A diatomic ideal gas is heated at constant at constant volume until the pressure is doubled and again heated of constant pressure until the volume is doubled. The average molar heat capacity for the whole process is

Options

A. `(13R)/(6)`
B. `(19R)/(6)`
C. `(23R)/(6)`
D. `(17R)/(6)`

Explanation

To solve the problem, we will analyze the heating process of a diatomic ideal gas step by step.

### Step 1: Understand the Process
The gas undergoes two processes:
1. Heating at constant volume until the pressure is doubled.
2. Heating at constant pressure until the volume is doubled.

### Step 2: Initial Conditions
Let:
- Initial pressure \( P_0 \)
- Initial volume \( V_0 \)
- Initial temperature \( T_0 \)

From the ideal gas law, we have:
\[ P_0 V_0 = nRT_0 \]

### Step 3: First Process (Constant Volume)
In the first process, the volume remains constant, and the pressure doubles:
- Final pressure \( P_B = 2P_0 \)
- Since the volume is constant, the temperature at point B can be found using the ideal gas law:
\[ P_B V_0 = nRT_B \implies 2P_0 V_0 = nRT_B \]
Substituting \( P_0 V_0 = nRT_0 \):
\[ 2(nRT_0) = nRT_B \implies T_B = 2T_0 \]

### Step 4: Calculate Heat for Process AB
The heat added during this process at constant volume is given by:
\[ \Delta Q_{AB} = \Delta U_{AB} = nC_V \Delta T \]
Where \( C_V \) for a diatomic gas is \( \frac{5}{2}R \):
\[ \Delta Q_{AB} = nC_V (T_B - T_0) = n \left(\frac{5}{2}R\right) (2T_0 - T_0) = n \left(\frac{5}{2}R\right) T_0 = \frac{5}{2}nRT_0 \]

### Step 5: Second Process (Constant Pressure)
In the second process, the pressure remains constant, and the volume doubles:
- Final volume \( V_C = 2V_0 \)
- The temperature at point C can be found using the ideal gas law:
\[ P_B V_C = nRT_C \implies 2P_0 (2V_0) = nRT_C \implies 4P_0 V_0 = nRT_C \]
Using \( P_0 V_0 = nRT_0 \):
\[ 4(nRT_0) = nRT_C \implies T_C = 4T_0 \]

### Step 6: Calculate Heat for Process BC
The heat added during this process at constant pressure is given by:
\[ \Delta Q_{BC} = nC_P \Delta T \]
Where \( C_P \) for a diatomic gas is \( \frac{7}{2}R \):
\[ \Delta Q_{BC} = nC_P (T_C - T_B) = n \left(\frac{7}{2}R\right) (4T_0 - 2T_0) = n \left(\frac{7}{2}R\right) (2T_0) = 7nRT_0 \]

### Step 7: Total Heat for the Process
The total heat added in the entire process (from A to C) is:
\[ \Delta Q_{ABC} = \Delta Q_{AB} + \Delta Q_{BC} = \frac{5}{2}nRT_0 + 7nRT_0 = \frac{5}{2}nRT_0 + \frac{14}{2}nRT_0 = \frac{19}{2}nRT_0 \]

### Step 8: Calculate Average Molar Heat Capacity
The average molar heat capacity \( C \) for the whole process can be defined as:
\[ \Delta Q_{ABC} = nC \Delta T_{total} \]
Where \( \Delta T_{total} = T_C - T_0 = 4T_0 - T_0 = 3T_0 \):
\[ \frac{19}{2}nRT_0 = nC (3T_0) \]
Dividing both sides by \( nT_0 \):
\[ \frac{19}{2}R = 3C \implies C = \frac{19}{6}R \]

### Final Answer
The average molar heat capacity for the whole process is:
\[ C = \frac{19}{6}R \]