Three moles ( $3.00 mol$ ) of an ideal gas at $300 K$ expand reversibly and isothermally from $5.00 L$ to $20.0 L$ . Calculate the magnitude of the work done by the gas in $kJ$ (two decimals). Use $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Numerical answer type, answer in kJ, tolerance 0.01.

Answer: 10.3731 kJ

For reversible isothermal expansion

∣ w ∣ = n RT ln (V_{2} / V_{1})

. Here

∣ w ∣ = 3 \times 8.314 \times 300 \times ln 4 = 3 \times 8.314 \times 300 \times 1.3863 J = 10.373 kJ