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ICSE Class 9 Chemistry Question 25 of 29

Study of Gas Laws — Question 10

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Question

Question 10

Certain amount of a gas occupies a volume of 0.4 litre at 17°C. Calculate the temperature at which the volume of gas is (a) doubled (b) reduced to half, pressure remaining constant?

Answer

(a) To find the temperature so that volume gets doubled :

V1 = Initial volume of the gas = 0.4 litre
T1 = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V2 = Final volume of the gas = double of initial = 2 x 0.4 = 0.8 litre
T2 = Final temperature of the gas = ?

By Charles' Law:

V1T1=V2T2\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}

Substituting the values :

0.4290=0.8T2T2=0.8×2900.4T2=580 K\dfrac{\text{0.4}}{290} = \dfrac{\text{0.8}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{0.8 \times 290}{0.4} \\[1em] \text{T}_2 = 580 \text{ K}

∴ Final temperature in °C = 580 - 273 = 307°C

(b) To find the temperature so that volume gets reduced to half :

V1 = Initial volume of the gas = 0.4 litre
T1 = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V2 = Final volume of the gas = reduced to half of initial = 0.2 litre
T2 = Final temperature of the gas = ?

By Charles' Law:

V1T1=V2T2\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}

Substituting the values :

0.4290=0.2T2T2=0.2×2900.4T2=145\dfrac{\text{0.4}}{290} = \dfrac{\text{0.2}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{0.2\times 290}{0.4} \\[1em] \text{T}_2 = 145

∴ Final temperature in °C = 145 - 273 = -128°C