If `Q` amount of heat is given to a diatomic ideal gas in a process in which the gas perform a work `(2Q)/(3)` on its surrounding. Find the molar heat capacity (in terms of `R`) for the process.

Explanation

To find the molar heat capacity \( C \) for the process involving a diatomic ideal gas, we can follow these steps:

### Step 1: Understand the relationship between heat, work, and internal energy
According to the first law of thermodynamics, we have:
\[ Q = \Delta U + W \]
Where:
- \( Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the system.

### Step 2: Substitute the given values
From the problem, we know:
- \( Q \) is the amount of heat given to the gas,
- The work done \( W = \frac{2Q}{3} \).

Substituting this into the first law equation gives:
\[ Q = \Delta U + \frac{2Q}{3} \]

### Step 3: Solve for change in internal energy \( \Delta U \)
Rearranging the equation to solve for \( \Delta U \):
\[ \Delta U = Q - \frac{2Q}{3} = \frac{Q}{3} \]

### Step 4: Relate internal energy change to temperature change
For an ideal gas, the change in internal energy is also given by:
\[ \Delta U = n C_V \Delta T \]
Where:
- \( n \) is the number of moles,
- \( C_V \) is the molar heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

### Step 5: Use the value of \( C_V \) for a diatomic gas
For a diatomic ideal gas, the molar heat capacity at constant volume is:
\[ C_V = \frac{5R}{2} \]
Thus, we can write:
\[ \Delta U = n \left(\frac{5R}{2}\right) \Delta T \]

### Step 6: Set the two expressions for \( \Delta U \) equal to each other
Now we can equate the two expressions for \( \Delta U \):
\[ \frac{Q}{3} = n \left(\frac{5R}{2}\right) \Delta T \]

### Step 7: Solve for \( n \Delta T \)
Rearranging gives:
\[ n \Delta T = \frac{Q}{3} \cdot \frac{2}{5R} = \frac{2Q}{15R} \]

### Step 8: Relate heat \( Q \) to molar heat capacity \( C \)
The heat added can also be expressed as:
\[ Q = n C \Delta T \]
Substituting \( n \Delta T \) from the previous step:
\[ Q = C \left(\frac{2Q}{15R}\right) \]

### Step 9: Solve for \( C \)
Rearranging gives:
\[ C = \frac{Q}{\frac{2Q}{15R}} = \frac{15R}{2} \]

### Final Answer
Thus, the molar heat capacity \( C \) in terms of \( R \) is:
\[ C = \frac{15R}{2} \quad \text{or} \quad C = 7.5R \]