The compressibility factor for a van der Waal | vedclass

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Question

(B) For $1 \ mol$ of a van der Waal gas, the equation is $(P+\frac{a}{V^2})(V-b)=RT$. At high pressure, the term $\frac{a}{V^2}$ is negligible compare

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$1+\frac{Pb}{RT}$

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(B) For $1 \ mol$ of a van der Waal gas, the equation is $(P+\frac{a}{V^2})(V-b)=RT$. At high pressure, the term $\frac{a}{V^2}$ is negligible compared to $P$, so $P+\frac{a}{V^2} \approx P$. Substituting this into the equation, we get $P(V-b)=RT$. Expanding this gives $PV-Pb=RT$, which rearranges to $PV=RT+Pb$. The compressibility factor $z$ is defined as $z=\frac{PV}{RT}$. Dividing the equation $PV=RT+Pb$ by $RT$, we get $z=\frac{RT}{RT}+\frac{Pb}{RT}$. Therefore, $z=1+\frac{Pb}{RT}$.

The compressibility factor for a van der Waal gas at high pressure is

Similar Questions

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