At high pressure, van der Waals' equation becomes:

P V = R T

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P V = R T + \frac{a}{V}

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P V = R T - \frac{a}{V}

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P V = R T + P b

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Solution

The correct option is D $P V = R T + P b$
For one mole, van der Waals' equation becomes:
$(P + \frac{a}{V^{2}}) (V - b) = R T$ , where $a$ , $b$ are van der Waals constants.
At high pressure, $(P + \frac{a}{V^{2}}) = P$ as value of $\frac{a}{V^{2}}$ << $P$ .
$∴ P (V - b) = R T$
= $P V - P b = R T$
$\Rightarrow P V = R T + P b$

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Similar questions

[P + \frac{a}{V^{2}}] (V - b) = R T

Z

Column A	Column B
(A) High pressure	(i) $P V = R T + P b$
(B) Low pressure	(ii) $P V = R T - \frac{a}{V}$
(C) $P \to 0$	(iii) $P V = R T$
(D) Neither $a$ nor $b$ is negligible	(iv) $(P + \frac{a}{V^{2}}) (V - b) = R T$

[P + \frac{a}{V^{2}}] V = R T

[P + \frac{a}{V^{2}}] V = R T

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