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States of Matter
Ideal Gas Equation
3 questions, each with its answer.
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A 10.0 L flask contains 64 g of oxygen at $27^\circ C$. (Assume $O_2$ gas is behaving ideally). The pressure inside the flask in bar is (Given $R=0.0831\ L\ bar\ K^{-1}mol^{-1}$)$n_{O_2} = \frac{64}{32} = 2$ mol, V = 10 L, T = 300 K $PV = nRT \Rightarrow P = \frac{2 \times 0.0831 \times 300}{10}$ P = 4.9 bar
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The volume occupied by 1.8 g of water vapour at $374^\circ C$ and 1 bar pressure will be: [Use R = 0.083 bar L K$^{-1}$ mol$^{-1}$]$n=\frac{1.8}{18}=0.1$ mol, $T=374+273=647$ K. $V=\frac{nRT}{P}=\frac{0.1\times0.083\times647}{1}=5.37$ L.
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The pressure exerted by 6.0 g of methane gas in a 0.03 $m^3$ vessel at $129^\circ C$ is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J $K^{-1}mol^{-1}$)Molar mass of $CH_4$ = 12.01 + 4(1.01) = 16.05 g $mol^{-1}$
Moles of methane: $n = \frac{6.0}{16.05} = 0.3738$ mol
T = 129 + 273 = 402 K; V = 0.03 $m^3$
$P = \frac{nRT}{V} = \frac{6.0}{16.05}\times\frac{8.314\times402}{0.03}$
$P = 0.3738\times111408$
P = 41648 Pa
That is every question in Ideal Gas Equation.
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