Question

In a cyclic process, a gas is taken from state $A$ to $B$ via path$-I$ as shown in the indicator diagram and taken back to state $A$ from state $B$ via path$-II$. In the complete cycle:

• A. Work is done by the gas
• B. Heat is ejected by the gas
• C. No work is done by the gas
• D. Nothing can be said about work as data is insufficient

Answer 1

Answer: (C)

No work is done by the gas

Thermodynamics first law for a closed system,

$\delta Q=\delta U+\delta W$

For a cyclic path ABA, $\delta U=0$, because internal energy is a state function, so it becomes zero for any closed cycle.

$\delta Q=\delta W$............................(1)

Also, the work done through path 2 is taken as negative because cycle goes anticlockwise.

Similarly, the work done by the path 1 is positive(clockwise).

Here, $W2>W1$

so, from eqn. (1),

$\delta Q=-ve$

that means no work is done by the gas, work is done on the gas.