A gas expands in volume from 2L to 5L against a pressure of 1 atm at constant temperature. The work done by the gas will be

Options

A. 3 J
B. `-303.9` J
C. `-303.9L*` atm
D. `303.9 L*`atm

Answer

To solve the problem of calculating the work done by a gas expanding from 2L to 5L against a pressure of 1 atm at constant temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume (V1) = 2 L
- Final volume (V2) = 5 L
- External pressure (P_external) = 1 atm

2. **Calculate the Change in Volume (ΔV):**
\[ \Delta V = V2 - V1 = 5 \, \text{L} - 2 \, \text{L} = 3 \, \text{L} \]

3. **Use the Formula for Work Done (W):**
The work done by the gas during expansion against an external pressure is given by:
\[ W = -P_{\text{external}} \Delta V \]
Here, the negative sign indicates that work is done by the system (the gas) on the surroundings.

4. **Substitute the Values into the Formula:**
\[ W = - (1 \, \text{atm}) \times (3 \, \text{L}) \]
\[ W = -3 \, \text{L atm} \]

5. **Convert Work Done from L atm to Joules:**
We know that:
\[ 1 \, \text{L atm} = 101.3 \, \text{J} \]
Therefore,
\[ W = -3 \, \text{L atm} \times 101.3 \, \text{J/L atm} = -303.9 \, \text{J} \]

6. **Final Result:**
The work done by the gas during the expansion is:
\[ W = -303.9 \, \text{J} \]