Question

At 61 K, one mole of an ideal gas of 1.0 L volume expands isothermally and reversibly to a final volume of 10.0 L. What is the work done in the expansion?

• A. \( -11.52 \) L atm
• B. \( -23.04 \) L atm
• C. \( -46.08 \) L atm
• D. \( -5.76 \) L atm

Hint:

In isothermal reversible expansion, work is calculated using \( W = -nRT \ln (V_f / V_i) \).
- Always use natural logarithm (\(\ln\)) in these calculations.

Answer 1

The Correct Option is A

The work done in an isothermal reversible expansion of an ideal gas is given by: \[ W = - nRT \ln \left( \frac{V_f}{V_i} \right) \] where: - \( n = 1 \) mole (given), - \( R = 0.0821 \) L atm K\(^{-1}\) mol\(^{-1}\), - \( T = 61 \) K, - \( V_i = 1.0 \) L, \( V_f = 10.0 \) L. \[ W = - (1) (0.0821) (61) \ln \left( \frac{10.0}{1.0} \right) \] \[ W = - (5.0081) \ln (10) \] Since \( \ln (10) = 2.302 \), \[ W = - (5.0081 \times 2.302) \] \[ W = -11.52 \text{ L atm} \] Thus, the correct answer is \(\boxed{-11.52 \text{ L atm}}\).