An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of P bar (2

Question

An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of P bar (2 < P < 8), and then against constant external pressure of 8 bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is W. Considering all possible values of P (2 < P < 8) and taking the gas constant as R (in J K^ -1 mol^ -1 ), the minimum value of |W| (in J) is

Quantrex Academy · Accepted Answer

Correct answer: B. 600R

Let the initial, intermediate, and final volumes of the gas be V_1, V_2, and V_3 respectively. Using the ideal gas equation V = nRT P , we have: V_1 = nRT 2 V_2 = nRT P V_3 = nRT 8 The work done on the gas during the first step against a constant external pressure P is: W_1 = -P_ ext,1 (V_2 - V_1) = -P ( nRT P - nRT 2 ) = nRT ( P 2 - 1 ) The work done on the gas during the second step against a constant external pressure of 8 bar is: W_2 = -P_ ext,2 (V_3 - V_2) = -8 ( nRT 8 - nRT P ) = nRT ( 8 P - 1 ) The total work done on the gas is: W = W_1 + W_2 = nRT ( P 2 - 1 ) + nRT ( 8 P - 1 ) = nRT ( P 2 + 8 P - 2 ) To find the minimum value of W, we need to minimize the expression ( P 2 + 8 P ). Using the AM-GM inequality: P 2 + 8 P 2 P 2 8 P = 2 4 = 4 The minimum value is 4, which occurs when P 2 = 8 P P^2 = 16 P = 4 bar. This value of P lies within the given range (2 Substituting this minimum value back into the work equation: W_ min = nRT (4 - 2) = 2nRT Given n = 0.5 mol and T = 600 K: W_ min = 2 0.5 R 600 = 600R Answer: 600R

JEE Advanced Chemistry Question (2026) — Solution

Question

Options

Answer

Step-by-step solution