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 $\Delta$ H_ABD.

Question

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

\Delta

H_ABD.

Quizplus · Accepted Answer

We can use the equation: ΔH = ΔU + Δ(PV) to calculate ΔH for each pathway. Then, we can use the equation: ΔH = Q_p at constant pressure to calculate ΔH for each pathway. Finally, we can use the equation: ΔS = Q/T (where T is in Kelvin) to calculate ΔS for each pathway. 
Here are the calculations:

Pathway 1: 
ΔH = ΔU + Δ(PV) = 0 + (1.00 atm - 3.00 atm) x (20.0 L) = -40.0 L•atm 
ΔH = Q_p = -PΔV = -(1.00 atm) x (-30.0 L) = 30.0 L•atm 
ΔS = Q/T = 30.0 L•atm / (293 K) = 0.102 L•atm/K

Pathway 2: 
ΔH = ΔU + Δ(PV) = 0 + 0 = 0 
ΔH = Q_p = -PΔV = -(3.00 atm) x (30.0 L) = -90.0 L•atm 
ΔS = Q/T = 90.0 L•atm / (293 K) = 0.308 L•atm/K

Now, we can use the equation: ΔG = ΔH - TΔS to find ΔG for each pathway:

Pathway 1: 
ΔG = -40.0 L•atm - (293 K) x (0.102 L•atm/K) = -67.7 L•atm

Pathway 2: 
ΔG = 0 - (293 K) x (0.308 L•atm/K) = -90.2 L•atm

Therefore, ΔG is more negative for pathway 1, meaning it is the spontaneous pathway and the best choice.

Consider a Process Carried Out on 1 Δ\DeltaΔ HABD

Consider a Process Carried Out on 1 $\Delta$ H_ABD