JEE Advanced 2026 — Paper 1 — PHY-II solution

JEE Advanced2026Paper 1PHY-II

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process ( $ab$ ), followed by an isochoric process ( $bc$ ) and an adiabatic process ( $ca$ ) as shown in the figure. The volumes of the gas are $V_{1}$ and $V_{2}$ at $a$ and $b$ , respectively. If the cycle has heat input $Q_{in}$ and output $Q_{out}$ , then the efficiency of the cycle is defined as $η = \frac{Q _{in} - Q _{out}}{Q _{in}}$ . The correct statement(s) is/are:

[Given : $ln 2 \approx 0.7$ ]

A
If $V_{2} / V_{1} = 8$ , the heat released in the process $bc$ is smaller than the heat absorbed in the process $ab$ .
B
For a given value of $V_{2} / V_{1}$ , $η$ does not depend on the temperature of the isothermal process.
C
If $V_{2} / V_{1} = 8$ , then the temperature of the gas at $a$ is 4 times the temperature of the gas at $c$ .
D
If $V_{2} / V_{1} = 8$ , then the pressure of the gas at $a$ is 4 times the pressure of the gas at $b$ .

Solution

Setup. Monoatomic gas: $C_{v} = \frac{3 R}{2}$ , $γ = \frac{5}{3}$ . Process $ab$ is isothermal at $T_{a}$ ; process $bc$ is isochoric (cooling from $T_{a}$ to $T_{c}$ ); process $ca$ is adiabatic ( $T_{c} V_{2}^{γ - 1} = T_{a} V_{1}^{γ - 1}$ ).

Adiabatic relation. $T_{a} V_{1}^{2/3} = T_{c} V_{2}^{2/3} \Rightarrow \frac{T _{a}}{T _{c}} = (V_{2} / V_{1})^{2/3}$ . For $V_{2} / V_{1} = 8$ : $T_{a} / T_{c} = 8^{2/3} = 4$ . (C) correct.

Energy balance per cycle. The cycle is clockwise on the $P$ – $V$ diagram so net work is positive, hence $Q_{in} - Q_{out} > 0$ . $Q_{in} = Q_{ab} = n R T_{a} ln (V_{2} / V_{1})$ . $Q_{out} = ∣ Q_{b c} ∣ = n C_{v} (T_{a} - T_{c})$ . The cyclic condition forces $Q_{out} < Q_{in}$ . (A) correct.

Efficiency. $η = 1 - \frac{n C _{v} ( T _{a} - T _{c} )}{n R T _{a} ln ( V _{2} / V _{1} )} = 1 - \frac{3}{2} \cdot \frac{1 - ( V _{1} / V _{2} ) ^{2/3}}{ln ( V _{2} / V _{1} )}$ . The expression depends only on $V_{2} / V_{1}$ , not on $T_{a}$ . (B) correct.

Option (D). Along the isotherm $ab$ : $P_{a} V_{1} = P_{b} V_{2}$ , so $P_{a} / P_{b} = V_{2} / V_{1} = 8$ , not 4. Incorrect.

Correct answers: (A), (B), (C).

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