JEE Advanced 2026 — Paper 1 — CHEM-I solution

JEE Advanced2026Paper 1CHEM-I

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
$600 R$
C
$630 R$
D
$900 R$

Solution

Set up the two-step compression. With $n = 0.5$ , $T = 600$ K, initial state $(2 bar, V_{0})$ . Step 1: external pressure $P$ until gas reaches $P$ bar, giving volume $V_{1} = n R T / P$ . Step 2: external pressure $8$ bar until gas reaches $8$ bar, giving final volume $V_{2} = n R T /8$ .

Work done on the gas (positive sign convention for compression): $W = - P_{ext} Δ V$ for each step.

Step 1: $W_{1} = - P (V_{1} - V_{0}) = - P (\frac{n R T}{P} - \frac{n R T}{2}) = - n R T (1 - \frac{P}{2})$ .

Step 2: $W_{2} = - 8 (V_{2} - V_{1}) = - 8 (\frac{n R T}{8} - \frac{n R T}{P}) = - n R T (1 - \frac{8}{P})$ .

Total: $W = - n R T (2 - \frac{P}{2} - \frac{8}{P}) = - 300 R (2 - \frac{P}{2} - \frac{8}{P})$ .

Minimise $∣ W ∣$ over $P \in (2, 8)$ . Differentiate with respect to $P$ : $\frac{d W}{d P} = - 300 R (- \frac{1}{2} + \frac{8}{P ^{2}}) = 0 \Rightarrow P^{2} = 16 \Rightarrow P = 4$ bar.

Substitute $P = 4$ : $W = - 300 R (2 - 2 - 2) = 600 R$ , so $∣ W ∣_{m i n} = 600 R$ . Option (B).

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