A thermally insulated rigid container contain an ideal gas heated by a filament of resistance 30Ω through a current of 2A for 3 min, then change in internal energy is

Options

A. 20.6KJ
B. 21.4KJ
C. 20KJ
D. 21.6KJ

Answer

To find the change in internal energy of the ideal gas in a thermally insulated rigid container, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a thermally insulated rigid container with an ideal gas inside. This means no heat can enter or leave the system (Q = 0).
- The container is rigid, so the volume does not change (ΔV = 0), which implies that no work is done (W = 0).

2. **Apply the First Law of Thermodynamics**:
- The first law of thermodynamics states:
\[ \Delta U = Q - W \]
- Since the container is insulated, \( Q = 0 \) (no heat exchange).
- Since the volume is constant, \( W = 0 \) (no work done).
- Therefore, we have:
\[ \Delta U = 0 - 0 = 0 \]

3. **Calculate the Heat Produced by the Resistor**:
- The heat produced by the resistor can be calculated using the formula:
\[ Q = I^2 R t \]
- Where:
- \( I = 2 \, \text{A} \) (current)
- \( R = 30 \, \Omega \) (resistance)
- \( t = 3 \, \text{min} = 3 \times 60 = 180 \, \text{s} \) (time in seconds)

4. **Substituting Values**:
- Substitute the values into the heat formula:
\[ Q = (2)^2 \times 30 \times 180 \]
\[ Q = 4 \times 30 \times 180 \]
\[ Q = 120 \times 180 = 21600 \, \text{J} \]

5. **Convert to Kilojoules**:
- To express the heat in kilojoules:
\[ Q = 21600 \, \text{J} = 21.6 \, \text{kJ} \]

6. **Final Result**:
- Since all the heat produced by the resistor goes into changing the internal energy of the gas, we have:
\[ \Delta U = Q = 21.6 \, \text{kJ} \]

### Conclusion:
The change in internal energy of the ideal gas is \( \Delta U = 21.6 \, \text{kJ} \).