Question
If $$Z$$ is a compressibility factor, van der Waals equation at low pressure can be written as :
A.
$$Z = 1 + \frac{{RT}}{{Pb}}$$
B.
$$Z = 1 - \frac{a}{{VRT}}$$
C.
$$Z = 1 - \frac{{Pb}}{{RT}}$$
D.
$$Z = 1 + \frac{{Pb}}{{RT}}$$
Answer :
$$Z = 1 - \frac{a}{{VRT}}$$
Solution :
$$\eqalign{ & {\text{Compressibility factor}}\,\left( Z \right) = \frac{{PV}}{{RT}} \cr & {\text{(For one mole of real gas)}} \cr & {\text{van der Waals equation}} \cr & \left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT \cr} $$
At low pressure, volume is very large and hence correction term $$b$$ can be neglected in comparison to very large volume of $$V.$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,V - b \approx V \cr & \left( {P + \frac{a}{{{V^2}}}} \right)V = RT\,;\,PV + \frac{a}{V} = RT \cr & PV = RT - \frac{a}{V};\,\frac{{PV}}{{RT}} = 1 - \frac{a}{{VRT}} \cr} $$
$${\text{Hence}},$$
$$\eqalign{ & {\text{Compressibility factor}}\,\left( Z \right) = \frac{{PV}}{{RT}} \cr & {\text{(For one mole of real gas)}} \cr & {\text{van der Waals equation}} \cr & \left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT \cr} $$
At low pressure, volume is very large and hence correction term $$b$$ can be neglected in comparison to very large volume of $$V.$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,V - b \approx V \cr & \left( {P + \frac{a}{{{V^2}}}} \right)V = RT\,;\,PV + \frac{a}{V} = RT \cr & PV = RT - \frac{a}{V};\,\frac{{PV}}{{RT}} = 1 - \frac{a}{{VRT}} \cr} $$
$${\text{Hence}},$$