An `HCl` molecule has rolational, translational and vibrational motions. If the rms velocity is `bar(v)`, m is its mass and `k_(B)` is Boltzmann constant, then its temperature will be :

Options

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

Answer

To find the temperature of an HCl molecule given its root mean square (rms) velocity, we can use the relationship between rms velocity, temperature, and mass. Here’s a step-by-step solution:

### Step 1: Understand the relationship for rms velocity
The rms velocity (\( \bar{v} \)) of a gas molecule can be expressed in terms of temperature (T) and mass (m) using the formula:
\[ \bar{v} = \sqrt{\frac{3k_B T}{m}} \]
where \( k_B \) is the Boltzmann constant.

### Step 2: Rearrange the formula to solve for temperature
To find the temperature (T), we will rearrange the formula. First, square both sides:
\[ \bar{v}^2 = \frac{3k_B T}{m} \]

### Step 3: Isolate T
Now, we can isolate T by multiplying both sides by m and then dividing by \( 3k_B \):
\[ T = \frac{m \bar{v}^2}{3k_B} \]

### Step 4: Write the final expression for temperature
Thus, the expression for the temperature of the HCl molecule is:
\[ T = \frac{m \bar{v}^2}{3k_B} \]

### Conclusion
This equation gives us the temperature of the HCl molecule based on its mass, rms velocity, and the Boltzmann constant.

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