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NCERT Class XI Chemistry States of Matter Solutions

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Question : 15 of 23
Marks: +1, -0
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3\mathrm{dm}^3 at 27°C. R = 0.083 bar dm3 K−1 mol−1\mathrm{dm}^3\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}
Solution:  
Partial pressure of oxygen gas,
P = nRTV\frac{nRT}{V} , n = 832\frac{8}{32} mol , V = 1 dm3\mathrm{dm}^3 , T = 300 K
PO2P_{\mathrm{O}_2} = 8×0.083×30032×1\frac{8\times0.083\times300}{32\times1} = 6.255 bar
Partial pressure of hydrogen gas, P = nRTV\frac{nRT}{V} , n = 42\frac{4}{2} = 2 mol
PH2P_{\mathrm{H}_2} = 2×0.083×3001\frac{2\times0.083\times300}{1} = 49.8 bar
Total pressure = PO2+PH2P_{\mathrm{O}_2} + P_{\mathrm{H}_2} = 6.225 + 49.8 = 56.025 bar
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