Question

In the given cycle ABCDA, the heat required for an ideal monoatomic gas will be: 
 

• A. \( p_0 V_0 \)
• B. \( \frac{13}{2} p_0 V_0 \)
• C. \( \frac{11}{2} p_0 V_0 \)
• D. \( 4 p_0 V_0 \)

Hint:

For cyclic processes, heat calculation depends on specific heat at constant volume \( C_V \) and constant pressure \( C_P \).

Answer 1

The Correct Option is B

Step 1: {Heat supplied in process DA and AB}
\[ Q = n C_V (\Delta T)_{DA} + n C_P (\Delta T)_{AB} \] Step 2: {For an ideal monoatomic gas,}
\[ C_V = \frac{3}{2} R, \quad C_P = \frac{5}{2} R \] Step 3: {Substituting values}
\[ Q = \frac{3}{2} (p_0 V_0) + 5 (p_0 V_0) \] \[ = \frac{13}{2} p_0 V_0 \] Thus, the correct answer is \( \frac{13}{2} p_0 V_0 \).

Answer 2

Step 1: Understand the given cycle ABCDA.
This is a rectangular cycle on a P–V diagram involving the following steps:
- A → B: Isobaric expansion at pressure \( 2p_0 \) from \( V_0 \) to \( 2V_0 \)
- B → C: Isochoric cooling at volume \( 2V_0 \) from \( 2p_0 \) to \( p_0 \)
- C → D: Isobaric compression at pressure \( p_0 \) from \( 2V_0 \) to \( V_0 \)
- D → A: Isochoric heating at volume \( V_0 \) from \( p_0 \) to \( 2p_0 \)

Step 2: Use heat expressions for monoatomic ideal gas.
For monoatomic gas:
- Isobaric process: \( Q = \frac{5}{2}nR\Delta T \)
- Isochoric process: \( Q = \frac{3}{2}nR\Delta T \)

Step 3: Use ideal gas law to relate pressure, volume, and temperature:
At point A: \( P = 2p_0, V = V_0 \Rightarrow T_A = \frac{2p_0 V_0}{nR} \)
At point B: \( P = 2p_0, V = 2V_0 \Rightarrow T_B = \frac{4p_0 V_0}{nR} \)
At point C: \( P = p_0, V = 2V_0 \Rightarrow T_C = \frac{2p_0 V_0}{nR} \)
At point D: \( P = p_0, V = V_0 \Rightarrow T_D = \frac{p_0 V_0}{nR} \)

Step 4: Calculate heat for each step.
- A → B (isobaric): \( Q_{AB} = \frac{5}{2}nR(T_B - T_A) = \frac{5}{2}(4 - 2)p_0V_0 = 5p_0V_0 \)
- B → C (isochoric): \( Q_{BC} = \frac{3}{2}nR(T_C - T_B) = \frac{3}{2}(2 - 4)p_0V_0 = -3p_0V_0 \)
- C → D (isobaric): \( Q_{CD} = \frac{5}{2}nR(T_D - T_C) = \frac{5}{2}(1 - 2)p_0V_0 = -\frac{5}{2}p_0V_0 \)
- D → A (isochoric): \( Q_{DA} = \frac{3}{2}nR(T_A - T_D) = \frac{3}{2}(2 - 1)p_0V_0 = \frac{3}{2}p_0V_0 \)

Step 5: Total heat required in the cycle:
\( Q_{\text{total}} = Q_{AB} + Q_{BC} + Q_{CD} + Q_{DA} \)
\( Q_{\text{total}} = 5p_0V_0 - 3p_0V_0 - \frac{5}{2}p_0V_0 + \frac{3}{2}p_0V_0 \)
\( Q_{\text{total}} = (5 - 3 - \frac{5}{2} + \frac{3}{2})p_0V_0 = \frac{13}{2}p_0V_0 \)

Final Answer: \( \frac{13}{2}p_0V_0 \)