Q 11-11-013JEE MainEAMCET 2013 (Engineering)Hard
Three moles of an ideal monatomic gas undergo a cyclic process 1 → 2 → 3 → 4 → 1. In processes 1 → 2 and 3 → 4 the pressure is proportional to the volume, while processes 2 → 3 and 4 → 1 are at constant pressure. The temperatures in states 1, 2, 3 and 4 are $400$ K, $700$ K, $2500$ K and $1100$ K respectively. The work done by the gas in the cycle is ($R$ = universal gas constant)
Answer: (A) $1650R$
Concept. Polytropic work $W = \dfrac{nR\Delta T}{1 - x}$: for $P \propto V$ ($x = -1$), $W = \dfrac{nR\Delta T}{2}$; isobaric, $W = nR\Delta T$.
Step 1. Each step ($n = 3$).
- $1 \to 2$: $\dfrac{3R(300)}{2} = 450R$.
- $2 \to 3$: $3R(1800) = 5400R$.
- $3 \to 4$: $\dfrac{3R(-1400)}{2} = -2100R$.
- $4 \to 1$: $3R(-700) = -2100R$.
Step 2. Total. $W = 1650R$.
Answer. $1650R$ (Option A).
Solution by Sreeraj P, M.Sc Physics