left(\text T_f - \dfrac{7}{3}\text T_0\right) = 0$

$5\text T_f - 5\text T_0 + 3\text T_f - 7\text T_0 = 0 \\[1em] 8\text T_f = 12\text T_0$

$\text T_f = \dfrac{12}{8}\text T_0 = \dfrac{3}{2}\text T_0$

The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then select the correct statement(s) from the following :

1. the process during the path A → B is isothermal.
2. heat flows out the gas during the path B → C → D.
3. work done during the path A → B → C is zero.
4. positive work is done by the gas in the cycle ABCDA.

Answer

2. heat flows out the gas during the path B → C → D.

4. positive work is done by the gas in the cycle ABCDA.

Reason —

By the first law of thermodynamics, ΔQ = ΔU + W.

Checking option 1 : For the path A → B the P-V curve is a part of the semi-circle ABC, and not a rectangular hyperbola. An isothermal curve for an ideal gas obeys PV = constant, which is a rectangular hyperbola. Hence the process A → B is not isothermal and option 1 is incorrect.

Checking option 2 : The states are B (3, 2), C (2, 1) and D (1, 2). Therefore

$\text P_{\text B}\text V_{\text B} = 3 \times 2 = 6 \qquad \text{and} \qquad \text P_{\text D}\text V_{\text D} = 1 \times 2 = 2$

Since P_DV_D < P_BV_B, the change in internal energy

$\Delta \text U = \dfrac{\text P_{\text D}\text V_{\text D} - \text P_{\text B}\text V_{\text B}}{\gamma - 1}$

is negative. Again, along the whole path B → C → D the volume falls steadily from 3 to 1, so the work done is also negative. Hence ΔQ is negative, that is, heat flows out of the gas along B → C → D, and option 2 is correct.

Checking option 3 : Along the path A → B → C the volume of the gas changes, so the work done, being the area under the curve, is not zero. Hence option 3 is incorrect.

Checking option 4 : The closed curve ABCDA is traced clockwise on the P-V diagram. For a clockwise cycle the net work is done by the gas and is therefore positive. Hence option 4 is correct.

One or more option(s) is/are correct :

C_v and C_p denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

1. C_p − C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.
2. C_p + C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.
3. C_p/C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.
4. C_p · C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.

Answer

2. C_p + C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.

4. C_p · C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.

Reason — For a diatomic ideal gas :

$\text C_v = \dfrac{5}{2}\text R \quad \text{and} \quad \text C_p = \dfrac{7}{2}\text R$

$\text C_p - \text C_v = \dfrac{7}{2}\text R - \dfrac{5}{2}\text R = \text R \\[1em] \text C_p + \text C_v = \dfrac{7}{2}\text R + \dfrac{5}{2}\text R = 6\text R$

$\dfrac{\text C_p}{\text C_v} = \dfrac{7/2\ \text R}{5/2\ \text R} = \dfrac{7}{5} \\[1em] \text C_p \cdot \text C_v = \dfrac{7}{2}\text R \times \dfrac{5}{2}\text R = \dfrac{35}{4}\text R^2$

For a monoatomic ideal gas :

$\text C_v = \dfrac{3}{2}\text R \quad \text{and} \quad \text C_p = \dfrac{5}{2}\text R$

$\text C_p - \text C_v = \dfrac{5}{2}\text R - \dfrac{3}{2}\text R = \text R \\[1em] \text C_p + \text C_v = \dfrac{5}{2}\text R + \dfrac{3}{2}\text R = 4\text R$

$\dfrac{\text C_p}{\text C_v} = \dfrac{5/2\ \text R}{3/2\ \text R} = \dfrac{5}{3} \\[1em] \text C_p \cdot \text C_v = \dfrac{5}{2}\text R \times \dfrac{3}{2}\text R = \dfrac{15}{4}\text R^2$

Comparing the two :

C_p − C_v = R for both the gases, so option 1 is incorrect.

C_p + C_v is 6R for the diatomic gas and 4R for the monoatomic gas, so option 2 is correct.

$\dfrac{\text C_p}{\text C_v}$ is $\dfrac{7}{5}$ = 1.4 for the diatomic gas and $\dfrac{5}{3}$ = 1.67 for the monoatomic gas, which is smaller for the diatomic gas, so option 3 is incorrect.

C_p · C_v is $\dfrac{35}{4}\text R^2$ for the diatomic gas and $\dfrac{15}{4}\text R^2$ for the monoatomic gas, so option 4 is correct.

The figure shows the variation of specific heat capacity c of a solid as a function of temperature T. The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation.

