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WBJEE · Physics · Thermodynamics

One mole of a van der Waals' gas obeying the equation \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) undergoes the quasi-static cyclic process which is shown in the \(p \cdot V\) diagram. The net heat absorbed by the gas in this process is

  1. A \(\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)\)
  2. B \(\frac{1}{2}\left(p_{1}+p_{2}\right)\left(V_{1}-V_{2}\right)\)
  3. C \(\frac{1}{2}\left(A_{1}+\frac{a}{V_{1}^{2}}-p_{2}-\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)\)
  4. D \(\frac{1}{2}\left(\rho_{1}+\frac{a}{V_{1}^{2}}+p_{2}+\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)\)

Step-by-step Solution

Detailed explanation

For the cyclic process Heat absorbed = Work done \[ \begin{array}{l} =\text { Area }=\frac{1}{2}(\Delta p) \times \Delta V \\ =\frac{1}{2}\left(p_{1}-p_{2}\right) \times\left(V_{1}-V_{2}\right) \\ =\frac{1}{2}\left(p_{1}-p_{2}\right)\left(V_{1}-V_{2}\right) \end{array} \]