JEE Advanced · Physics · 13. Thermodynamics
One mole of a monatomic ideal gas undergoes the cyclic process \(\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}\), as shown in the P-T diagram.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
[\(\mathcal{R}\) is the gas constant.]
| List-I | List-II |
| (P) Work done in the complete cyclic process | (1) \(R T_0-4 R T_0 \ln 2\) |
| (Q) Change in the internal energy of the gas in the process JK | (2) \(0\) |
| (R) Heat given to the gas in the process KL | (3) \(3 K T_0\) |
| (S) Change in the internal energy of the gas in the process MJ | (4) \(-2 R T_0 \ln 2\) |
| \((5)-3 R T_0 \ln 2\) |
- A \(\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4\)
- B \(\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2\)
- C \(\mathrm{P} \rightarrow 4\); \(\mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 2\)
- D \(\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4\)
Answer & Solution
Correct Answer
(B) \(\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right) \)
\( \mathrm{K}\left(\mathrm{P}_0, 3 \mathrm{~V}_0, 3 \mathrm{~T}_0\right) \)
\( \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \)
\( \mathrm{L}\left(2 \mathrm{P}_0, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \)
\( \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \)
\( \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \)
\( \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \)
\( \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ln \left(\frac{1}{2}\right)\) \(=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \)
\( \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}{ }_0 \ln 2 \)
\( \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT} \mathrm{e}_0 \)
\( \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT} \mathrm{n}_0 \ln 2 ; \Delta \mathrm{U}=0 \)
\( \mathrm{WD} \text { net }=-2 \mathrm{nRT} \mathrm{e}_0 \ln 2 \)
\( \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2\)
\( \mathrm{K}\left(\mathrm{P}_0, 3 \mathrm{~V}_0, 3 \mathrm{~T}_0\right) \)
\( \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \)
\( \mathrm{L}\left(2 \mathrm{P}_0, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \)
\( \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \)
\( \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \)
\( \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \)
\( \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ln \left(\frac{1}{2}\right)\) \(=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \)
\( \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}{ }_0 \ln 2 \)
\( \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT} \mathrm{e}_0 \)
\( \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT} \mathrm{n}_0 \ln 2 ; \Delta \mathrm{U}=0 \)
\( \mathrm{WD} \text { net }=-2 \mathrm{nRT} \mathrm{e}_0 \ln 2 \)
\( \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2\)
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