Consider a process carried out on 1.00 mol of a monatomic ideal gas by the following two different pathways. The first pathway is A (3.00 atm, 20.0 L) to C (1.00 atm, 20.0 L) to D (1.00 atm, 50.0 L) ; and the second pathway is A (3.00 atm, 20.0 L) to B (3.00 atm, 50.0 L) to D (1.00 atm, 50.0 L) . In each case, the gas is taken from state A to state D.
-Calculate q_AC.

Question

Quizplus · Accepted Answer

We can use the first law of thermodynamics, which states that the change in internal energy (ΔU) equals the heat added to the system (q) minus the work done by the system (w): ΔU = q - w. For pathway 1: ΔU1 = q1 - w1 = 0 (since internal energy is a state function and does not depend on the path taken) w1 = -PΔV = -(3.00 atm)(20.0 L - 50.0 L) = 90.0 L•atm q1 = ΔU1 + w1 = 90.0 L•atm For pathway 2: w2 = -PΔV = -(3.00 atm)(50.0 L - 20.0 L) = -90.0 L•atm (since work is done by the system in expanding from B to D) ΔU2 = q2 - w2 = 0 (since the internal energy only depends on the initial and final states) q2 = ΔU2 + w2 = -90.0 L•atm Now we can calculate the desired quantity: q_AC = [q1 - q2] - [w1 - w2] = 90.0 L•atm - (-90.0 L•atm) = -60 L•atm Therefore, the answer is (C) -60 L•atm.

Consider a Process Carried Out on 1