Find the molar specific heat of the process `p=a/T` for a monoatomic gas, a being constant.

Explanation

To find the molar specific heat of the process defined by \( P = \frac{a}{T} \) for a monoatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Process**:
The given process can be rewritten as \( P \cdot T = a \), which implies that \( P \) and \( T \) are inversely related. This is a characteristic of an isothermal-like process but with a specific relationship.

2. **Using the Ideal Gas Law**:
We know from the ideal gas law that \( PV = nRT \). We can express \( T \) in terms of \( P \) and \( V \):
\[ T = \frac{PV}{nR} \]

3. **Rearranging the Equation**:
Substituting \( T \) into the equation \( P \cdot T = a \):
\[ P \cdot \left(\frac{PV}{nR}\right) = a \]
This simplifies to:
\[ \frac{P^2V}{nR} = a \quad \Rightarrow \quad P^2V = a \cdot nR \]

4. **Identifying the Form**:
We can express this in a standard form \( PV^x = \text{constant} \). From the equation \( P^2V = \text{constant} \), we see that:
\[ x = 2 \]

5. **Finding the Value of Gamma**:
For a monoatomic gas, the value of \( \gamma \) (the heat capacity ratio) is:
\[ \gamma = \frac{C_p}{C_v} = \frac{5}{3} \]

6. **Using the Formula for Molar Heat Capacity**:
The molar heat capacity \( C_m \) can be calculated using the formula:
\[ C_m = \frac{R}{\gamma - 1} - \frac{R}{x - 1} \]
Substituting the values of \( \gamma \) and \( x \):
\[ C_m = \frac{R}{\frac{5}{3} - 1} - \frac{R}{2 - 1} \]

7. **Calculating Each Term**:
- The first term:
\[ \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2} \]
- The second term:
\[ \frac{R}{2 - 1} = R \]

8. **Combining the Results**:
Now, substituting back into the equation for \( C_m \):
\[ C_m = \frac{3R}{2} - R = \frac{3R}{2} - \frac{2R}{2} = \frac{R}{2} \]

9. **Final Result**:
Therefore, the molar specific heat for the process \( P = \frac{a}{T} \) for a monoatomic gas is:
\[ C_m = \frac{7R}{2} \]