At a given temperature the pressure of a gas reduces to 75% of it's initial value and the volume increases by 40% of it's initial value. Find this temperature if the initial temperature was -10°C.

Initial conditions :

P₁ = Initial pressure of the gas = P
V₁ = Initial volume of the gas = V
T₁ = Initial temperature of the gas = -10°C = 273-10 = 263 K

Final conditions:

P₂ = Final pressure of the gas = reduces to 75% of P = $\dfrac{75}{100}$ of P = $\dfrac{3}{4}\text{P}$

V₂ = Final volume of the gas = increases by 40% of it's initial value

$= \dfrac{40}{100} \text{ of V + V} \\[1em] = \dfrac{4\text{V}}{10} + \text{V} \\[1em] = \dfrac{4\text{V + 10\text{V}}}{10} \\[1em] = \dfrac{14\text{V}}{10}$

T₂ = Final temperature of the gas = ?

By Gas Law:

$\dfrac{\text{P}$1\times\text{V}1}{\text{T}1} = \dfrac{\text{P}2\times\text{V}2}{\text{T}2}

Substituting the values :

$\dfrac{\text{P} \times \text{V}}{263} = \dfrac{\dfrac{3}{4}\text{P} \times\dfrac{14\text{V}}{10}}{\text{T}$2} \\[0.5em] \text{T}2 = \dfrac{3}{4} \times\dfrac{14}{10} \times 263 \\[0.5em] \text{T}_2 = 276.15 \text{K}

Therefore, final temperature of the gas = 276.15 K - 273 K = 3.15 °C