Question

One mole of a vander Waals’ gas obeying the equation
$\left(P+\frac{a}{V^{2}}\right)\left(V-b\right)=RT$
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. $\frac{1}{2}\left(P_{1}-P_{2}\right)\left(V_{1}-V_{2}\right)$
• B. $\frac{1}{2}\left(P_{1}+P_{2}\right)\left(V_{1}-V_{2}\right)$
• C. $\frac{1}{2}\left(P_{1}+\frac{a}{v^{2}_{1}}-P_{2}-\frac{a}{v^{2}_{2}}\right)\left(V_{1}-V_{2}\right)$
• D. $\frac{1}{2}\left(P_{1}+\frac{a}{v^{2}_{1}}+P_{2}+\frac{a}{v^{2}_{2}}\right)\left(V_{1}-V_{2}\right)$

Answer 1

Answer: (A)

For the cyclic process

Heat absorbed $=$ Work done
$=$ Area $=\frac{1}{2}(\Delta p) \times \Delta V$
$=\frac{1}{2}\left(p_{1}-p_{2}\right) \times\left(V_{1}-V_{2}\right)$
$=\frac{1}{2}\left(p_{1}-p_{2}\right)\left(V_{1}-V_{2}\right)$