Question

Two moles of helium gas are taken over the cycle $ABCDA$, as shown (below) in the $P-T$ diagram. (Assume the gas to be ideal and $R$ is gas constant).
Now, match List I with List II and select the options given below:

	List - I		List - II
(P)	Magnitude of work done on the gas in taking from $A$ to $B$	(1)	693 R
(Q)	Magnitude of work done on the gas in taking from $B\to C$	(2)	277 R
(R)	Magnitude of work done on the gas in taking from $D\to A$	(3)	400 R
(S)	Magnitude of the net heat absorbed/evolved in the cycle $ABCDA$	(4)	416 R

• A. P-1, Q-2, R-3, S-4
• B. P-3, Q-2, R-4, S-3
• C. P-3, Q-1, R-4, S-2
• D. P-4, Q-3, R-1, S-2

Answer 1

The correct option is C P-3, Q-1, R-4, S-2

$W_{A\to B}=-nRT(T_2-T_1)=-2\times R\times (500-300)=-400R$ (Isobaric process)

$W_{B\to C}=-2.30nRTlog\frac{P_1}{P_2}=-2.303\times 2\times R\times 500log2=-693R$ (Isothermal process)

$W_{C\to D}=-nRT(T_1-T_2)=+400R$ (Isobaric process)

$W_{D\to A}=-2.303nRTlog\frac{P_1}{P_2}$ (Isothermal process) $=-2.303\times 2\times R\times 300log\frac{1}{2}=+416R$

Note: The magnitude of the work done will be considered positive.
Hence, option $\left(C\right)$ is correct.