Question

During an isothermal expansion, 2 moles of an ideal gas expand from 1 L to 3 L at a constant temperature of 300 K. Using R=8.314 J/mol.K calculate the work done by the gas.

• A. 477.8 J
• B. 5478.4 J
• C. 95(4)6 J
• D. -95(4)6 J

Hint:

Work in Isothermal Expansion. For reversible isothermal expansion of n moles of ideal gas from \(V_i\) to \(V_f\) at temperature T: \(W_{by = nRT \ln(V_f/V_i)\). Work done is positive for expansion.

Answer 1

The Correct Option is B

The formula for work done during an isothermal expansion of an ideal gas is given by:

\(\text{Work} = nRT \ln \left(\frac{V_f}{V_i}\right)\)

Where:

• n = number of moles = 2 moles
• R = gas constant = 8.314 J/mol·K
• T = temperature = 300 K
• V_f = final volume = 3 L
• V_i = initial volume = 1 L

Now, plug the values into the formula:

\(\text{Work} = 2 \times 8.314 \times 300 \times \ln \left(\frac{3}{1}\right)\)

Calculating the natural logarithm:

\(\ln \left(\frac{3}{1}\right) = \ln(3) \approx 1.0986\)

Substitute into the equation:

\(\text{Work} = 2 \times 8.314 \times 300 \times 1.0986 \approx 5478.4 \text{ J}\)

Thus, the work done by the gas is approximately 5478.4 J.