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GATE 2022 CH – Question 37

Thermodynamics · Laws of thermodynamics, open and closed systems, entropy and chemical potential · 2 marks · Multiple choice

$N$ moles of an ideal gas undergo a two-step process as shown in the figure. Let $P$, $V$ and $T$ denote the pressure, volume and temperature of the gas, respectively. The gas, initially at state-1 ($P_1$, $V_1$, $T_1$), undergoes an isochoric (constant volume) process to reach state-A, and then undergoes an isobaric (constant pressure) expansion to reach state-2 ($P_2$, $V_2$, $T_2$). For an ideal gas, $C_P - C_V = NR$, where $C_P$ and $C_V$ are the heat capacities at constant pressure and constant volume, respectively, and assumed to be temperature independent. The heat gained by the gas in the two-step process is given by

a P-V diagram. State 1 is at the top left, the isochoric line goes down to state A, and the isobaric line goes right from A to state 2.
  1. $P_2(V_2 - V_1) + C_V(T_2 - T_1)$
  2. $P_2(V_2 - V_1) + C_P(T_2 - T_1)$
  3. $C_P(T_2 - T_1) + C_V(T_2 - T_1)$
  4. $P_2V_2 - P_1V_1$

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Correct answer: (A) $P_2(V_2 - V_1) + C_V(T_2 - T_1)$

Explanation

By the first law, $Q = \Delta U + W$. For an ideal gas $\Delta U = C_V(T_2 - T_1)$ for any path. Work is done only in the isobaric step, at the pressure $P_A = P_2$: $W = P_2(V_2 - V_1)$. So $Q = C_V(T_2 - T_1) + P_2(V_2 - V_1)$.