Calculate the volume of dry air at $S T P$ that occupies $28 c m^{3}$ at $14^{o} C$ and $750 m m H g$ pressure when saturated with water vapour. The vapour pressure of water at $14^{o} C$ is $12 m m H g$ .

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Solution

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Here, $P_{1} = 750 - 12 = 738 m m, V_{1} = 28 c m^{3}, T_{1} = 14 + 273 = 287 K, P_{2} = 760 m m, T_{2} = 273 K$
Then, $V_{2} = \frac{738 \times 28 \times 273}{287 \times 760} = 25.9 c m^{3}$

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

P1V1T1=P2V2T2Here, P1=750−12=738mm,V1=28cm3,T1=14+273=287K,P2=760mm,T2=273KThen, V2=738×28×273287×760=25.9cm3

Calculate the volume of dry air at $S T P$ that occupies $28 c m^{3}$ at $14^{o} C$ and $750 m m H g$ pressure when saturated with water vapour. The vapour pressure of water at $14^{o} C$ is $12 m m H g$ .

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Here, $P_{1} = 750 - 12 = 738 m m, V_{1} = 28 c m^{3}, T_{1} = 14 + 273 = 287 K, P_{2} = 760 m m, T_{2} = 273 K$
Then, $V_{2} = \frac{738 \times 28 \times 273}{287 \times 760} = 25.9 c m^{3}$