Considering all type of `D.O.F` for HCl molecule of mass `m` having `V_(rms)` as `overline(v)`. Then the temperature of gas will be

Options

A. `(moverline(v)^(2))/(3k)`
B. `(moverline(v)^(2))/(5k)`
C. `(moverline(v)^(2))/(7k)`
D. `(moverline(v)^(2))/(6k)`

Answer

To find the temperature of the HCl gas based on the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between Vrms and temperature**:
The root mean square velocity (Vrms) of a gas is related to its temperature by the formula:
\[ V_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- \( V_{rms} \) is the root mean square velocity,
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature,
- \( m \) is the mass of the gas molecule.

2. **Substituting the given values**:
We know \( V_{rms} = V \) (as given in the problem). Therefore, we can write:
\[ V = \sqrt{\frac{3kT}{m}} \]

3. **Square both sides**:
To eliminate the square root, we square both sides of the equation:
\[ V^2 = \frac{3kT}{m} \]

4. **Rearranging the equation to solve for T**:
We want to isolate \( T \) on one side. Multiply both sides by \( m \):
\[ mV^2 = 3kT \]
Now, divide both sides by \( 3k \):
\[ T = \frac{mV^2}{3k} \]

5. **Final expression for temperature**:
Thus, the temperature \( T \) of the gas is given by:
\[ T = \frac{mV^2}{3k} \]

### Conclusion:
The temperature of the HCl gas is \( \frac{mV^2}{3k} \).