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) $1650R$
For $P \propto V$ (i.e. $PV^{-1} = $ constant), $W = \dfrac{nR\Delta T}{1 - (-1)} = \dfrac{nR\Delta T}{2}$. For an isobaric process, $W = nR\Delta T$. With $n = 3$:
$1 \to 2$: $\dfrac{3R(300)}{2} = 450R$. $\quad 2 \to 3$: $3R(1800) = 5400R$.
$3 \to 4$: $\dfrac{3R(-1400)}{2} = -2100R$. $\quad 4 \to 1$: $3R(-700) = -2100R$.
$$W = 450R + 5400R - 2100R - 2100R = 1650R$$
Solution by Sreeraj P, M.Sc Physics