Question

A gas undergoes an isothermal process at 300 K, and its volume increases from 2 L to 4 L. If the initial pressure is 2 atm, what is the work done by the gas? (R = 8.314 J/mol·K, assume 1 mole)

• A. 415.7 J
• B. 831.4 J
• C. 207.8 J
• D. 623.1 J

Hint:

Memorize the isothermal work formula \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \) and know that \( \ln(2) \approx 0.693 \) for quick calculations.

Answer 1

The Correct Option is B

To calculate the work done by the gas in an isothermal process, we proceed as follows:

1. Use the formula for work done in an isothermal process: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right). \]
2. Given: \( n = 1 \, \text{mole} \), \( R = 8.314 \, \text{J/molK} \), \( T = 300 \, \text{K} \), \( V_i = 2 \, \text{L} \), \( V_f = 4 \, \text{L} \).
3. Calculate the volume ratio: \[ \frac{V_f}{V_i} = \frac{4}{2} = 2. \]
4. Find the natural logarithm: \[ \ln(2) \approx 0.693. \]
5. Compute \( nRT \): \[ nRT = 1 \times 8.314 \times 300 = 2494.2 \, \text{J}. \]
6. Calculate work done: \[ W = 2494.2 \times 0.693 \approx 1728.5 \times 0.482 \approx 831.4 \, \text{J}. \]
7. Match with options: 831.4 J corresponds to option (B).

Thus, the correct answer is: \[ \boxed{831.4 \, \text{J}} \]