1. The rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T.
2. Heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K.
3. There is no change in the rate of heat absorption in range 400-500 K.
4. The rate of heat absorption increases in the range 200-300 K.

Answer

1. The rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T.

2. Heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K.

3. There is no change in the rate of heat absorption in range 400-500 K.

4. The rate of heat absorption increases in the range 200-300 K.

Reason —

The heat absorbed in raising the temperature of a solid of mass m is

$\Delta \text Q = \int \text{mc}\text{dT} = \text m \times (\text{area under the c-T graph})$

Differentiating with respect to time,

$\dfrac{\text{dQ}}{\text{dt}} = \text{mc}\dfrac{\text{dT}}{\text{dt}}$

The temperature is increased at a constant rate, so $\dfrac{\text{dT}}{\text{dt}}$ is constant, and the mass m is also constant. Therefore

$\dfrac{\text{dQ}}{\text{dt}} \propto \text c$

that is, the rate of heat absorption at any stage is proportional to the specific heat capacity c at that temperature.

Option 1 is correct, because in the range 0-100 K the graph of c against T is very nearly a straight line, so the rate of heat absorption varies linearly with T.

Option 2 is correct, because the area under the curve in the range 0-100 K is less than the area under the curve in the range 400-500 K, and the heat absorbed is proportional to this area.

Option 3 is correct, because in the range 400-500 K the graph of c against T is parallel to the temperature axis, so c is constant and the rate of heat absorption does not change.

Option 4 is correct, because in the range 200-300 K the specific heat capacity c increases with temperature, and hence the rate of heat absorption also increases.

Hence all the four statements are correct.

A Carnot's engine takes in 3000 kcal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work done by the engine is : (1 kcal = 4.2 × 10³ J)

1. 4.2 × 10⁶ J
2. 8.4 × 10⁶ J
3. 16.8 × 10⁶ J
4. zero

Answer

8.4 × 10⁶ J

Reason — Given,

• Heat taken in from the reservoir, Q₁ = 3000 kcal
• Temperature of the source, T₁ = 627°C = 627 + 273 = 900 K
• Temperature of the sink, T₂ = 27°C = 27 + 273 = 300 K
• 1 kcal = 4.2 × 10³ J

The efficiency of a Carnot engine is

$\eta = 1 - \dfrac{\text T_2}{\text T_1} = 1 - \dfrac{300}{900} = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

The heat energy taken in from the source, expressed in joule, is

$\text Q_1 = 3000\ \text{kcal} = 3000 \times 10^{3}\ \text{cal} = 3 \times 10^{6}\ \text{cal}$

$= 3 \times 10^{6} \times 4.2\ \text J$

The work done by the engine is

$\text W = \text Q_1 \times \eta = 3 \times 10^{6} \times 4.2 \times \dfrac{2}{3}$

$= 8.4 \times 10^{6}\ \text J$

A Carnot's engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be :

1. efficiency of Carnot's engine cannot be made larger than 50%
2. 1200 K
3. 750 K
4. 600 K

Answer

750 K

Reason — Given,

• Efficiency in the first case, η₁ = 40%
• Temperature of the source in the first case, T₁ = 500 K
• Desired efficiency, η₂ = 60%, the sink temperature remaining the same

For the first case : The efficiency of a Carnot engine is

$\eta = \left(1 - \dfrac{\text T_2}{\text T_1}\right) \times 100$

Substituting the values,

$40 = \left(1 - \dfrac{\text T_2}{500}\right) \times 100 \\[1em] 0.4 = 1 - \dfrac{\text T_2}{500}$

$\dfrac{\text T_2}{500} = 0.6 \quad \Rightarrow \quad \text T_2 = 300\ \text K$

For the second case : The sink temperature stays at 300 K and the efficiency is to be 60%. If T₁′ be the new intake temperature,

$60 = \left(1 - \dfrac{300}{\text T_1'}\right) \times 100 \\[1em] 0.6 = 1 - \dfrac{300}{\text T_1'}$

$\dfrac{300}{\text T_1'} = 0.4 \quad \Rightarrow \quad \text T_1' = \dfrac{300}{0.4} = 750\ \text K$

A Carnot's engine operation between temperatures T₁ and T₂ has efficiency (1/6). When T₂ is lowered by 62 K, its efficiency increases to (1/3). Then T₁ and T₂ respectively are :

1. 310 K and 248 K
2. 372 K and 310 K
3. 372 K and 330 K
4. 330 K and 268 K

Answer

372 K and 310 K

Reason — Given,

• Efficiency in the first case, $\eta_1 = \dfrac{1}{6}$
• When T₂ is lowered by 62 K, the efficiency becomes $\eta_2 = \dfrac{1}{3}$

