It is found, on heating a gas, its volume increases by 50% and pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated?

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = P
V₁ = Initial volume of the gas = V
T₁ = Initial temperature of the gas = -15°C = -15 + 273 = 258 K

Final conditions :

P₂ (Final pressure) = pressure decreases to 60% of its original value

= $\dfrac{60}{100}$ of P

= $\dfrac{3}{5}$P

V₂ (Final volume) = volume increases by 50% of its original value

= 1 + $\dfrac{50}{100}$ of V

= $\dfrac{100 + 50}{100}$ of V

= $\dfrac{150}{100}$ of V

= $\dfrac{3}{2}$V

T₂ (Final temperature) = ?

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{PV }}{258} = \dfrac{\dfrac{3}{5} \text{P}\times\dfrac{3}{2}\text{V}}{\text{T}$2} \\[1em] \text{T}2 = \dfrac{3\times 3\times 258}{5 \times 2} = \dfrac{2322}{10} \\[1em] \text{T}_2 = 232.2 \text{K} \\[1em]

∴ Final temperature of the gas = 232.2 - 273 = -40.8°C