If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands from 6.00 L to 18.00 L and a final pressure of 1.20 atm in two steps: (1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm,And (2)It is heated and allowed to expand against a constant pressure of 1.20 atm until its volume reaches 18.00 L,Which of the following is correct?

Question

Quizplus · Accepted Answer

In step (1), the volume does not change, so work done (w) is 0. In step (2), the work done by the gas is calculated using w = -PΔV, where P = 1.20 atm and ΔV = 18.00 L - 6.00 L = 12.00 L. Converting to joules, w = -1.20 atm * 12.00 L * 101.3 J/L·atm = -1.46 kJ.

If 200 Mol of an Ideal Gas at 300 K and and 3.00