A container of ideal gas at 0°C and 1 atm undergoes an isothermal expansion, and its entropy changes by 3.7 J/K. How much work does it do? A) zero B) 1.0 × 10³ J C) −1.0 × 10³ J D) None of these choices are correct.

Question

Quizplus · Accepted Answer

For an isothermal process involving an ideal gas, the change in entropy (ΔS) can be related to the work done (W) by the equation ΔS = W/T, where T is the temperature in Kelvin. Given ΔS = 3.7 J/K and T = 273 K (0°C in Kelvin), we can solve for W: W = ΔS * T = 3.7 J/K * 273 K = 1010.1 J, which is approximately 1.0 × 10^3 J.

A Container of Ideal Gas at 0°C and 1 Atm