JEE PYQ: Thermodynamics Question 4
Question 4 - 2021 (18 Mar Shift 1)
The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use $\gamma = 1.4$):
[P-V diagram showing: A(200 N/m$^2$, 1 m$^3$) — D(200 N/m$^2$, 4 m$^3$) at top; B(100 N/m$^2$, 1 m$^3$) — C(100 N/m$^2$, 3 m$^3$) at bottom. CD is an adiabatic process.]
(1) $-500$ J
(2) $-400$ J
(3) 400 J
(4) 200 J
Show Answer
Answer: (1)
Solution
Adiabatic process is from C to D.
$WD = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma} = \frac{P_D V_D - P_C V_C}{1 - \gamma} = \frac{200(3) - (100)(4)}{1 - 1.4} = -500$ J
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