A rigid container with thermally insulated walls contains a gas and a coil of resistance `50 Omega`, carrying a current of 1A. The change in internal energy of the gas after 2 min will be

Options

A. 6kJ
B. 10kJ
C. 3kJ
D. 12kJ

Explanation

To solve the problem, we need to determine the change in internal energy of the gas in a thermally insulated rigid container after a certain period of time, given the resistance of a coil and the current flowing through it.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = 50 Ω
- Current (I) = 1 A
- Time (T) = 2 minutes = 2 × 60 seconds = 120 seconds

2. **Understand the First Law of Thermodynamics:**
The first law of thermodynamics states that:
\[ \Delta U = Q - W \]
Where:
- \(\Delta U\) = Change in internal energy
- \(Q\) = Heat added to the system
- \(W\) = Work done by the system

3. **Analyze the Work Done (W):**
Since the container is rigid, the volume of the gas does not change. Therefore, the work done (W) is zero:
\[ W = P \Delta V = 0 \quad (\text{because } \Delta V = 0) \]

4. **Relate Heat to Change in Internal Energy:**
Since \(W = 0\), the first law simplifies to:
\[ \Delta U = Q \]
This means that all the heat generated will contribute to the change in internal energy.

5. **Calculate the Heat Generated (Q):**
The heat generated by the coil can be calculated using the formula:
\[ Q = I^2 R T \]
Substituting the known values:
\[ Q = (1 \, \text{A})^2 \times 50 \, \Omega \times 120 \, \text{s} \]
\[ Q = 1 \times 50 \times 120 = 6000 \, \text{J} \]

6. **Determine the Change in Internal Energy:**
Since \(Q = \Delta U\):
\[ \Delta U = 6000 \, \text{J} \]

7. **Convert Joules to Kilojoules:**
To express the change in internal energy in kilojoules:
\[ \Delta U = \frac{6000 \, \text{J}}{1000} = 6 \, \text{kJ} \]

### Final Answer:
The change in internal energy of the gas after 2 minutes is **6 kJ**.