Question

In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process A $\rightarrow$ B and C $\rightarrow$ D are $T_1$ and $T_2$ ($T_1>T_2$) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic. 

 

• A. $W_{BC} + W_{DA}>0$
• B. $W_{AB}<W_{CD}$
• C. $W_{AB} = W_{DC}$
• D. $W_{AD} = W_{BC}$

Hint:

In cyclic P–V diagrams, compare areas instead of formulas. Larger area at higher pressure always dominates work contribution.

Answer 1

The Correct Option is A

In a P–V diagram, the work done by a gas is equal to the area under the curve. Work is:
positive for expansion,
negative for compression.
The processes $BC$ and $DA$ are adiabatic. During an adiabatic process, the pressure falls (or rises) more steeply than in an isothermal process. Process $BC$: It is an adiabatic expansion occurring at higher pressures and over a larger volume range. Hence, the magnitude of work done $|W_{BC}|$ is large and $W_{BC}>0$. Process $DA$: It is an adiabatic compression occurring at lower pressures and over a smaller volume range. Hence, the magnitude of work done $|W_{DA}|$ is smaller and $W_{DA}<0$. Since the area under $BC$ is greater than the area under $DA$, \[ |W_{BC}|>|W_{DA}| \] Therefore, \[ W_{BC} + W_{DA}>0 \] \[ \boxed{W_{BC} + W_{DA}>0} \] Hence, option (A) is correct.