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ICSE Class 9 Chemistry Question 18 of 29

Study of Gas Laws — Question 27

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Question

Question 27

Calculate the volume of dry air at S.T.P. that occupies 28 cm3 at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.

Answer

Initial conditions :

P1 = 750 - water vapour pressure = 750 - 12 = 738 mm Hg

V1 = Initial volume of the gas = 28 cm3

T1 = Initial temperature of the gas = 14°C + 273 = 287 K

Final conditions [S.T.P.] :

P2 (Final pressure) = 760 mm of Hg

T2 (Final temperature) = 273 K

V2 (Final volume) = ?

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 :

738×28287=760×V2273V2=738×28×273287×760V2=25.86 cm3\dfrac{738 \times 28 }{287} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{738 \times 28 \times 273}{287 \times 760} \\[1em] \text{V}_2 = 25.86 \text{ cm}^3 \\[1em]

∴ Volume of dry gas = 25.9 cm3