Temperature of diatomic gas is `300 K`. If moment of intertia of its molecules is `8.28 xx 10^-38 g- cm^2`. Calculate their root mean square angular velocity.

Explanation

To solve the problem of calculating the root mean square angular velocity of a diatomic gas at a temperature of 300 K, we can follow these steps:

### Step 1: Understand the formulas
We need to use the relationship between kinetic energy and angular velocity. The kinetic energy of rotation is given by:
\[ KE_{angular} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

For a diatomic gas, the rotational kinetic energy can also be expressed as:
\[ KE_{angular} = \frac{f}{2} k T \]
where \( f \) is the number of rotational degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.

### Step 2: Identify the degrees of freedom
For a diatomic gas, the number of rotational degrees of freedom \( f \) is 2.

### Step 3: Use the Boltzmann constant
The value of the Boltzmann constant \( k \) is:
\[ k = 1.38 \times 10^{-23} \, \text{J/K} \]

### Step 4: Set up the equation
Equate the two expressions for kinetic energy:
\[ \frac{1}{2} I \omega^2 = \frac{f}{2} k T \]

### Step 5: Solve for angular velocity \( \omega \)
Rearranging the equation gives:
\[ I \omega^2 = f k T \]
\[ \omega^2 = \frac{f k T}{I} \]
\[ \omega = \sqrt{\frac{f k T}{I}} \]

### Step 6: Substitute the values
Now, substitute the known values:
- \( f = 2 \)
- \( k = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T = 300 \, \text{K} \)
- \( I = 8.28 \times 10^{-38} \, \text{g cm}^2 \)

First, convert the moment of inertia from \( \text{g cm}^2 \) to \( \text{kg m}^2 \):
\[ I = 8.28 \times 10^{-38} \, \text{g cm}^2 = 8.28 \times 10^{-38} \times 10^{-3} \, \text{kg} \times 10^{-4} \, \text{m}^2 = 8.28 \times 10^{-45} \, \text{kg m}^2 \]

Now substitute these values into the equation for \( \omega \):
\[ \omega = \sqrt{\frac{2 \times (1.38 \times 10^{-23}) \times 300}{8.28 \times 10^{-45}}} \]

### Step 7: Calculate \( \omega \)
Calculating the numerator:
\[ 2 \times (1.38 \times 10^{-23}) \times 300 = 8.28 \times 10^{-21} \]

Now calculate \( \omega \):
\[ \omega = \sqrt{\frac{8.28 \times 10^{-21}}{8.28 \times 10^{-45}}} \]
\[ \omega = \sqrt{10^{24}} = 10^{12} \, \text{rad/s} \]

### Final Answer
Thus, the root mean square angular velocity \( \omega \) is:
\[ \omega = 10^{12} \, \text{rad/s} \]

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