If ` gamma ` is the ratio of specific heats and R is the universal gas constant, then the molar specific heat at constant volume `C_V ` is given by

Options

A. ` gamma R `
B. ` ((gamma -1)R)/(gamma )`
C. ` ( R ) /( gamma - 1)`
D. ` (gamma R ) /(gamma -1)`

Explanation

To find the molar specific heat at constant volume \( C_V \) in terms of the ratio of specific heats \( \gamma \) and the universal gas constant \( R \), we can follow these steps:

### Step 1: Understand the relationship between specific heats
We know from thermodynamics that the difference between the molar specific heats at constant pressure \( C_P \) and constant volume \( C_V \) is given by:
\[ C_P - C_V = R \]

### Step 2: Express \( C_P \) in terms of \( C_V \) and \( \gamma \)
The ratio of specific heats \( \gamma \) is defined as:
\[ \gamma = \frac{C_P}{C_V} \]
From this equation, we can express \( C_P \) in terms of \( C_V \):
\[ C_P = \gamma C_V \]

### Step 3: Substitute \( C_P \) into the first equation
Now, we can substitute the expression for \( C_P \) into the equation from Step 1:
\[ \gamma C_V - C_V = R \]

### Step 4: Factor out \( C_V \)
This simplifies to:
\[ (\gamma - 1) C_V = R \]

### Step 5: Solve for \( C_V \)
Now, we can solve for \( C_V \):
\[ C_V = \frac{R}{\gamma - 1} \]

### Conclusion
Thus, the molar specific heat at constant volume \( C_V \) is given by:
\[ C_V = \frac{R}{\gamma - 1} \]