JEE Advanced 2026 · Paper 2 · Physics Q9

← JEE Advanced 2026 · Paper 2

PhysicsMultiple correct+4 / -1Question 27 of 54

Ten moles of an ideal monoatomic gas, initially in state $\boldsymbol a$ at atmospheric pressure and temperature $T_a=27^\circ$ C, is enclosed in a metal cylinder of volume $V_0$ fitted with a frictionless piston. The gas is suddenly compressed to state $\boldsymbol b$ with volume $V_0/3$ . Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature $11^\circ$ C until the gas reaches the temperature of the water bath, which is denoted as state $\boldsymbol c$ . Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state $\boldsymbol f$ . If $R$ is universal gas constant, then the correct option(s) is/are:
[Given: $9^{1/3}=2.08$ ]

The schematic P-V diagram of the processes described above is:

P-V diagram: a at large volume and low pressure; a steep curve rises to b at one third of the volume; a vertical drop at that volume leads down to c; a flatter curve from c returns to f, just below a at the starting volume. Dashed verticals mark the two volumes.

The change in internal energy in going from state

\boldsymbol a

\boldsymbol b

4860R

The net change in the internal energy in the whole process is

-240R

The pressure and temperature of the state

\boldsymbol b

are 2.08 times the atmospheric pressure and 624 K, respectively.

Answer and worked solution

Official answer: A, B, C

$n=10$ , $C_V=\frac32R$ , $\gamma=\frac53$ , $T_a=300$ K.

$a\to b$ (sudden, so adiabatic), $V_0\to V_0/3$ :
$T_b=300\cdot3^{2/3}=300\cdot9^{1/3}=624\text{ K},\qquad P_b=P_a\cdot3^{5/3}=3\times2.08\,P_a=6.24\,P_a.$
(D) false (the pressure is 6.24, not 2.08, times atmospheric).

$\Delta U_{ab}=10\cdot\frac32R\cdot(624-300)=4860R$ . (B) true.

$b\to c$ : isochoric cooling to 284 K. $c\to f$ : slow isothermal expansion at 284 K back to $V_0$ .

Net. $\Delta U=10\cdot\frac32R\cdot(284-300)=-240R$ . (C) true.

(A) The diagram shows: a steep adiabat rising from $a$ to $b$ ; a vertical drop $b\to c$ ; and a flatter isotherm from $c$ out to $f$ at $V_0$ , just below $a$ (since 284 K < 300 K). True.

Topics:Thermodynamic processes First law and internal energy

← Previous question Next question →

JEE Advanced 2026 · Paper 2, Physics question 9