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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 $K^{-1}mol^{-1}$)
A
215216 Pa
B
13409 Pa
C
41648 Pa
D
31684 Pa
Detailed Solution
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
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