A diatomic ideal gas is heated at constant volume until its pressure becomes three times. It is again heated at constant pressure until its volume is doubled. Find the molar heat capacity for the whole process.

Explanation

To find the molar heat capacity for the entire process involving a diatomic ideal gas, we will break down the problem into two parts: the heating at constant volume and the heating at constant pressure.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial temperature of the gas be \( T \).
- The initial pressure is \( P \) and the initial volume is \( V \).

2. **First Process (Constant Volume)**:
- The gas is heated at constant volume until its pressure becomes three times the initial pressure.
- According to the ideal gas law, at constant volume, pressure is directly proportional to temperature:
\[ P \propto T \implies \frac{P_f}{P_i} = \frac{T_f}{T_i} \]
- Since \( P_f = 3P_i \), we have:
\[ \frac{3P}{P} = \frac{T_f}{T} \implies T_f = 3T \]
- The change in temperature for the first process (\( \Delta T_1 \)) is:
\[ \Delta T_1 = T_f - T_i = 3T - T = 2T \]

3. **Second Process (Constant Pressure)**:
- The gas is then heated at constant pressure until its volume doubles.
- At constant pressure, volume is directly proportional to temperature:
\[ V \propto T \implies \frac{V_f}{V_i} = \frac{T_f}{T_i} \]
- Since \( V_f = 2V_i \), we have:
\[ \frac{2V}{V} = \frac{T_f}{3T} \implies T_f = 2 \times 3T = 6T \]
- The change in temperature for the second process (\( \Delta T_2 \)) is:
\[ \Delta T_2 = T_f - T_i = 6T - 3T = 3T \]

4. **Calculate Total Change in Temperature**:
- The total change in temperature for the entire process (\( \Delta T \)):
\[ \Delta T = T_f - T_i = 6T - T = 5T \]

5. **Calculate Heat for Each Process**:
- For the first process (constant volume), the heat added (\( Q_1 \)):
\[ Q_1 = nC_v \Delta T_1 = n \left(\frac{5}{2}R\right)(2T) = 5nRT \]
- For the second process (constant pressure), the heat added (\( Q_2 \)):
\[ Q_2 = nC_p \Delta T_2 = n \left(\frac{7}{2}R\right)(3T) = \frac{21}{2}nRT \]

6. **Total Heat Added**:
- The total heat added (\( Q \)):
\[ Q = Q_1 + Q_2 = 5nRT + \frac{21}{2}nRT = \left(5 + \frac{21}{2}\right)nRT = \frac{10}{2}nRT + \frac{21}{2}nRT = \frac{31}{2}nRT \]

7. **Calculate Molar Heat Capacity**:
- The molar heat capacity \( C \) for the whole process is given by:
\[ C = \frac{Q}{n \Delta T} = \frac{\frac{31}{2}nRT}{n(5T)} = \frac{31}{10}R \]

### Final Answer:
The molar heat capacity for the whole process is:
\[ C = 3.1R \]