One mole of an ideal gas expands reversibly and isothermally at 300 K from `5dm^(3)" to "50dm^(3)`. The work done by the gas for the process is equal to

Options

A. `-1.382` kcal
B. `+1.382` kcal
C. `-1381.8` kcal
D. `+1382` kcal

Answer

To solve the problem of calculating the work done by one mole of an ideal gas during a reversible isothermal expansion from 5 dm³ to 50 dm³ at a temperature of 300 K, we can follow these steps:

### Step 1: Identify the formula for work done in isothermal expansion
The work done (W) by an ideal gas during a reversible isothermal expansion can be calculated using the formula:
\[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
where:
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin
- \( V_1 \) = initial volume
- \( V_2 \) = final volume

### Step 2: Substitute the known values into the formula
Given:
- \( n = 1 \) mole
- \( R = 2 \) cal/(K·mol) (since we want the answer in calories)
- \( T = 300 \) K
- \( V_1 = 5 \) dm³
- \( V_2 = 50 \) dm³

Substituting these values into the formula gives:
\[ W = -1 \times 2 \times 300 \times \ln \left( \frac{50}{5} \right) \]

### Step 3: Calculate the ratio of volumes
Calculate the ratio \( \frac{V_2}{V_1} \):
\[ \frac{V_2}{V_1} = \frac{50}{5} = 10 \]

### Step 4: Calculate the natural logarithm
Now, we need to calculate \( \ln(10) \). Using the approximation:
\[ \ln(10) \approx 2.303 \]

### Step 5: Substitute back to find work done
Now substituting back into the work formula:
\[ W = -1 \times 2 \times 300 \times 2.303 \]
\[ W = -600 \times 2.303 \]
\[ W \approx -1381.8 \text{ cal} \]

### Step 6: Convert to kilocalories
To convert calories to kilocalories, we divide by 1000:
\[ W \approx -1.3818 \text{ kcal} \]

### Final Answer
The work done by the gas during the isothermal expansion is approximately:
\[ W \approx -1.38 \text{ kcal} \]