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

Explanation

To solve the problem, we need to calculate the heat supplied to the gas when n moles of an ideal diatomic gas dissociate into atoms while maintaining a constant temperature. Here’s a step-by-step solution:

### Step 1: Understand the System
We have N moles of an ideal diatomic gas at temperature T. When we supply heat, the temperature remains constant, and n moles of the gas dissociate into monatomic atoms.

### Step 2: Identify Work Done
Since the process occurs at constant volume (the gas is in a cylinder), the work done (W) on or by the gas is zero. This is because work done is defined as:
\[ W = P \Delta V \]
where \( \Delta V = 0 \) (no change in volume), thus:
\[ W = 0 \]

### Step 3: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Since \( W = 0 \), we have:
\[ Q = \Delta U \]

### Step 4: Calculate Change in Internal Energy (\( \Delta U \))
The change in internal energy for an ideal gas is given by:
\[ \Delta U = U_f - U_i \]
where \( U_f \) is the final internal energy and \( U_i \) is the initial internal energy.

### Step 5: Calculate Initial Internal Energy (\( U_i \))
For N moles of a diatomic gas, the initial internal energy is:
\[ U_i = N C_v T \]
For a diatomic gas, \( C_v = \frac{5}{2} R \), so:
\[ U_i = N \left(\frac{5}{2} R\right) T = \frac{5}{2} NRT \]

### Step 6: Calculate Final Internal Energy (\( U_f \))
After n moles dissociate, we have:
- \( (N - n) \) moles of diatomic gas remaining.
- \( 2n \) moles of monatomic gas formed (since each diatomic molecule produces 2 atoms).

The internal energy for the remaining diatomic gas is:
\[ U_{diatomic} = (N - n) \left(\frac{5}{2} R\right) T \]

For the monatomic gas, \( C_v = \frac{3}{2} R \), so:
\[ U_{monatomic} = 2n \left(\frac{3}{2} R\right) T = 3nRT \]

Thus, the total final internal energy is:
\[ U_f = (N - n) \left(\frac{5}{2} R\right) T + 3nRT \]

### Step 7: Combine and Simplify
Now, substituting \( U_f \):
\[ U_f = \left(\frac{5}{2} (N - n) R T + 3nRT\right) \]
\[ = \frac{5}{2} NRT - \frac{5}{2} nRT + 3nRT \]
\[ = \frac{5}{2} NRT + \left(3 - \frac{5}{2}\right)nRT \]
\[ = \frac{5}{2} NRT + \frac{1}{2} nRT \]

### Step 8: Calculate Change in Internal Energy
Now, we can find \( \Delta U \):
\[ \Delta U = U_f - U_i \]
\[ = \left(\frac{5}{2} NRT + \frac{1}{2} nRT\right) - \frac{5}{2} NRT \]
\[ = \frac{1}{2} nRT \]

### Step 9: Final Result for Heat Supplied
Since \( Q = \Delta U \):
\[ Q = \frac{1}{2} nRT \]

Thus, the heat supplied to the gas is:
\[ Q = \frac{1}{2} nRT \]

### Conclusion
The correct answer is:
\[ Q = \frac{1}{2} nRT \]