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One mole of a van der Waals gas obeying the equation \[\left( p+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\]Undergoes the quasi-static cyclic process which is shown in the p-V diagram. The net heat absorbed by the gas in this process is

A)  \[\frac{1}{2}({{\rho }_{1}}-{{\rho }_{2}})({{V}_{1}}-{{V}_{2}})\]

B)  \[\frac{1}{2}({{\rho }_{1}}+{{\rho }_{2}})({{V}_{1}}-{{V}_{2}})\]

C)  \[\frac{1}{2}\left( {{p}_{1}}+\frac{a}{V_{1}^{2}}-{{p}_{2}}-\frac{a}{V_{2}^{2}} \right)({{V}_{1}}-{{V}_{2}})\]

D)  \[\frac{1}{2}\left( {{p}_{1}}+\frac{a}{V_{1}^{2}}+{{p}_{2}}+\frac{a}{V_{2}^{2}} \right)({{V}_{1}}-{{V}_{2}})\]

Correct Answer: A

Solution :

 For the cyclic process Heat absorbed = Work done \[=Area=\frac{1}{2}(\Delta p)\times \Delta V\] \[=\frac{1}{2}({{p}_{1}}-{{p}_{2}})\times ({{V}_{1}}-{{V}_{2}})\] \[=\frac{1}{2}({{p}_{1}}-{{p}_{2}})({{V}_{1}}-{{V}_{2}})\]