If force of attraction between the molecules is negligible, van der Waals; equation (for one mole) will become:

P V = R T + P b

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

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P V = R T + a / V

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P V = R T - a / V

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Solution

The correct option is A $P V = R T + P b$
If force of attraction between the molecules is negligible, van der Waals; equation (for one mole) will become
$P V = R T + P b$
The van der Waals; equation (for one mole) is $(P + \frac{a}{V^{2}}) (V - b) = R T$
The force of attraction between the molecules is negligible. The term $\frac{a}{V^{2}}$ can be neglected in comparison to P.
$(P) (V - b) = R T$
$P V - P b = R T$
$P V = R T + P b$
Note:
'a' is a proportionality constant called coefficient of attraction. It refers to the pressure correction. Its value depends on the nature of the gas. It corrects for the attractive forces present in a real gas in the van der Waals' equation.
The term $^{'} n b^{'}$ refers to volume correction.

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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 - b) = R T

(a / V^{2})

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

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

P ¯ ¯¯ ¯ V = R T [1 + \frac{B}{¯ ¯¯ ¯ V} + . . . .]

b - \frac{n a}{R T}

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