N moles of an ideal gas undergo a two-step process as shown in the figure. Let \( P, V, T \) denote the pressure, volume and temperature of the gas. 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 heat capacities at constant pressure and volume. The heat gained by the gas in the two-step process is given by
N moles of an ideal gas undergo a two-step process as shown in the figure. Let \( P, V, T \) denote the pressure, volume and temperature of the gas. 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 heat capacities at constant pressure and volume. The heat gained by the gas in the two-step process is given by
Show Hint
- \( P_2 (V_2 - V_1) + C_V (T_2 - T_1) \)
- \( P_2 (V_2 - V_1) + C_P (T_2 - T_1) \)
- \( C_P (T_2 - T_1) + C_V (T_2 - T_1) \)
- \( P_2 V_2 - P_1 V_1 \)
The Correct Option is A
Solution and Explanation
Step 1: Heat added during isochoric process.
At constant volume, the heat gained is
\[
Q_{\text{isochoric}} = C_V (T_A - T_1).
\]
Since state-A lies vertically below state-2 in the figure and the final temperature is \( T_2 \), we have \( T_A = T_2 \). Therefore,
\[
Q_{\text{isochoric}} = C_V (T_2 - T_1).
\]
Step 2: Heat added during isobaric process.
At constant pressure \( P_2 \), heat added is
\[
Q_{\text{isobaric}} = P_2 (V_2 - V_1).
\]
Step 3: Total heat added.
Adding both contributions gives
\[
Q = C_V (T_2 - T_1) + P_2 (V_2 - V_1).
\]
This matches option (A).
Final Answer: \( P_2 (V_2 - V_1) + C_V (T_2 - T_1) \)
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