Q1 / 48 · JEE Advanced 2026 (Paper 1) · Chemistry
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$207R$
- Correct answerB$600R$
- C$630R$
- D$900R$
Solution
Let the initial, intermediate, and final volumes of the gas be $V_1$, $V_2$, and $V_3$ respectively.
Using the ideal gas equation $V = \dfrac{nRT}{P}$, we have:
$V_1 = \dfrac{nRT}{2}$
$V_2 = \dfrac{nRT}{P}$
$V_3 = \dfrac{nRT}{8}$
The work done on the gas during the first step against a constant external pressure $P$ is:
$W_1 = -P_{ext,1}(V_2 - V_1) = -P \left( \dfrac{nRT}{P} - \dfrac{nRT}{2} \right) = nRT \left( \dfrac{P}{2} - 1 \right)$
The work done on the gas during the second step against a constant external pressure of $8$ bar is:
$W_2 = -P_{ext,2}(V_3 - V_2) = -8 \left( \dfrac{nRT}{8} - \dfrac{nRT}{P} \right) = nRT \left( \dfrac{8}{P} - 1 \right)$
The total work done on the gas is:
$W = W_1 + W_2 = nRT \left( \dfrac{P}{2} - 1 \right) + nRT \left( \dfrac{8}{P} - 1 \right) = nRT \left( \dfrac{P}{2} + \dfrac{8}{P} - 2 \right)$
To find the minimum value of $W$, we need to minimize the expression $\left( \dfrac{P}{2} + \dfrac{8}{P} \right)$.
Using the AM-GM inequality:
$\dfrac{P}{2} + \dfrac{8}{P} \ge 2 \sqrt{\dfrac{P}{2} \times \dfrac{8}{P}} = 2 \sqrt{4} = 4$
The minimum value is $4$, which occurs when $\dfrac{P}{2} = \dfrac{8}{P} \Rightarrow P^2 = 16 \Rightarrow P = 4$ bar. This value of $P$ lies within the given range $(2 < P < 8)$.
Substituting this minimum value back into the work equation:
$W_{min} = nRT (4 - 2) = 2nRT$
Given $n = 0.5$ mol and $T = 600$ K:
$W_{min} = 2 \times 0.5 \times R \times 600 = 600R$
Answer: $600R$
Using the ideal gas equation $V = \dfrac{nRT}{P}$, we have:
$V_1 = \dfrac{nRT}{2}$
$V_2 = \dfrac{nRT}{P}$
$V_3 = \dfrac{nRT}{8}$
The work done on the gas during the first step against a constant external pressure $P$ is:
$W_1 = -P_{ext,1}(V_2 - V_1) = -P \left( \dfrac{nRT}{P} - \dfrac{nRT}{2} \right) = nRT \left( \dfrac{P}{2} - 1 \right)$
The work done on the gas during the second step against a constant external pressure of $8$ bar is:
$W_2 = -P_{ext,2}(V_3 - V_2) = -8 \left( \dfrac{nRT}{8} - \dfrac{nRT}{P} \right) = nRT \left( \dfrac{8}{P} - 1 \right)$
The total work done on the gas is:
$W = W_1 + W_2 = nRT \left( \dfrac{P}{2} - 1 \right) + nRT \left( \dfrac{8}{P} - 1 \right) = nRT \left( \dfrac{P}{2} + \dfrac{8}{P} - 2 \right)$
To find the minimum value of $W$, we need to minimize the expression $\left( \dfrac{P}{2} + \dfrac{8}{P} \right)$.
Using the AM-GM inequality:
$\dfrac{P}{2} + \dfrac{8}{P} \ge 2 \sqrt{\dfrac{P}{2} \times \dfrac{8}{P}} = 2 \sqrt{4} = 4$
The minimum value is $4$, which occurs when $\dfrac{P}{2} = \dfrac{8}{P} \Rightarrow P^2 = 16 \Rightarrow P = 4$ bar. This value of $P$ lies within the given range $(2 < P < 8)$.
Substituting this minimum value back into the work equation:
$W_{min} = nRT (4 - 2) = 2nRT$
Given $n = 0.5$ mol and $T = 600$ K:
$W_{min} = 2 \times 0.5 \times R \times 600 = 600R$
Answer: $600R$
Hence, the correct answer is option: 'B'.