Calculate pressure exerted by 1 mole of a van der Waals' gas at a temperature of 80.0821 K in 12 Lcontainer if volume of the molecule is assumed to be negligible and van der Waal's constant a=2 atm-lit2mole−2

Use the following formula;P+n2aV2(V)=nRTPVm+aVm=R×80.0821PVm2−8Vm+a=0According to the question, Vm=12L=0.5LSo, P(0.5)2−8(0.5)+2=0P=8 atm

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Calculate pressure exerted by 1 mole of a van der Waals' gas at a temperature of $\frac{8}{0.0821} K$ in $\frac{1}{2} L$
container if volume of the molecule is assumed to be negligible and van der Waal's constant $a = 2$ atm- ${lit}^{2} {mole}^{- 2}$

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Use the following formula;

$[P + \frac{n^{2} a}{V^{2}}] (V) = nRT$

${PV}_{m} + \frac{a}{V_{m}} = R \times \frac{8}{0 .0821}$

${PV}_{m}^{2} - 8 V_{m} + a = 0$

According to the question,

$V_{m} = \frac{1}{2} L = 0 .5 L$

So, $\begin{array}{l} P (0 .5)^{2} - 8 (0 .5) + 2 = 0 \\ P = 8 atm \end{array}$

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