Find the work done when 2 moles of hydrogen expand isothermally from 15 to 50 litres against a constant pressure of 1atm at `25^(@)C` .

Options

A. `847.0cal`
B. `847Kcal`
C. `84.7cal`
D. `84.7Kcal`

Explanation

To find the work done when 2 moles of hydrogen expand isothermally from 15 to 50 liters against a constant pressure of 1 atm at 25°C, we can follow these steps:

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

### Step 2: Calculate the change in volume
We need to calculate the change in volume (\( \Delta V \)):
\[ \Delta V = V_2 - V_1 \]
Given:
- \( V_1 = 15 \) liters
- \( V_2 = 50 \) liters

So,
\[ \Delta V = 50 \, \text{liters} - 15 \, \text{liters} = 35 \, \text{liters} \]

### Step 3: Substitute the values into the work done formula
Now, substitute the values into the work done formula:
- \( P = 1 \, \text{atm} \)
- \( \Delta V = 35 \, \text{liters} \)

Thus,
\[ W = -1 \, \text{atm} \times 35 \, \text{liters} = -35 \, \text{liter atm} \]

### Step 4: Convert liter atm to joules
Next, we need to convert the work done from liter atm to joules. The conversion factor is:
\[ 1 \, \text{liter atm} = 101.3 \, \text{joules} \]

So,
\[ W = -35 \, \text{liter atm} \times 101.3 \, \text{joules/liter atm} = -3545.5 \, \text{joules} \]

### Step 5: Convert joules to calories
Now, we convert joules to calories. The conversion factor is:
\[ 1 \, \text{calorie} = 4.18 \, \text{joules} \]
Thus,
\[ 1 \, \text{joule} = \frac{1}{4.18} \, \text{calories} \]

Now, convert the work done:
\[ W = \frac{-3545.5 \, \text{joules}}{4.18 \, \text{joules/calorie}} = -848.20 \, \text{calories} \]

### Step 6: Present the final answer
The work done during the isothermal expansion of hydrogen is approximately:
\[ W \approx -848.20 \, \text{calories} \]
or
\[ W \approx -0.848 \, \text{kilo calories} \]