A poly-atomic molecule ( $\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}, \mathrm{C}_{\mathrm{P}}=4 \mathrm{R}$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^{5} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^{3}\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^{4} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^{3}\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^{4} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^{3}\right)$. 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 :

(B) $450 \mathrm{R}(\ln 4-\ln 3)$

(A) $500 \mathrm{R}(\ln 3+\ln 4)$

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Heat & Thermodynamics

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 :

Answer: (B)

The path A→B is adiabatic, so no heat is exchanged along it:

Δ Q_{A B} = 0

The path B→C is isothermal, lying along the 450 K isotherm marked on the given

P

V

diagram, so all the heat absorbed goes into work done by the gas:

Δ Q_{B C} = Δ W_{B C} = n R T ln (\frac{V _{C}}{V _{B}})

With

n = 1

mole and

T = 450

Δ Q_{B C} = 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)

The total heat absorbed per mole over the full path is

Δ Q = Δ Q_{A B} + Δ Q_{B C} = 450 R (ln 4 - ln 3)

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option