At low pressure,the van der Waals equation be | vedclass

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(C) The van der Waals equation for $1 \ mol$ of a gas is given by: $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$. At low pressure,the molar volu

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$\left( P + \frac{a}{V_m^2} \right) V_m = RT$

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(C) The van der Waals equation for $1 \ mol$ of a gas is given by: $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$. At low pressure,the molar volume $V_m$ is very large,so $V_m \gg b$. Therefore,the term $(V_m - b)$ can be approximated as $V_m$. Substituting this into the equation: $\left( P + \frac{a}{V_m^2} \right) V_m = RT$. Expanding this,we get: $P V_m + \frac{a}{V_m} = RT$.

At low pressure,the van der Waals equation becomes:

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