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 expands isothermally against a constant pressure of 1 atm from 15 L to 50 L, we can follow these steps:

### Step 1: Identify the formula for work done
The work done (W) during an isothermal expansion against a constant pressure can be calculated using the formula:
\[ W = -P \Delta V \]
where:
- \( W \) is the work done,
- \( P \) is the constant pressure,
- \( \Delta V \) is the change in volume.

### Step 2: Calculate the change in volume (\( \Delta V \))
The change in volume (\( \Delta V \)) can be calculated as:
\[ \Delta V = V_2 - V_1 \]
where:
- \( V_2 \) is the final volume (50 L),
- \( V_1 \) is the initial volume (15 L).

Substituting the values:
\[ \Delta V = 50 \, \text{L} - 15 \, \text{L} = 35 \, \text{L} \]

### Step 3: Substitute the values into the work done formula
Now, substituting \( P = 1 \, \text{atm} \) and \( \Delta V = 35 \, \text{L} \) into the work done formula:
\[ W = - (1 \, \text{atm}) \times (35 \, \text{L}) \]
\[ W = -35 \, \text{L atm} \]

### Step 4: Final answer
The work done during the expansion is:
\[ W = -35 \, \text{L atm} \]

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

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