If the graph is plotted for $1 m o l e$ gas in such a way that $P V$ is plotted against $^{'} P^{'}$ , then intercept of the graph for real gas will be

R T + P b + a

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R T

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R T - P b + a

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R T \div P b + a b + a

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Solution

The correct option is B $R T$
If the graph is plotted for 1 mole gas in such a way that $P V$ is plotted against $P$ , then intercept of the graph for real gas will be $R T$ .
For 1 mole of real gas, the van der waals equation is $(P + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T$
$P V_{m} - P b + \frac{a}{V_{m}} - \frac{a}{V_{m}^{2}} = R T$
$P V_{m} = R T + P b - \frac{a}{V_{m}} + \frac{a b}{V_{m}^{2}}$
To calculate the intercept, $P \to 0$ , so $V_{m} \to \infty$ .
The last two terms on the right side of the equation can be neglected.
$P V_{m} = R T + P b$
Thus, when $P = 0$ , intercept $= R T$

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If the graph is plotted for 1 mole gas in such a way that PV is plotted against ′P′, then intercept of the graph for real gas will be

If the graph is plotted for $1 m o l e$ gas in such a way that $P V$ is plotted against $^{'} P^{'}$ , then intercept of the graph for real gas will be