Van der Waal's equation at high pressure for $1 m o l e$ is:

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

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P V = R T

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P (V + b) = R T

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P (V - b) = R T

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Solution

The correct option is A $P V + \frac{a}{V} = R T$
For $1$ mole of a gas, ideal gas equation is $P V = R T ⟶ (1)$
On applying volume and pressure correction equation can be written as , $(P + \frac{a}{V^{2}}) (V - b) = R T ⟶ (2)$
Where $a$ is constant depending on the nature of the gas.
$b$ is the excluded volume.
At high pressure volume of the gas is very small hence the pressure correction $\frac{a}{V^{e}}$ can be neglected Thus equation $(2)$ can be written as $P (V - b) = R T$

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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 (V - b) = R T

b

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}

\frac{R T}{P V}

\frac{R T}{P V}

P V

NTP

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