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ICSE Class 9 Chemistry Question 12 of 29

Study of Gas Laws — Question 21

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Question

Question 21

Calculate the temperature at which the volume of a gas is doubled if its pressure at the same time increases from 70 cm of Hg to 80 cm of Hg.

Answer

Initial conditions :

P1 = Initial pressure of the gas = 70 cm of Hg

V1 = Initial volume of the gas = V

T1 = Initial temperature of the gas = 273 K

Final conditions:

P2 (Final pressure) = 80 cm of Hg

V2 (Final volume) = 2V

T2 (Final temperature) = ?

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 :

70×V273=80×2VT2T2=80×2V×27370×VT2=16×2737T2=624 K\dfrac{70 \times \text{V}}{273} = \dfrac{80 \times \text{2V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{80 \times \text{2V} \times 273}{70 \times \text{V}} \\[1em] \text{T}_2 = \dfrac{16 \times 273}{7} \\[1em] \text{T}_2 = 624 \text{ K} \\[1em]

∴ Final temperature of the gas = 624 - 273 = 351°C