An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is `vec(v), m` is its mass and `k_(s)` is Bolzmann constant, then its temperature will be `(mv^(2))/(nk_(B))` , where n is __________ .

Explanation

To solve the problem, we need to determine the value of \( n \) in the equation for temperature given by:

\[ T = \frac{m v^2}{n k_B} \]

where:
- \( T \) is the temperature,
- \( m \) is the mass of the HCl molecule,
- \( v \) is the root mean square (rms) velocity of the HCl molecules,
- \( k_B \) is the Boltzmann constant.

### Step-by-Step Solution:

1. **Identify Degrees of Freedom**:
- For a diatomic molecule like HCl, it has translational, rotational, and vibrational motions.
- The translational motion contributes 3 degrees of freedom (movement along x, y, and z axes).
- The rotational motion contributes 2 degrees of freedom (rotation about two perpendicular axes).
- The vibrational motion contributes 1 degree of freedom (vibration along the bond).
- Therefore, the total degrees of freedom \( F \) for HCl is:
\[ F = 3 \text{ (translational)} + 2 \text{ (rotational)} + 1 \text{ (vibrational)} = 6 \]

2. **Relate Kinetic Energy to Temperature**:
- The average kinetic energy of a molecule can be expressed in two ways:
- Using degrees of freedom:
\[ \text{K.E.} = \frac{F}{2} k_B T \]
- Using rms velocity:
\[ \text{K.E.} = \frac{1}{2} m v^2 \]

3. **Set the Two Expressions Equal**:
- Since both expressions represent the average kinetic energy, we can set them equal to each other:
\[ \frac{F}{2} k_B T = \frac{1}{2} m v^2 \]

4. **Substitute the Value of \( F \)**:
- We already found that \( F = 6 \):
\[ \frac{6}{2} k_B T = \frac{1}{2} m v^2 \]
- Simplifying this gives:
\[ 3 k_B T = \frac{1}{2} m v^2 \]

5. **Solve for Temperature \( T \)**:
- Rearranging the equation to solve for \( T \):
\[ T = \frac{m v^2}{6 k_B} \]

6. **Compare with Given Equation**:
- The problem states that:
\[ T = \frac{m v^2}{n k_B} \]
- By comparing both expressions for \( T \), we can see that:
\[ n = 6 \]

### Conclusion:
The value of \( n \) is \( 6 \).