Question

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

• A. $82.82\, atm$
• B. $152.51\, atm$
• C. $190.52\, atm$
• D. $70.52\, atm$

Answer 1

Answer: (A)

$\left(P+\frac{n^{2}a}{V^{2}}\right) \left(V-nb\right)=nRT$
$\left(P+\frac{2.253}{0.25\times0.25}\right)\left(0.25-0.0428\right)=0.0821\times300$
or $\left(P+36.048\right)\left(0.2072\right)=24.63$
$\Rightarrow P+36.075=118.87$
$\Rightarrow P=82.82 \,atm$