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 ΔH_ABD.

Question

Quizplus · Accepted Answer

First, we need to calculate the change in internal energy (ΔU) and entropy (ΔS) for each pathway using the ideal gas law and assuming the gas is undergoing reversible processes: For pathway 1: ΔU1 = (3.00 atm)(20.0 L) - (1.00 atm)(20.0 L) = 40.0 L•atm ΔU2 = (1.00 atm)(50.0 L) - (1.00 atm)(20.0 L) = 30.0 L•atm ΔU = ΔU1 + ΔU2 = 70.0 L•atm ΔS1 = nRln(V2/V1) = (1.00 mol)(8.31 J/mol•K)ln(50.0 L/20.0 L) = 11.4 J/K ΔS2 = nRln(V2/V1) = (1.00 mol)(8.31 J/mol•K)ln(50.0 L/1.00 L) = 58.8 J/K ΔS = ΔS1 + ΔS2 = 70.2 J/K For pathway 2: ΔU1 = (3.00 atm)(20.0 L) - (3.00 atm)(50.0 L) = -90.0 L•atm ΔU2 = (1.00 atm)(50.0 L) - (3.00 atm)(50.0 L) = -100.0 L•atm ΔU = ΔU1 + ΔU2 = -190.0 L•atm ΔS1 = nRln(V2/V1) = (1.00 mol)(8.31 J/mol•K)ln(50.0 L/20.0 L) = 11.4 J/K ΔS2 = nRln(V2/V1) = (1.00 mol)(8.31 J/mol•K)ln(50.0 L/1.00 L) = 58.8 J/K ΔS = ΔS1 + ΔS2 = 70.2 J/K Next, we can calculate the change in enthalpy (ΔH) for each pathway using the equation ΔH = ΔU + Δ(PV): For pathway 1: ΔH = ΔU + Δ(PV) = 70.0 L•atm + (1.00 atm)(30.0 L) = 100.0 L•atm For pathway 2: ΔH = ΔU + Δ(PV) = -190.0 L•atm + (1.00 atm)(100.0 L) = -90.0 L•atm Finally, we can use the equation ΔH = ΔS*T to calculate the change in Gibbs free energy (ΔG) for each pathway at room temperature (298 K): For pathway 1: ΔG = ΔH - TΔS = 100.0 L•atm - (298 K)(70.2 J/K)(1 atm/101.3 kPa)(1 L/atm) = -106.5 kJ For pathway 2: ΔG = ΔH - TΔS = -90.0 L•atm - (298 K)(70.2 J/K)(1 atm/101.3 kPa)(1 L/atm) = -296.1 kJ The desired quantity is then: ΔH_ABD = ΔH - TΔS = (-90.0 L•atm) - (298 K)(70.2 J/K)(1 atm/101.3 kPa)(1 L/atm) - (100.0 L•atm) + (298 K)(70.2 J/K)(1 atm/101.3 kPa)(1 L/atm) = -25 L•atm Therefore, the best choice is B, as it is the only one that matches the calculated value.

Consider a Process Carried Out on 1