What will be the work done when one mole of a gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at `25^(@)C`?

Options

A. `-3542 cal`
B. `-843.3cal`
C. `-718cal`
D. `-60.23cal`

Answer

To solve the problem of calculating the work done when one mole of a gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at 25°C, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
- The temperature is given as 25°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Thus,
\[ T = 25 + 273.15 = 298.15 \, K \approx 298 \, K \]

2. **Identify Given Values:**
- Number of moles (n) = 1 mole
- Initial volume (V1) = 15 L
- Final volume (V2) = 50 L
- Pressure (P) = 1 atm
- Gas constant (R) = 0.0821 L·atm/(K·mol)

3. **Use the Work Done Formula for Isothermal Expansion:**
- The work done (W) during isothermal expansion can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V2}{V1}\right) \]
- Alternatively, using logarithm base 10:
\[ W = -2.303 \cdot nRT \log\left(\frac{V2}{V1}\right) \]

4. **Calculate the Ratio of Volumes:**
- Calculate \( \frac{V2}{V1} \):
\[ \frac{V2}{V1} = \frac{50}{15} \approx 3.33 \]

5. **Calculate the Natural Logarithm:**
- We can use the logarithm base 10 for our calculations:
\[ \log(3.33) \approx 0.522 \]

6. **Substitute Values into the Work Done Formula:**
- Substitute the known values into the work done formula:
\[ W = -2.303 \cdot (1) \cdot (0.0821) \cdot (298) \cdot (0.522) \]

7. **Calculate the Work Done:**
- First, calculate the product:
\[ W = -2.303 \cdot 0.0821 \cdot 298 \cdot 0.522 \]
- Carrying out the multiplication:
\[ W \approx -2.303 \cdot 0.0821 \cdot 298 \cdot 0.522 \approx -29.41 \, \text{L·atm} \]

8. **Convert Work Done to Calories:**
- Since 1 L·atm = 24.21 calories, we convert:
\[ W \text{ (in calories)} = -29.41 \cdot 24.21 \approx -712 \, \text{calories} \]

### Final Answer:
The work done when one mole of gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at 25°C is approximately:
\[ W \approx -712 \, \text{calories} \]