Pressure exerted by 1 , mole of methane in a | vedclass

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(A) The van der Waals equation is $(P + \frac{n^2a}{V^2})(V - nb) = nRT$. Given: $n = 1 \, mol$,$V = 0.25 \, L$,$T = 300 \, K$,$a = 2.253 \, atm \, L^

Vedclass · Accepted Answer

$82.82$

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(A) The van der Waals equation is $(P + \frac{n^2a}{V^2})(V - nb) = nRT$. Given: $n = 1 \, mol$,$V = 0.25 \, L$,$T = 300 \, K$,$a = 2.253 \, atm \, L^2 \, mol^{-2}$,$b = 0.0428 \, L \, mol^{-1}$,and $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$. Substituting the values: $(P + \frac{1^2 	imes 2.253}{0.25^2})(0.25 - 1 	imes 0.0428) = 1 	imes 0.0821 	imes 300$. $(P + \frac{2.253}{0.0625})(0.2072) = 24.63$. $(P + 36.048)(0.2072) = 24.63$. $P + 36.048 = \frac{24.63}{0.2072} \approx 118.87$. $P = 118.87 - 36.048 = 82.822 \, atm$.

Gas	$Ar$	$Ne$	$Kr$	$Xe$
$a \ (atm \ dm^6 \ mol^{-2})$	$1.3$	$0.2$	$5.1$	$4.1$
$b \ (10^{-2} \ dm^3 \ mol^{-1})$	$3.2$	$1.7$	$1.0$	$5.0$

Pressure exerted by $1 \, mole$ of methane in a $0.25 \, L$ container at $300 \, K$ using van der Waals equation is ............... $atm$ (given $a = 2.253 \, atm \, L^2 \, mol^{-2}, b = 0.0428 \, L \, mol^{-1}$):

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