At low pressure and high temperature, the vander Waal's equation is finally reduced (simplified) to:

(p + \frac{a}{v_{m}^{2}}) (V_{m} - b) = R T

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ρ (V_{m} - b) = R T

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(p + \frac{a}{v_{m}^{2}}) V_{m} = R T

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p V_{m} = R T

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Solution

The correct option is B $p V_{m} = R T$
At low pressure and high temperature, van der Waal's equation is modified as:

$p V_{m} = R T$ (for $1$ mole)

$∵$ At low pressure and high temperature, the terms $\frac{a}{V^{2}}$ and $b$ becomes very small as compared to $p$ and $V$ .

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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

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

\frac{a}{v^{2}}]

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

a

b

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