Question:

The pressure exerted by a non-reactive gaseous mixture of 6.4g of methane and 8.8 g of carbon dioxide in a 10L vessel at 27\(^\circ\)C is __________ kPa. (Round off to the Nearest Integer).

[Assume gases are ideal, R = 8.314 J \(\mathrm{mol^{-1}\ K^{-1}}\)

Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]

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Add the moles of CH4 and CO2 and use \(P=\frac{nRT}{V}\) with T = 300 K, V = 10 L = 0.010 m3.
Updated On: Nov 4, 2026
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Correct Answer: 150

Solution and Explanation

\(n_{\mathrm{CH_4}}=\frac{6.4}{16}=0.4\) mol; \(n_{\mathrm{CO_2}}=\frac{8.8}{44}=0.2\) mol; total \(n=0.6\) mol.
\(T=27+273=300\) K, \(V=10\ \mathrm{L}=10\times10^{-3}\ \mathrm{m^3}\).
\(P=\frac{nRT}{V}=\frac{0.6\times8.314\times300}{10\times10^{-3}}=1.4965\times10^{5}\ \mathrm{Pa}\approx149.65\ \mathrm{kPa}\).
Rounded off: 150 kPa.
Answer: 150
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