When two moles of hydrogen expands isothermally against a constant pressure of 1 atm , at `25^@C` from 15 L to 50 L , the work done (in litre atm) will be

Options

A. `-17.5`
B. `-35`
C. `-51.5`
D. `-70`

Answer

To solve the problem of calculating the work done when two moles of hydrogen expand isothermally against a constant pressure of 1 atm from 15 L to 50 L, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Number of moles (n) = 2 moles (not directly needed for work calculation)
- Initial volume (V_initial) = 15 L
- Final volume (V_final) = 50 L
- Constant pressure (P) = 1 atm

2. **Calculate the Change in Volume (ΔV)**:
\[ \Delta V = V_{final} - V_{initial} = 50 \, \text{L} - 15 \, \text{L} = 35 \, \text{L} \]

3. **Use the Formula for Work Done (W)**:
The work done during an isothermal expansion against a constant pressure is given by:
\[ W = -P \times \Delta V \]
Here, the negative sign indicates that work is done by the system (expansion).

4. **Substitute the Values into the Formula**:
\[ W = - (1 \, \text{atm}) \times (35 \, \text{L}) = -35 \, \text{L atm} \]

5. **Final Answer**:
The work done during the isothermal expansion is \(-35 \, \text{L atm}\).

### Summary:
The work done when two moles of hydrogen expand isothermally against a constant pressure of 1 atm from 15 L to 50 L is \(-35 \, \text{L atm}\).