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. (a) `0`
B. (b)`5/2` nRT
C. (c)`3/2` nRT
D. (d)`1/2` nRT

Answer

To solve the problem, we need to analyze the kinetic energy of the gas before and after the conversion from diatomic to monatomic gas.

### Step 1: Calculate the initial kinetic energy of the diatomic gas.
The kinetic energy (KE) of an ideal gas is given by the formula:
\[ KE = \frac{3}{2} nRT \]
For diatomic gas, we have:
\[ KE_{\text{initial}} = \frac{5}{2} nRT \]
This is because a diatomic gas has 5 degrees of freedom (3 translational + 2 rotational).

### Step 2: Calculate the final kinetic energy after conversion to monatomic gas.
When n moles of the diatomic gas are converted to monatomic gas, the number of moles of the gas becomes:
\[ n_{\text{final}} = 2n \]
The kinetic energy of a monatomic gas is given by:
\[ KE = \frac{3}{2} nRT \]
Thus, the final kinetic energy for the 2n moles of monatomic gas is:
\[ KE_{\text{final}} = \frac{3}{2} (2n) RT = 3nRT \]

### Step 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:
\[ \Delta KE = 3nRT - \frac{5}{2} nRT \]
To simplify this, we can express \(\frac{5}{2} nRT\) with a common denominator:
\[ \Delta KE = 3nRT - \frac{5}{2} nRT = \frac{6}{2} nRT - \frac{5}{2} nRT = \frac{1}{2} nRT \]

### Conclusion
Thus, the change in the total kinetic energy of the gas is:
\[ \Delta KE = \frac{1}{2} nRT \]