For 1 mol of gas,the plot of pV vs p is show | vedclass

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(A) For $1$ mole of a real gas,the van der Waals equation is $(P + \frac{a}{V_m^2})(V_m - b) = RT$.At low pressure,the attractive forces are dominant,

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

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(A) For $1$ mole of a real gas,the van der Waals equation is $(P + \frac{a}{V_m^2})(V_m - b) = RT$.At low pressure,the attractive forces are dominant,so $Z The compressibility factor $Z$ is given by $Z = \frac{PV_m}{RT}$.For low pressure,the approximate form of the van der Waals equation is $Z = 1 - \frac{a}{V_mRT}$.At point $A$,which lies in the region where $Z < 1$ (below the ideal gas line),the compressibility factor is $1 - \frac{a}{V_mRT}$ (or $1 - \frac{a}{RTV}$).

For $1 \ mol$ of gas,the plot of $pV$ vs $p$ is shown below. $p$ is the pressure and $V$ is the volume of the gas. What is the value of compressibility factor at point $A$?

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