Question

One mole of a monatomic ideal gas undergoes the cyclic process J→ K→ L→ M→ 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.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [ℛ is the gas constant.]

List-I		List-II
P	Work done in the complete cyclic process	I	\(ℛT_0 − 4ℛT_0 ln 2\)
Q	Change in the internal energy of the gas in the process JK	II	\(0\)
R	Heat given to the gas in the process KL	III	\(3ℛT_0\)
S	Change in the internal energy of the gas in the process MJ	IV	\(−2ℛT_0 ln 2\)
			\[−3ℛT_0 ln 2\]

 

• A. ) P → 1; Q → 3; R → 5; S → 4
• B. P → 4; Q → 3; R → 5; S → 2
• C. P → 4; Q → 1; R → 2; S → 2
• D. P → 2; Q → 5; R → 3; S → 4

Answer 1

The Correct Option is B

Step 1: Calculate \( \Delta U_{JK} \) 

The change in internal energy is given by:

\[ \Delta U = n C_V \Delta T. \]

For a monatomic gas, \( C_V = \frac{3R}{2} \), and given \( \Delta T = 3T_0 - T_0 = 2T_0 \):

\[ \Delta U_{JK} = 1 \cdot \frac{3R}{2} \cdot 2T_0 = 3RT_0. \]

Step 2: Calculate \( Q_{KL} \)

Heat transfer in an isothermal process is given by:

\[ Q = nRT \ln \frac{V_2}{V_1} = nRT \ln \frac{P_1}{P_2}. \]

Here, \( T = T_0 \), \( P_1 = 2P_0 \), and \( P_2 = P_0 \):

\[ Q_{KL} = 1 \cdot R \cdot T_0 \ln \frac{2P_0}{P_0} = 3RT_0 \ln 2. \]

Step 3: Work Done in the Cyclic Process

The work done is given by:

\[ W = RT_0 - 4RT_0 \ln 2. \]

Step 4: Calculate \( \Delta U_{MJ} \)

Since the process is **isochoric**, \( \Delta U = 0 \).

Conclusion

Based on the calculations:

• P → 4
• Q → 3
• R → 5
• S → 2

Final Answer:

The correct option is (B).