JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Chemical Thermodynamics

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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

(A)207 R
(B)600 R✓ Correct
(C)630 R
(D)900 R

Answer: (B)

Solution

For an isothermal compression in two steps, the total work is:

W = W_{1} + W_{2} = - n R T [1 - \frac{P}{2}] - n R T [1 - \frac{8}{P}]

With

n = 0.5

mol and

T = 600

W = - 300 R [2 - \frac{P}{2} - \frac{8}{P}]

To find the minimum value of

∣ W ∣

, differentiate with respect to P and set equal to zero:

\frac{d W}{d P} = - 300 R [- \frac{1}{2} + \frac{8}{P ^{2}}] = 0

Solving:

\frac{1}{2} = \frac{8}{P ^{2}}

gives

P = 4

bar.

Substituting

P = 4

back:

W = - 300 R [2 - 2 - 2] = 600 R

Therefore,

∣ W ∣_{m i n} = 600 R

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Chemical Thermodynamics

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