JEE Main 2025 Jan 29 Shift 2, Physics Q6: Thermodynamic Processes

JEE Main2025Jan 29, Shift 2Physics

A poly-atomic molecule ( $C_{V} = 3 R$ , $C_{P} = 4 R$ , where $R$ is gas constant) goes from phase space point $A (P_{A} = 1 0^{5}$ Pa, $V_{A} = 4 \times 1 0^{- 6} m^{3})$ to point $B$ ( $P_{B} = 5 \times 1 0^{4}$ Pa, $V_{B} = 6 \times 1 0^{- 6} m^{3}$ ) to point $C$ ( $P_{C} = 1 0^{4}$ Pa, $V_{C} = 8 \times 1 0^{- 6} m^{3}$ ). $A$ to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path.

The net heat absorbed per unit mole by the system is :

A
$500 R (ln 3 + ln 4)$
B
$450 R (ln 4 - ln 3)$
C
$500 R ln 2$
D
$400 R ln 4$

Solution

Heat exchange in each leg:

$A \to B$ (adiabatic): $Q_{A B} = 0$ .

$B \to C$ (isothermal at the lower isotherm, $T = 450$ K): for one mole,

$Q_{B C} = W_{B C} = R T ln \frac{V _{C}}{V _{B}} = 450 R ln \frac{8 \times 1 0 ^{- 6}}{6 \times 1 0 ^{- 6}} = 450 R ln \frac{4}{3} = 450 R (ln 4 - ln 3)$ .

Net heat absorbed per mole $= Q_{A B} + Q_{B C} = 450 R (ln 4 - ln 3)$ .

(Note: the numerical data in the question do not strictly satisfy $P V =$ constant on $B \to C$ , but the answer is obtained from the stated isothermal temperature.)

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JEE Main 2025 Jan 29 Shift 2, Physics Q6: Thermodynamic Processes

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