What is the work done when 1 mole of a gas expands isothermally from 25 L to 250 L at a constant pressure of 1 atm and a temperature of 300 K ?

Options

A. (a) `-3542 J`
B. (b) `-5744J`
C. (c) `-2657J`
D. (d) `-4890J`

Answer

To solve the problem of calculating the work done when 1 mole of a gas expands isothermally from 25 L to 250 L at a constant pressure of 1 atm and a temperature of 300 K, we can follow these steps:

### Step 1: Identify the Given Data
- Number of moles (n) = 1 mole
- Initial volume (V_initial) = 25 L
- Final volume (V_final) = 250 L
- Constant pressure (P) = 1 atm
- Temperature (T) = 300 K

### Step 2: Use the Formula for Work Done in Isothermal Expansion
For isothermal expansion of an ideal gas, the work done (W) can be calculated using the formula:
\[ W = -2.303 \, nRT \log\left(\frac{V_{\text{final}}}{V_{\text{initial}}}\right) \]

### Step 3: Convert the Gas Constant to Appropriate Units
Since the pressure is given in atm and we want the work in joules, we will use the gas constant \( R = 8.314 \, \text{J/(mol K)} \).

### Step 4: Calculate the Ratio of Volumes
Calculate the ratio of the final volume to the initial volume:
\[ \frac{V_{\text{final}}}{V_{\text{initial}}} = \frac{250 \, \text{L}}{25 \, \text{L}} = 10 \]

### Step 5: Calculate the Logarithm
Now calculate the logarithm:
\[ \log(10) = 1 \]

### Step 6: Substitute Values into the Work Formula
Now substitute all the values into the work formula:
\[ W = -2.303 \times 1 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 300 \, \text{K} \times 1 \]

### Step 7: Perform the Calculation
Calculate the work done:
\[ W = -2.303 \times 1 \times 8.314 \times 300 \]
\[ W = -5744.19 \, \text{J} \]

### Step 8: Final Answer
Thus, the work done when 1 mole of gas expands isothermally from 25 L to 250 L is approximately:
\[ W \approx -5744 \, \text{J} \]