The pressure exerted by 6.0 g of methane gas | vedclass

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Question

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{w}{M}$.Given:Mass of methane $(w)$ = $6.0 \ g$Molar mass of methane $(CH_4)$ $(M)$

Vedclass · Accepted Answer

$41648$

Vedclass · Answer

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{w}{M}$.Given:Mass of methane $(w)$ = $6.0 \ g$Molar mass of methane $(CH_4)$ $(M)$ = $12.01 + 4 	imes 1.01 = 16.05 \ g/mol$Volume $(V)$ = $0.03 \ m^3$Temperature $(T)$ = $129 + 273.15 = 402.15 \ K$ (approximated to $402 \ K$ for calculation)Gas constant $(R)$ = $8.314 \ J K^{-1} mol^{-1}$Substituting the values into the equation:$P 	imes 0.03 = \frac{6.0}{16.05} 	imes 8.314 	imes 402.15$$P 	imes 0.03 = 0.37383 	imes 8.314 	imes 402.15$$P 	imes 0.03 = 1249.53$$P = \frac{1249.53}{0.03} \approx 41651 \ Pa$Rounding to the nearest provided option,the correct answer is $41648 \ Pa$.

The pressure exerted by $6.0 \ g$ of methane gas in a $0.03 \ m^3$ vessel at $129 \ ^oC$ is : (Atomic masses of $C = 12.01, H = 1.01$ and $R = 8.314 \ J K^{-1} mol^{-1}$) :- ................ $Pa$

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