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ICSE Class 9 Chemistry Question 3 of 29

Study of Gas Laws — Question 8

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Question 8

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.

Answer

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 = 75100\dfrac{75}{100} of P = 34P\dfrac{3}{4}\text{P}

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

=40100 of V + V=4V10+V=4V + 10V10=14V10= \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:

P1×V1T1=P2×V2T2\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 :

P×V263=34P×14V10T2T2=34×1410×263T2=276.15K\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