JEE Challenger
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ChemMedAdv 2026 • Paper 1

Minimum Work Done in Two Step Isothermal Compression of Ideal Gas

An ideal gas (0.5 mol0.5\text{ mol}), initially at 2 bar2\text{ bar} pressure, is compressed at a constant temperature of 600 K600\text{ K} in two steps: first, against a constant external pressure of P barP\text{ bar} (2<P<82 < P < 8), and then against constant external pressure of 8 bar8\text{ 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 WW. Considering all possible values of PP (2<P<82 < P < 8) and taking the gas constant as RR (in J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}), the minimum value of ∣W∣|W| (in J\text{J}) is

Options

A

207R207R

B

600R600R

Correct
C

630R630R

D

900R900R

Topics & Concepts

ThermodynamicsFirst Law of Thermodynamics

Step-by-Step Solution

To find the minimum value of the total work done on the gas, ∣W∣|W|, during the two-step irreversible isothermal compression, we analyze the process step-by-step using the thermodynamics of an ideal gas.

1. Initial, Intermediate, and Final States

For n=0.5 moln = 0.5\text{ mol} of an ideal gas at constant temperature T=600 KT = 600\text{ K}:

  • Initial State: Pressure P0=2 barP_0 = 2\text{ bar}, Volume V0=nRTP0=nRT2V_0 = \frac{nRT}{P_0} = \frac{nRT}{2}
  • Intermediate State: Pressure P1=P barP_1 = P\text{ bar}, Volume V1=nRTP1=nRTPV_1 = \frac{nRT}{P_1} = \frac{nRT}{P}
  • Final State: Pressure P2=8 barP_2 = 8\text{ bar}, Volume V2=nRTP2=nRT8V_2 = \frac{nRT}{P_2} = \frac{nRT}{8}

2. Work Done in Each Step

The work done on the gas during an irreversible process against a constant external pressure PextP_{\text{ext}} is given by: W=−PextΔVW = - P_{\text{ext}} \Delta V

Step 1: Compression against external pressure Pext,1=P barP_{\text{ext}, 1} = P\text{ bar}: W1=−P(V1−V0)=−P(nRTP−nRT2)=nRT(P2−1)W_1 = - P (V_1 - V_0) = - P \left( \frac{nRT}{P} - \frac{nRT}{2} \right) = nRT \left( \frac{P}{2} - 1 \right)

Step 2: Compression against external pressure Pext,2=8 barP_{\text{ext}, 2} = 8\text{ bar}: W2=−8(V2−V1)=−8(nRT8−nRTP)=nRT(8P−1)W_2 = - 8 (V_2 - V_1) = - 8 \left( \frac{nRT}{8} - \frac{nRT}{P} \right) = nRT \left( \frac{8}{P} - 1 \right)

3. Total Work Done

The total work done WW on the gas is: W=W1+W2=nRT(P2−1+8P−1)=nRT(P2+8P−2)W = W_1 + W_2 = nRT \left( \frac{P}{2} - 1 + \frac{8}{P} - 1 \right) = nRT \left( \frac{P}{2} + \frac{8}{P} - 2 \right)

Since P>2 barP > 2\text{ bar}, the gas is compressed in both steps, so W>0W > 0 and ∣W∣=W|W| = W.

4. Minimizing the Work Done

To minimize ∣W∣|W|, we need to minimize the expression f(P)=P2+8P−2f(P) = \frac{P}{2} + \frac{8}{P} - 2 for 2<P<82 < P < 8.

By the AM-GM (Arithmetic Mean - Geometric Mean) inequality: P2+8P2≥P2⋅8P\frac{\frac{P}{2} + \frac{8}{P}}{2} \ge \sqrt{\frac{P}{2} \cdot \frac{8}{P}}

P2+8P≥24=4\frac{P}{2} + \frac{8}{P} \ge 2 \sqrt{4} = 4

The equality (minimum value) holds when: P2=8P  ⟹  P2=16  ⟹  P=4 bar\frac{P}{2} = \frac{8}{P} \implies P^2 = 16 \implies P = 4\text{ bar}

Since P=4 barP = 4\text{ bar} lies within the allowed range (2<P<8)(2 < P < 8), the minimum value of f(P)f(P) is: f(P)min=4−2=2f(P)_{\text{min}} = 4 - 2 = 2

5. Calculation of Minimum ∣W∣|W|

Substituting f(P)min=2f(P)_{\text{min}} = 2, n=0.5 moln = 0.5\text{ mol}, and T=600 KT = 600\text{ K}: ∣W∣min=nRT×2=(0.5)×R×600×2=600R|W|_{\text{min}} = nRT \times 2 = (0.5) \times R \times 600 \times 2 = 600R

Thus, the correct option is (B).

Minimum Work Done in Two Step Isothermal Compression of Ideal Gas | Chemistry PYQ Solution - JEE Challenger