Calculate the temperature of the gas if it obeys van der Waals' equation from the following data. A flask of $2.5 l i t r e$ contains $10 m o l e$ of a gas under $50 a t m$ . Given, $a = 5.46 a t m l i t r e^{2} m o l^{- 2}$ and $b = 0.031 l i t r e m o l^{- 1}$ .

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Solution

Given that,
$P = 50 a t m; V = 2.5 l i t r e, n = 10; a = 5.46 a t m l i t r e^{2} m o l^{- 2}$ ;
$b = 0.031 l i t r e m o l^{- 1}; R = 0.0821 l i t r e - a t m K^{- 1} m o l^{- 1}$ .
Applying van der Waals' equation,
$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$
or $T = \frac{(P + \frac{a n^{2}}{V^{2}}) (V - n b)}{n R}$
$= \frac{(50 + \frac{5.46 \times 100}{(2.5)^{2}}) (2.5 - 10 \times 0.031)}{10 \times 0.0821} = 366.4 K$

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