A current of 2A passes through a resistance of `50Omega`, which in turn in used to heat an ideal gas constained in an insulated rigid constainer. By What amount internal energy of the gas will increase in 3 minutes?

Options

A. 18kJ
B. 36kJ
C. 72kJ
D. 9kJ

Explanation

To solve the problem, we need to calculate the increase in internal energy of the gas when a current of 2 A passes through a resistance of 50 Ω for 3 minutes.

### Step-by-Step Solution:

1. **Identify the given values:**
- Current (I) = 2 A
- Resistance (R) = 50 Ω
- Time (t) = 3 minutes = 3 × 60 seconds = 180 seconds

2. **Calculate the heat generated (Q):**
The heat generated in a resistor can be calculated using the formula:
\[ Q = I^2 R t \]
Substituting the values:
\[ Q = (2 \, \text{A})^2 \times (50 \, \Omega) \times (180 \, \text{s}) \]

3. **Perform the calculations:**
- First, calculate \(I^2\):
\[ I^2 = 2^2 = 4 \]
- Next, calculate \(Q\):
\[ Q = 4 \times 50 \times 180 \]
\[ Q = 4 \times 9000 = 36000 \, \text{J} \]

4. **Convert the heat generated into kilojoules:**
\[ Q = 36000 \, \text{J} = 36 \, \text{kJ} \]

5. **Determine the increase in internal energy (ΔU):**
Since the gas is in a rigid container, there is no work done (W = 0), and according to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Therefore:
\[ \Delta U = Q \]
Thus, the increase in internal energy of the gas is:
\[ \Delta U = 36 \, \text{kJ} \]

### Final Answer:
The internal energy of the gas will increase by **36 kJ** in 3 minutes.

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