At low pressure,the Van der Waals equation is | vedclass

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(A) The Van der Waals equation for $1 \ mol$ of gas is: $(P + \frac{a}{V_m^2})(V_m - b) = RT$. At low pressure,$V_m$ is very large,so $b$ is negligibl

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$Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$

Vedclass · Answer

(A) The Van der Waals equation for $1 \ mol$ of gas is: $(P + \frac{a}{V_m^2})(V_m - b) = RT$. At low pressure,$V_m$ is very large,so $b$ is negligible compared to $V_m$ $(V_m - b \approx V_m)$. The equation becomes: $(P + \frac{a}{V_m^2})V_m = RT$. Expanding this: $PV_m + \frac{a}{V_m} = RT$. Dividing by $RT$: $\frac{PV_m}{RT} + \frac{a}{V_m RT} = 1$. Therefore,the compressibility factor $Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$.

At low pressure,the Van der Waals equation is reduced to:

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