A gas is at 1 atm pressure with a volume `800 cm^(3)`. When `100 J` of heat is supplied to the gas, it expands to `1L` at constant pressure. The change in its internal energy is

Options

A. `80 J`
B. `- 80 J`
C. `20 J`
D. `- 20 J`

Explanation

To solve the problem, we will use the first law of thermodynamics and the formula for work done during an isobaric (constant pressure) process. Here are the steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial pressure (P) = 1 atm = \(1 \times 10^5\) Pa
- Initial volume (V1) = 800 cm³ = \(800 \times 10^{-6}\) m³
- Final volume (V2) = 1 L = \(1000 \times 10^{-6}\) m³
- Heat supplied (Q) = 100 J

2. **Calculate the Change in Volume (ΔV):**
\[ \Delta V = V2 - V1 = (1000 \times 10^{-6}) - (800 \times 10^{-6}) = 200 \times 10^{-6} \, \text{m}^3 \]

3. **Calculate the Work Done (W) at Constant Pressure:**
The work done during expansion at constant pressure is given by:
\[ W = P \times \Delta V \]
Substituting the values:
\[ W = (1 \times 10^5 \, \text{Pa}) \times (200 \times 10^{-6} \, \text{m}^3) = 20 \, \text{J} \]

4. **Apply the First Law of Thermodynamics:**
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Substituting the known values:
\[ \Delta U = 100 \, \text{J} - 20 \, \text{J} = 80 \, \text{J} \]

5. **Conclusion:**
The change in internal energy (ΔU) is:
\[ \Delta U = 80 \, \text{J} \]

### Final Answer:
The change in internal energy is \(80 \, \text{J}\).

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