At high pressure, van der Waal's equation for one mole of the gas 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 $P V = R T + P b$
$(P + a \frac{n^{2}}{V^{2}}) (V - n b) = n R T (P + \frac{a}{V^{2}}) (V - b) = R T for 1 mol gas At high pressure, (P + \frac{a}{V^{2}}) = P ∴ P (V - b) = R T P V = R T + P b$

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Similar questions

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

R T = 2 \sqrt{a . P}

a

mole

2

1

\begin{matrix} Column 1 & Column 2 A & At low pressure & i & P (V - b) = R T B & At high pressure & i i & (P + \frac{a}{V^{2}}) V = R T C & At low pressure and high temperature & i i i & P V = R T D & At room temperature and pressure & i v & (P + \frac{a}{V^{2}}) (V - b) = R T \end{matrix}

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

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