A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at 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 find the average molar heat capacity for the process described, we will break down the problem step by step.

### Step 1: Define Initial Conditions
Let the initial state of the gas be defined as:
- Pressure at state 1: \( P_1 = P_0 \)
- Volume at state 1: \( V_1 = V_0 \)
- Temperature at state 1: \( T_1 = T_0 \)

### Step 2: Process 1 to 2 (Constant Volume)
In this process, the gas is heated at constant volume until the pressure is doubled.
- Final pressure at state 2: \( P_2 = 2P_0 \)
- Volume remains constant: \( V_2 = V_0 \)

Using the ideal gas law \( PV = nRT \):
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \implies \frac{2P_0}{P_0} = \frac{T_2}{T_0} \implies T_2 = 2T_0 \]

### Step 3: Process 2 to 3 (Constant Pressure)
In this process, the gas is heated at constant pressure until the volume is doubled.
- Pressure remains constant: \( P_3 = P_2 = 2P_0 \)
- Final volume at state 3: \( V_3 = 2V_0 \)

Using the ideal gas law again:
\[ \frac{V_3}{V_2} = \frac{T_3}{T_2} \implies \frac{2V_0}{V_0} = \frac{T_3}{2T_0} \implies T_3 = 4T_0 \]

### Step 4: Calculate Heat Added in Each Process
1. **Heat added in process 1 to 2 (constant volume)**:
\[ Q_{1 \to 2} = n C_V (T_2 - T_1) = n C_V (2T_0 - T_0) = n C_V T_0 \]
For a diatomic gas, \( C_V = \frac{5}{2} R \):
\[ Q_{1 \to 2} = n \left(\frac{5}{2} R\right) T_0 \]

2. **Heat added in process 2 to 3 (constant pressure)**:
\[ Q_{2 \to 3} = n C_P (T_3 - T_2) = n C_P (4T_0 - 2T_0) = n C_P (2T_0) \]
For a diatomic gas, \( C_P = \frac{7}{2} R \):
\[ Q_{2 \to 3} = n \left(\frac{7}{2} R\right) (2T_0) = n \left(7R\right) T_0 \]

### Step 5: Total Heat Added
The total heat added during the entire process is:
\[ Q_{\text{total}} = Q_{1 \to 2} + Q_{2 \to 3} = n \left(\frac{5}{2} R T_0\right) + n (7R T_0) \]
Combining the terms:
\[ Q_{\text{total}} = n \left(\frac{5}{2} R T_0 + 7R T_0\right) = n \left(\frac{5}{2} R T_0 + \frac{14}{2} R T_0\right) = n \left(\frac{19}{2} R T_0\right) \]

### Step 6: Average Molar Heat Capacity
The average molar heat capacity \( C_{\text{avg}} \) for the entire process is given by:
\[ C_{\text{avg}} = \frac{Q_{\text{total}}}{\Delta T} \]
Where \( \Delta T = T_3 - T_1 = 4T_0 - T_0 = 3T_0 \):
\[ C_{\text{avg}} = \frac{n \left(\frac{19}{2} R T_0\right)}{3T_0} = \frac{19}{6} R \]

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