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NCERT Class XI Chemistry States of Matter Solutions

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Question : 8 of 23
Marks: +1, -0
What will be the pressure of the gaseous mixture when 0.5 L of H2\mathrm{H}_2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution:  
Partial pressure of hydrogen gas
V1 = 0.5 L V2 = 1.0 L P1 = 0.8 bar P2 = ?
Applying Boyle’s law (constant T and n)
P1V1P_1V_1 = P2V2P_2V_2 or P2P_2 = P1V1V2\frac{P_1V_1}{V_2}
∴ P2P_2 = (0.8 bar)×(0.5 L)1.0 L\frac{(0.8\ \mathrm{bar})\times(0.5\ \mathrm{L})}{1.0\ \mathrm{L}} = 0.40 bar
Partial pressure of oxygen gas,
V1V_1 = 2.0 L, V2V_2 = 1.0 L, P1P_1 = 0.7 bar , P2P_2 = ?
or P2P_2 = P1V1V2\frac{P_1V_1}{V_2} = (0.7 bar)×(2.0 L)1.0 L\frac{(0.7\ \mathrm{bar})\times(2.0\ \mathrm{L})}{1.0\ \mathrm{L}} = 1.40 bar
Pressure of the gas mixture, PmixP_{\text{mix}} = PH2+PO2P_{\mathrm{H}_2} + P_{\mathrm{O}_2} = 0.40 + 1.40 = 1.80 bar
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