n moles of diatomic gas in a cylinder is at a temperature T . Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monatomic gas . The change in the total kinetic energy of the gas is

Options

A. 0
B. `5/2 nRT`
C. `3/2 nRT`
D. `1/2 nRT`

Answer

To solve the problem, we need to calculate the change in the total kinetic energy of the gas when n moles of diatomic gas are converted into monatomic gas at constant temperature.

### Step-by-step Solution:

1. **Identify the Initial Kinetic Energy of Diatomic Gas:**
- For n moles of diatomic gas, the total kinetic energy (KE_initial) can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{5}{2} nRT \]
- This is because diatomic gases have 5 degrees of freedom (3 translational and 2 rotational).

2. **Identify the Final Kinetic Energy of Monatomic Gas:**
- When n moles of diatomic gas are converted into monatomic gas, the number of moles of monatomic gas becomes 2n (since each diatomic molecule splits into two monatomic atoms).
- The total kinetic energy (KE_final) of 2n moles of monatomic gas is given by:
\[ KE_{\text{final}} = \frac{3}{2} (2n)RT = 3nRT \]
- Monatomic gases only have 3 translational degrees of freedom.

3. **Calculate the Change in Kinetic Energy:**
- The change in kinetic energy (ΔKE) is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
- Substituting the values we calculated:
\[ \Delta KE = 3nRT - \frac{5}{2} nRT \]

4. **Simplify the Expression:**
- To simplify, we can convert \(3nRT\) into a fraction with a common denominator:
\[ 3nRT = \frac{6}{2} nRT \]
- Now, substituting back into the equation:
\[ \Delta KE = \frac{6}{2} nRT - \frac{5}{2} nRT = \frac{1}{2} nRT \]

5. **Final Answer:**
- Therefore, the change in the total kinetic energy of the gas is:
\[ \Delta KE = \frac{1}{2} nRT \]