JEE Advanced Physics Question (2026) — Solution

Step-by-step solution

For state a: n = 10, T_a = 27^ C = 300 K , P_a = 1 atm , V_a = V_0. For a monoatomic gas, C_v = 3 2 R and = 5 3 . Process a b is a sudden compression, which is an adiabatic process. V_b = V_0 3 Using T_a V_a^ -1 = T_b V_b^ -1 : T_b = T_a ( V_a V_b )^ -1 = 300 (3)^ 5/3 - 1 = 300 3^ 2/3 = 300 (9)^ 1/3 Given 9^ 1/3 = 2.08, we get T_b = 300 2.08 = 624 K . Using P_a V_a^ = P_b V_b^ : P_b = P_a ( V_a V_b )^ = 1 (3)^ 5/3 = 3 3^ 2/3 = 3 2.08 = 6.24 atm . Thus, the pressure at state b is 6.24 times the atmospheric pressure. Option (D) is incorrect. The change in internal energy for process a b is: U_ ab = n C_v (T_b - T_a) = 10 3 2 R (624 - 300) = 15R 324 = 4860R. Option (B) is correct. Process b c is isochoric (stationary piston), so V_c = V_0 3 . The gas cools to the water bath temperature, T_c = 11^ C = 284 K . Process c f is isothermal (slow expansion in water bath), so T_f = T_c = 284 K and V_f = V_0. The net change in internal energy for the entire process a f depends only on the initial and final temperatures: U_ net = n C_v (T_f - T_a) = 10 3 2 R (284 - 300) = 15R (-16) = -240R. Option (C) is correct. For the P-V diagram: Process a b is an adiabatic compression (curve upwards and to the left). Process b c is an isochoric cooling (vertical line downwards). Process c f is an isothermal expansion (curve downwards and to the right). Comparing pressures at volume V_0: P_a = nR(300) V_0 and P_f = nR(284) V_0 Since P_a > P_f, point a lies above point f. The given schematic diagram correctly represents all these features. Option (A) is correct. Answer: The schematic P-V diagram of the processes described above is: ; The change in internal energy in going from state a to b is 4860R.; The net change in the internal energy in the whole process is -240R.