The compressibility factor (Z) of one mole of | vedclass

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Question

(C) The van der Waals' equation for $1$ mole of gas is $(p+\frac{a}{V^2})(V-b) = RT$. Since '$a$' is negligible,the equation simplifies to $p(V-b) = R

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$1+\frac{b p}{R T}$

Vedclass · Answer

(C) The van der Waals' equation for $1$ mole of gas is $(p+\frac{a}{V^2})(V-b) = RT$. Since '$a$' is negligible,the equation simplifies to $p(V-b) = RT$. Expanding this,we get $pV - pb = RT$. Dividing both sides by $RT$,we get $\frac{pV}{RT} - \frac{pb}{RT} = 1$. Since the compressibility factor $Z = \frac{pV}{RT}$,the equation becomes $Z - \frac{pb}{RT} = 1$. Therefore,$Z = 1 + \frac{pb}{RT}$.

The compressibility factor $(Z)$ of one mole of a van der Waals' gas of negligible '$a$' value is

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