A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 `dm^(3)` to a volume of 20 `dm^(3)`. It absorbs 800 J of thermal energy from its surroundings. The `Delta`U is

Options

A. `-312` J
B. `+123` J
C. `+312` J
D. `-213` J

Answer

To solve the problem, we will follow the steps outlined in the video transcript and apply the first law of thermodynamics.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial Volume (V1) = 10 dm³
- Final Volume (V2) = 20 dm³
- External Pressure (P) = 1 atm
- Heat absorbed (Q) = +800 J (since it is absorbed)

2. **Calculate the Change in Volume (ΔV)**:
\[ \Delta V = V2 - V1 = 20 \, \text{dm}^3 - 10 \, \text{dm}^3 = 10 \, \text{dm}^3 \]

3. **Calculate Work Done (W)**:
- Work done by the gas during expansion against a constant external pressure is given by:
\[ W = -P \Delta V \]
- Convert pressure from atm to joules:
- 1 atm = 101.325 J/L
- Thus,
\[ W = -1 \, \text{atm} \times 10 \, \text{dm}^3 = -1 \times 10 \, \text{L} \times 101.325 \, \text{J/L} = -1013.25 \, \text{J} \]

4. **Apply the First Law of Thermodynamics**:
- The first law states:
\[ Q = \Delta U + W \]
- Rearranging gives:
\[ \Delta U = Q - W \]
- Substitute the values of Q and W:
\[ \Delta U = 800 \, \text{J} - (-1013.25 \, \text{J}) = 800 \, \text{J} + 1013.25 \, \text{J} \]
\[ \Delta U = 800 \, \text{J} + 1013.25 \, \text{J} = 1813.25 \, \text{J} \]

5. **Final Calculation**:
- Since the work done is negative, we should have:
\[ \Delta U = 800 \, \text{J} - 1013.25 \, \text{J} = -213.25 \, \text{J} \]

### Conclusion:
The change in internal energy (ΔU) is approximately:
\[ \Delta U \approx -213 \, \text{J} \]