N moles of an ideal diatomic gas are in a cylinder at temperature T. suppose on supplying heat to the gas, its temperature remain constant but n moles get dissociated into atoms. Heat supplied to the gas is

Options

A. `zero`
B. `(1)/(2) nRT`
C. `(3)/(2) nRT`
D. `(3)/(2)(N - n) RT`

Explanation

To solve the problem, we need to understand the dissociation of a diatomic gas into its atomic components and how this affects the heat supplied to the system.

### Step-by-Step Solution:

1. **Understanding the Initial State**:
- We have \( N \) moles of a diatomic gas at temperature \( T \).
- The internal energy \( U_1 \) of the diatomic gas can be calculated using the formula:
\[ U_1 = \frac{f}{2} NRT \]
where \( f \) (the degrees of freedom for a diatomic gas) is 5. Thus,
\[ U_1 = \frac{5}{2} NRT \]

2. **Dissociation of the Gas**:
- When heat is supplied, \( n \) moles of the diatomic gas dissociate into \( n \) moles of monoatomic gas.
- After dissociation, the remaining moles of diatomic gas are \( N - n \) and the moles of monoatomic gas are \( n \).

3. **Calculating the Final Internal Energy**:
- The internal energy \( U_2 \) after dissociation can be calculated as:
\[ U_2 = \text{(Internal energy of remaining diatomic gas)} + \text{(Internal energy of monoatomic gas)} \]
- The internal energy of the remaining \( N - n \) moles of diatomic gas is:
\[ U_{diatomic} = \frac{5}{2} (N - n) RT \]
- The internal energy of \( n \) moles of monoatomic gas is:
\[ U_{monoatomic} = \frac{3}{2} n RT \]
- Therefore, the total internal energy after dissociation is:
\[ U_2 = \frac{5}{2} (N - n) RT + \frac{3}{2} n RT \]

4. **Simplifying the Expression for \( U_2 \)**:
- Expanding \( U_2 \):
\[ U_2 = \frac{5}{2} NRT - \frac{5}{2} nRT + \frac{3}{2} nRT \]
- Combining the terms:
\[ U_2 = \frac{5}{2} NRT - \frac{2}{2} nRT = \frac{5}{2} NRT - nRT = \frac{5}{2} NRT - \frac{2}{2} nRT \]

5. **Calculating the Change in Internal Energy**:
- The change in internal energy \( \Delta U \) is given by:
\[ \Delta U = U_2 - U_1 \]
- Substituting the values:
\[ \Delta U = \left( \frac{5}{2} NRT - nRT \right) - \frac{5}{2} NRT \]
- This simplifies to:
\[ \Delta U = -nRT \]

6. **Relating Heat Supplied to Internal Energy**:
- Since the temperature is constant, the work done \( W = 0 \) (because \( \Delta V = 0 \)).
- Therefore, the heat supplied \( Q \) is equal to the change in internal energy:
\[ Q = \Delta U \]
- Thus, the heat supplied is:
\[ Q = -nRT \]

### Final Answer:
The heat supplied to the gas is \( Q = -nRT \).