Calculate the work done during isothermal reversible expansion expansion of one mol of an ideal gas from 10 atm to 1 atm at 300K.

Options

A. 4938 .8 J
B. 4138.8 J
C. 5744.1 J
D. 6257.2 J

Answer

To calculate the work done during the isothermal reversible expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K, we can use the formula for work done in an isothermal reversible process:

### Step-by-Step Solution:

**Step 1: Identify the formula for work done.**
The work done (W) during an isothermal reversible expansion is given by the formula:
\[ W = -nRT \ln \left( \frac{P_f}{P_i} \right) \]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin
- \( P_f \) = final pressure
- \( P_i \) = initial pressure

**Step 2: Substitute the known values.**
Given:
- \( n = 1 \) mol
- \( R = 8.314 \) J/mol·K
- \( T = 300 \) K
- \( P_f = 1 \) atm
- \( P_i = 10 \) atm

Substituting these values into the formula:
\[ W = -1 \times 8.314 \times 300 \times \ln \left( \frac{1}{10} \right) \]

**Step 3: Calculate the natural logarithm.**
Calculate \( \ln \left( \frac{1}{10} \right) \):
\[ \ln \left( \frac{1}{10} \right) = \ln(0.1) \approx -2.3026 \]

**Step 4: Substitute the logarithm value back into the equation.**
Now, substituting this back into the equation:
\[ W = -1 \times 8.314 \times 300 \times (-2.3026) \]

**Step 5: Calculate the work done.**
Now calculate:
\[ W = 8.314 \times 300 \times 2.3026 \]
Calculating this gives:
\[ W \approx 5744.1 \text{ J} \]

### Final Answer:
The work done during the isothermal reversible expansion is approximately **5744.1 J**.