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JEE Main levelSingle correct optionDifficulty 2 of 3rg181

Rearranging the van der Waals equation, the pressure of $n$ moles of a real gas is given by

A $P = \frac{n RT}{V - nb} - \frac{a n ^{2}}{V ^{2}}$
B $P = \frac{n RT}{V - nb} + \frac{a n ^{2}}{V ^{2}}$
C $P = \frac{n RT}{V + nb} - \frac{a n ^{2}}{V ^{2}}$
D $P = \frac{n RT}{V} - \frac{an}{V ^{2}}$

Answer: A

From

(P + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n RT

,

P + \frac{a n ^{2}}{V ^{2}} = \frac{n RT}{V - nb}

, so

P = \frac{n RT}{V - nb} - \frac{a n ^{2}}{V ^{2}}

. The first term exceeds the ideal pressure and the second lowers it.

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