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. (a)`(5R)/(3)`
B. (b)`(3)/(5)R`
C. (c)`(7)/(5)R`
D. (d)`(R )/(4)`

Answer

To solve the problem of finding the molar heat capacity of a monoatomic gas undergoing a process where \( PV^{-3} = \text{constant} \), we can follow these steps:

### Step 1: Identify the type of process
The relationship \( PV^{-3} = \text{constant} \) can be rewritten as \( P \cdot V^n = \text{constant} \) where \( n = -3 \). This indicates that the process is a polytropic process.

### Step 2: Use the formula for molar heat capacity
For a polytropic process, the molar heat capacity \( C \) can be expressed as:
\[ C = C_V + \frac{R}{1 - n} \]
where:
- \( C_V \) is the molar heat capacity at constant volume,
- \( R \) is the gas constant,
- \( n \) is the polytropic index.

### Step 3: Determine \( C_V \) for a monoatomic gas
For a monoatomic ideal gas, the molar heat capacity at constant volume is given by:
\[ C_V = \frac{3}{2} R \]

### Step 4: Substitute values into the heat capacity formula
Substituting \( C_V \) and \( n = -3 \) into the heat capacity formula:
\[ C = C_V + \frac{R}{1 - (-3)} = \frac{3}{2} R + \frac{R}{1 + 3} \]
\[ C = \frac{3}{2} R + \frac{R}{4} \]

### Step 5: Find a common denominator and simplify
To add these fractions, we need a common denominator. The least common multiple of 2 and 4 is 4:
\[ C = \frac{6}{4} R + \frac{1}{4} R = \frac{6R + 1R}{4} = \frac{7R}{4} \]

### Step 6: Final result
Thus, the molar heat capacity \( C \) for this process is:
\[ C = \frac{7R}{4} \]

### Conclusion
The correct answer is \( \frac{7R}{4} \).

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