The molar specific heats of an ideal gas at constant pressure and volume are denotes by `C_(P)` and `C_(upsilon)` respectively. If `gamma = (C_(P))/(C_(upsilon))` and `R` is the universal gas constant, then `C_(upsilon)` is equal to

Options

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

Answer

To solve the problem, we need to find the expression for the molar specific heat at constant volume \( C_{\upsilon} \) in terms of the universal gas constant \( R \) and the ratio \( \gamma \) (which is defined as \( \gamma = \frac{C_{P}}{C_{\upsilon}} \)).

### Step-by-step Solution:

1. **Understand the relationship between specific heats**:
We know that for an ideal gas, the relationship between the molar specific heats at constant pressure \( C_{P} \) and constant volume \( C_{\upsilon} \) is given by:
\[ C_{P} - C_{\upsilon} = R \]

2. **Express \( C_{P} \) in terms of \( C_{\upsilon} \) and \( \gamma \)**:
From the definition of \( \gamma \):
\[ \gamma = \frac{C_{P}}{C_{\upsilon}} \]
We can rearrange this to express \( C_{P} \):
\[ C_{P} = \gamma C_{\upsilon} \]

3. **Substitute \( C_{P} \) into the specific heat relationship**:
Now, substitute \( C_{P} \) in the equation \( C_{P} - C_{\upsilon} = R \):
\[ \gamma C_{\upsilon} - C_{\upsilon} = R \]

4. **Factor out \( C_{\upsilon} \)**:
This can be factored as:
\[ (\gamma - 1) C_{\upsilon} = R \]

5. **Solve for \( C_{\upsilon} \)**:
Now, isolate \( C_{\upsilon} \):
\[ C_{\upsilon} = \frac{R}{\gamma - 1} \]

### Final Answer:
Thus, the expression for the molar specific heat at constant volume \( C_{\upsilon} \) is:
\[ C_{\upsilon} = \frac{R}{\gamma - 1} \]