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Pressure exerted by 1 mole of methane in a 0.25 litre container at $300 \mathrm{~K}$ using vander Waal's equation (given $\left.1=2.253 \mathrm{atml}^2 \mathrm{~mol}^{-2}, b=0.0428 / \mathrm{litmol}^{-1}\right)$ is
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The correct answer is:
$82.82 \mathrm{~atm}$
$\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$
$\left(P+\frac{2.253}{0.25 \times 0.25}\right)(0.25-0.0428)=0.0821 \times 300$
or $(P+36.048)(0.2072)=24.63$
$\Rightarrow P+36.048=118.87 \Rightarrow P=82.82 \mathrm{~atm}$.
$\left(P+\frac{2.253}{0.25 \times 0.25}\right)(0.25-0.0428)=0.0821 \times 300$
or $(P+36.048)(0.2072)=24.63$
$\Rightarrow P+36.048=118.87 \Rightarrow P=82.82 \mathrm{~atm}$.
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