At low pressures,van der Waals' equation is w | vedclass

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Question

(A) The van der Waals' equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$. At low pressures,the volume $V$ is large,so $b$ is negligibl

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$(1 - \frac{a}{RTV})$

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(A) The van der Waals' equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$. At low pressures,the volume $V$ is large,so $b$ is negligible compared to $V$. The equation simplifies to $(P + \frac{a}{V^2})V = RT$. Expanding this,we get $PV + \frac{a}{V} = RT$. Dividing both sides by $RT$,we get $\frac{PV}{RT} + \frac{a}{VRT} = 1$. Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z + \frac{a}{VRT} = 1$. Therefore,$Z = 1 - \frac{a}{VRT}$.

At low pressures,van der Waals' equation is written as $(P + \frac{a}{V^2})V = RT$. The compressibility factor $Z$ will be

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