Study of Gas Laws — Question 22

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 75% of its original value

$= \dfrac{75 }{100} \text{ of P} \\[1em] = \dfrac{3 }{4} \text{P} \\[1em] = 0.75 \text {P}$

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

$= \text {1} + \dfrac{50}{100} \text{ of V} \\[1em] = \dfrac{100 + 50}{100} \text{ of V} \\[1em] = \dfrac{150}{100} \text{of V} \\[1em] = \dfrac{3}{2} \text{V} = 1.5 \text{V}$

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 :

$\dfrac{\text{P}\text{V}}{258} = \dfrac{0.75\text{P}\times\text{1.5V}}{\text{T}_2}$

Solve for T₂:

$\text {T}_2 = 1.125 \times 258 \\[1em] \text {T}_2 = 290.25 \text {K}$

∴ Final temperature of the gas = 290.25 - 273 = 17.25°C