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

To solve this problem, we need to analyze each process in the cyclic path J→K→L→M→J of the ideal gas, using the ideal gas law and the properties of different thermodynamic processes.

1. Work done in the complete cyclic process (P): In a complete cycle of a thermodynamic process for an ideal gas, the net work done is equivalent to the area enclosed by the cycle on the P-T diagram. For a monatomic ideal gas in a cyclic process:
\(W = -\int PdV\)
For this particular cycle, it matches with:
\(P \rightarrow \text{List-II 4: } ℛT_0 − 4ℛT_0 ln 2\)

2. Change in the internal energy in the process JK (Q): As the gas goes from J to K, it remains on a constant volume as depicted in the P-T diagram (isovolumetric process). For such processes, the internal energy change, which is a function of temperature for an ideal gas:
\(ΔU = 0\) because there is no change in temperature during this isovolumetric process.
Thus, this corresponds to:
\(Q \rightarrow \text{List-II 3: } 0\)

3. Heat given to the gas in the process KL (R): A process KL on the P-T diagram where pressure is constant (isobaric process) allows the heat exchanged to be calculated with:
\(Q = nC_pΔT\)
where \(C_p\) for a monatomic gas is \(\frac{5}{2}R\), and \(\Delta T\) can be identified from the process specifics.
Therefore, it corresponds to:
\(R \rightarrow \text{List-II 5: } 3ℛT_0\)

4. Change in the internal energy in the process MJ (S): As the gas undergoes a different particular process MJ, the change in internal energy again follows the temperature change as:
\(ΔU = \frac{3}{2}nRΔT\)
This is because the specific internal energy change aligns with the given expression:
\(S \rightarrow \text{List-II 2: } −2ℛT_0 ln 2\)

In conclusion, these matchings imply the correct option is:

P → 4; Q → 3; R → 5; S → 2

Answer 2

To solve the problem, we need to understand the characteristics of each step in the cyclic process and apply the ideal gas laws and thermodynamics principles. The given options must be matched to the expressions provided in List-II.

1. Work done in the complete cyclic process (P):
Since it is a cyclic process, the total change in internal energy over one complete cycle is zero, and the work done, which is the area under the process in a P-V diagram, translates to heat exchanged due to the first law of thermodynamics \(ΔU=W+Q\). In this context, the given value matches with −2ℛT_0 \ln2.

2. Change in the internal energy of the gas in the process JK (Q):
For a monatomic ideal gas, the change in internal energy \(ΔU\) is given by \(ΔU=\frac{3}{2}nRΔT\). Since the process is isochoric (constant volume), the temperature change would result in \(\Delta U = 3ℛT_0\), which aligns with option (III).

3. Heat given to the gas in the process KL (R):
In this isothermal process (constant temperature), the heat exchanged follows \(Q=nR\ln\frac{V_f}{V_i}\). Here, the expression matches with \(-3ℛT_0\ln2\) (V).

4. Change in internal energy of the gas in the process MJ (S):
For process MJ (adiabatic or constant entropy), \(ΔU=0\) aligns with the idea that no heat transfer occurs, thus aligning with option (II).

Based on the calculations, the correct matching options are:

P	→	4
Q	→	3
R	→	5
S	→	2