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 of finding the average molar heat capacity for the whole process involving a diatomic ideal gas, we will break down the steps as follows:

### Step 1: Understand the Process
The process consists of two parts:
1. Heating at constant volume until the pressure is doubled.
2. Heating at constant pressure until the volume is doubled.

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

### Step 3: Analyze the First Process (A to B)
In the first process (constant volume):
- The pressure is doubled, so \( P_B = 2P_0 \).
- Using the ideal gas law \( PV = nRT \), since volume is constant, we have:
\[ \frac{P_A}{T_A} = \frac{P_B}{T_B} \]
This implies:
\[ \frac{P_0}{T_0} = \frac{2P_0}{T_B} \Rightarrow T_B = 2T_0 \]

- The change in internal energy \( \Delta U_{AB} \) is given by:
\[ Q_{AB} = \Delta U_{AB} = n C_{V_m} \Delta T \]
where \( C_{V_m} \) for a diatomic gas is \( \frac{5}{2}R \). The change in temperature is:
\[ \Delta T = T_B - T_A = 2T_0 - T_0 = T_0 \]
Thus:
\[ Q_{AB} = n \left(\frac{5}{2}R\right) T_0 = \frac{5}{2} n R T_0 \]

### Step 4: Analyze the Second Process (B to C)
In the second process (constant pressure):
- The volume is doubled, so \( V_C = 2V_B \).
- Since pressure is constant, we can use the ideal gas law again:
\[ \frac{P_B}{T_B} = \frac{P_C}{T_C} \]
Here, \( P_C = P_B = 2P_0 \), and we need to find \( T_C \):
\[ \frac{2P_0}{2T_0} = \frac{2P_0}{T_C} \Rightarrow T_C = 4T_0 \]

- The heat added in this process is:
\[ Q_{BC} = n C_{P_m} \Delta T \]
where \( C_{P_m} \) for a diatomic gas is \( \frac{7}{2}R \). The change in temperature is:
\[ \Delta T = T_C - T_B = 4T_0 - 2T_0 = 2T_0 \]
Thus:
\[ Q_{BC} = n \left(\frac{7}{2}R\right) (2T_0) = 7nRT_0 \]

### Step 5: Total Heat for the Whole Process
The total heat \( Q_{ABC} \) for the entire process is:
\[ Q_{ABC} = Q_{AB} + Q_{BC} = \frac{5}{2} n R T_0 + 7 n R T_0 = \left(\frac{5}{2} + 7\right) n R T_0 = \frac{19}{2} n R T_0 \]

### Step 6: Calculate Average Molar Heat Capacity
The average molar heat capacity \( C_{m} \) is defined as:
\[ Q_{ABC} = n C_m \Delta T \]
where \( \Delta T = T_C - T_A = 4T_0 - T_0 = 3T_0 \). Therefore:
\[ \frac{19}{2} n R T_0 = n C_m (3T_0) \]
Dividing both sides by \( n T_0 \):
\[ \frac{19}{2} R = C_m (3) \]
Thus:
\[ C_m = \frac{19}{6} R \]

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