Question

The compressibility factors for 1 mole of real gases at low pressure, high pressure and that of gases of very low molar masses are $Z_1,Z_2$ and $Z_3$. These are

• A. $Z_1=(1-\frac{a}{RTV}),Z_2=(1+\frac{Pb}{RT}),Z_3=1$
• B. $Z_1=(1-\frac{a}{RTV}),Z_2=1,Z_3=(1+\frac{Pb}{RT})$
• C. $Z_1=(1+\frac{Pb}{RT}),Z_2=(1-\frac{a}{RTV}),Z_3=(1+\frac{Pb}{RT})$
• D. $Z_1=(1-\frac{a}{RTV}),Z_2=(1+\frac{Pb}{RT}),Z_3=(1+\frac{Pb}{RT})$

Answer 1

The correct option is D $Z_1=(1-\frac{a}{RTV}),Z_2=(1+\frac{Pb}{RT}),Z_3=(1+\frac{Pb}{RT})$

Van der Waals equation is:
$(P+\frac{a}{V^2})(V-b)=RT\text{(for 1 mol gas)}$
Case 1: At low pressure, volume in high, hence taking
$(V-b)\approx V$
$(P+\frac{a}{V^2})V=RT$
$PV+\frac{a}{V}=RT$
$\frac{PV}{RT}+\frac{a}{RTV}=1$
$Z_1=\frac{PV}{RT}=(1-\frac{a}{RTV})$
Case II: At high pressure, $\frac{a}{V^2}$ becomes negligible, so
$(P+\frac{a}{V^2})\approx P$
$P(V-b)=RT$
$PV-Pb=RT$
$\frac{PV}{RT}-\frac{Pb}{RT}=1$
$Z_2=\frac{PV}{RT}=1+\frac{Pb}{RT}$
Case III: Gases are of low molar masses thus, forces of attraction are very small i.e. $a$ is negligible
Thus, $(P+\frac{a}{V^2})\approx P$
As in case II, $Z_3=1+\frac{Pb}{RT}$