The ratio of `P` to `V` at any instant is constant and is equal to `1`, for a monoatomic ideal gas under going a process. What is the molar heat capacity of the gas

Options

A. `(3R)/(2)`
B. `(4R)/(2)`
C. `(5R)/(2)`
D. 0

Explanation

To solve the problem, we need to determine the molar heat capacity of a monoatomic ideal gas undergoing a process where the ratio of pressure (P) to volume (V) is constant and equal to 1.

### Step-by-step Solution:

1. **Understanding the Process**:
Given that the ratio \( \frac{P}{V} = 1 \), we can express this as:
\[ P = V \]
This indicates that pressure and volume are directly proportional in this process.

2. **Using the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
Substituting \( P = V \) into the ideal gas law gives:
\[ V^2 = nRT \]
From this, we can express the temperature \( T \) in terms of volume \( V \):
\[ T = \frac{V^2}{nR} \]

3. **Calculating the Change in Internal Energy**:
The change in internal energy \( \Delta U \) for a monoatomic ideal gas is given by:
\[ \Delta U = \frac{3}{2} nR \Delta T \]
We need to find \( \Delta T \). Since \( T \) is a function of \( V \), we can differentiate:
\[ \Delta T = \frac{dT}{dV} \Delta V \]
From \( T = \frac{V^2}{nR} \):
\[ \frac{dT}{dV} = \frac{2V}{nR} \]
Therefore:
\[ \Delta T = \frac{2V}{nR} \Delta V \]

4. **Calculating the Molar Heat Capacity**:
The molar heat capacity \( C \) is defined as:
\[ C = \frac{\Delta Q}{\Delta T} \]
For an ideal gas, we can relate the heat added \( \Delta Q \) to the change in internal energy and work done:
\[ \Delta Q = \Delta U + P \Delta V \]
Substituting \( \Delta U \) and \( P \):
\[ \Delta Q = \frac{3}{2} nR \Delta T + P \Delta V \]
Since \( P = V \):
\[ \Delta Q = \frac{3}{2} nR \Delta T + V \Delta V \]

5. **Final Expression for Molar Heat Capacity**:
Rearranging the equation for \( C \):
\[ C = \frac{\Delta U + P \Delta V}{\Delta T} \]
Substituting the expressions we derived for \( \Delta U \) and \( \Delta T \):
\[ C = \frac{\frac{3}{2} nR \Delta T + V \Delta V}{\Delta T} \]
Simplifying gives:
\[ C = \frac{3}{2} nR + \frac{V \Delta V}{\Delta T} \]

6. **Conclusion**:
Since \( V \) is constant in this process, the molar heat capacity \( C \) can be expressed as:
\[ C = \frac{5}{2} R \]
Therefore, the molar heat capacity of the monoatomic ideal gas under the given conditions is:
\[ C = \frac{5}{2} R \]