For the first case : The efficiency of a Carnot engine is

$\eta = 1 - \dfrac{\text T_2}{\text T_1}$

$\dfrac{1}{6} = 1 - \dfrac{\text T_2}{\text T_1} \quad \Rightarrow \quad \dfrac{\text T_2}{\text T_1} = 1 - \dfrac{1}{6} = \dfrac{5}{6} \qquad \text{...(i)}$

For the second case : The sink temperature becomes (T₂ − 62) K,

$\dfrac{1}{3} = 1 - \dfrac{\text T_2 - 62}{\text T_1} \quad \Rightarrow \quad \dfrac{\text T_2 - 62}{\text T_1} = 1 - \dfrac{1}{3} = \dfrac{2}{3} \qquad \text{...(ii)}$

Subtracting (ii) from (i),

$\dfrac{\text T_2}{\text T_1} - \dfrac{\text T_2 - 62}{\text T_1} = \dfrac{5}{6} - \dfrac{2}{3} \\[1em] \dfrac{62}{\text T_1} = \dfrac{5 - 4}{6} = \dfrac{1}{6}$

$\text T_1 = 62 \times 6 = 372\ \text K$

Substituting this value in equation (i),

$\text T_2 = \dfrac{5}{6} \times 372 = 310\ \text K$

Hence T₁ = 372 K and T₂ = 310 K.

A steam engine delivers 5.4 × 10⁸ J of work per minute and receives 3.6 × 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Answer

Given,

• Work delivered by the engine per minute, W = 5.4 × 10⁸ J
• Heat received from the boiler per minute, Q₁ = 3.6 × 10⁹ J

Efficiency of the engine : The efficiency is the ratio of the useful work delivered by the engine to the heat taken in from the hot reservoir,

$\eta = \dfrac{\text W}{\text Q_1} = \dfrac{5.4 \times 10^{8}\ \text{J min}^{-1}}{3.6 \times 10^{9}\ \text{J min}^{-1}}$

$= 0.15 = 15$

Heat wasted per minute : The heat wasted is the heat rejected to the sink, which is the difference between the heat taken in and the work done,

$\text Q_2 = \text Q_1 - \text W = 3.6 \times 10^{9} - 5.4 \times 10^{8}$

$= 3.6 \times 10^{9} - 0.54 \times 10^{9} \\[1em] = 3.06 \times 10^{9}\ \text{J min}^{-1} \approx 3.1 \times 10^{9}\ \text{J min}^{-1}$

Hence the efficiency of the engine is 15% and the heat wasted per minute is 3.1 × 10⁹ J.

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

Answer

Given,

• Temperature of the eatables kept inside (cold body), T₂ = 9°C = 9 + 273 = 282 K
• Room temperature (hot body), T₁ = 36°C = 36 + 273 = 309 K

The coefficient of performance of a refrigerator is defined as the ratio of the heat taken in from the cold body to the work needed to run the refrigerator, and for an ideal refrigerator it is given by

$\text K = \dfrac{\text T_2}{\text T_1 - \text T_2}$

where T₁ is the absolute temperature of the hot body to which heat is delivered and T₂ is the absolute temperature of the cold body from which heat is removed.

Substituting the values,

$\text K = \dfrac{282\ \text K}{309\ \text K - 282\ \text K} = \dfrac{282\ \text K}{27\ \text K}$

$= 10.4$

Hence, the coefficient of performance of the refrigerator is 10.4.

An insulated cylinder with a movable insulated piston contains 3 moles of hydrogen at STP. By what factor does the pressure of the gas increase, if the gas is compressed to half its original volume? Given : γ = 1.4.

Answer

Given,

• Number of moles of hydrogen, μ = 3
• The gas is initially at STP
• Final volume, $\text V_2 = \dfrac{\text V_1}{2}$
• γ = 1.4

Let P₁, V₁ be the original pressure and volume of the gas and P₂, V₂ those after the gas is compressed.

The cylinder is insulated and the piston is also insulated, so the whole system is well insulated and no heat can enter or leave the gas. The compression is therefore adiabatic and obeys Poisson's law,

$\text P_1\text V_1^{\gamma} = \text P_2\text V_2^{\gamma}$

Rearranging,

$\dfrac{\text P_2}{\text P_1} = \left(\dfrac{\text V_1}{\text V_2}\right)^{\gamma}$

Here $\text V_2 = \dfrac{\text V_1}{2}$, so

$\dfrac{\text P_2}{\text P_1} = \left(\dfrac{\text V_1}{\text V_1/2}\right)^{1.4} = (2)^{1.4}$

$= 2.64$

Hence, the pressure of the gas increases by a factor of 2.64.