A monoatomic gas undergoes a process in which the pressure (P) and the volume (V) of the gas are related as `PV^(-3)=`constant. What will be the molar heat capacity of gas for this process?

Options

A. `(5R)/(3)`
B. `(3)/(5)R`
C. `(7)/(5)R`
D. `(R )/(4)`

Answer

To solve the problem, we need to find the molar heat capacity of a monoatomic gas undergoing a process defined by the equation \( PV^{-3} = \text{constant} \).

### Step-by-Step Solution:

1. **Identify the Type of Process**:
The given equation \( PV^{-3} = \text{constant} \) can be rewritten as \( PV^n = \text{constant} \) where \( n = -3 \). This indicates that the process is a polytropic process.

2. **Use the Formula for Molar Heat Capacity**:
The molar heat capacity \( C \) for a polytropic process can be given by the formula:
\[ C = C_v + \frac{R}{1 - n} \]
where \( C_v \) is the molar heat capacity at constant volume, \( R \) is the universal gas constant, and \( n \) is the polytropic index.

3. **Determine \( C_v \) for a Monoatomic Gas**:
For a monoatomic gas, the molar heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{3R}{2} \]

4. **Substitute Values into the Heat Capacity Formula**:
Now, substituting \( C_v \) and \( n \) into the heat capacity formula:
\[ C = \frac{3R}{2} + \frac{R}{1 - (-3)} \]
Simplifying the denominator:
\[ 1 - (-3) = 1 + 3 = 4 \]
Thus, the equation becomes:
\[ C = \frac{3R}{2} + \frac{R}{4} \]

5. **Find a Common Denominator**:
To add these fractions, we need a common denominator. The common denominator between 2 and 4 is 4:
\[ C = \frac{6R}{4} + \frac{R}{4} = \frac{6R + R}{4} = \frac{7R}{4} \]

6. **Final Answer**:
Therefore, the molar heat capacity of the gas for this process is:
\[ C = \frac{7R}{4} \]