The pressure exerted by 5 mol of a real gas | vedclass

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Question

(B) The van der Waals equation for a real gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$ Given: $n = 5 \ mol$,$V = 1 \ L$,$T = 47 + 273 = 320 \ K$,$a

Vedclass · Accepted Answer

$77.2$

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(B) The van der Waals equation for a real gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$ Given: $n = 5 \ mol$,$V = 1 \ L$,$T = 47 + 273 = 320 \ K$,$a = 3.592 \ atm \ L^2 \ mol^{-2}$,$b = 0.0427 \ L \ mol^{-1}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$ Substituting the values: $(P + \frac{3.592 	imes 5^2}{1^2})(1 - 5 	imes 0.0427) = 5 	imes 0.0821 	imes 320$ $(P + 89.8)(1 - 0.2135) = 131.36$ $(P + 89.8)(0.7865) = 131.36$ $P + 89.8 = \frac{131.36}{0.7865} \approx 167.02$ $P = 167.02 - 89.8 = 77.22 \ atm$ Thus,the pressure is approximately $77.2 \ atm$.

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

The pressure exerted by $5 \ mol$ of a real gas in a $1 \ L$ container at $47 \ ^oC$ is ......... $atm$. (Given: $a = 3.592 \ atm \ L^2 \ mol^{-2}$,$b = 0.0427 \ L \ mol^{-1}$)

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