Answer:
a.
ΔF = w = 2.94 kJ
ΔSuniv = 0
ΔG = 0
b.
ΔS = 0.1000×8.315×1.099 = 0.913 J/mol K
ΔQ =0
ΔH = 0
Explanation:
a. You have to find the ΔG, ΔF, who are two forms of free energy
G: Gibbs free energy
F: Helmholtz free energy
-G: Gibbs free energy:
For solve these, you have the following equation:
ΔG = ΔH – T ΔS with T constant (Eq 1)
Where:
ΔG = change in Gibbs free energy
ΔH= change in enthalpy
T = temperature
ΔS = change in entropy
This process is irreversible and isothermic, it last means that temperature doesn’t change.
For that reason:
ΔS = q/T with p constant. (Eq 2)
Where:
q = heat
And, with p constant, it just making P-V work, so:
ΔH = qp =q (Eq 3)
where:
qp = heat at constant pressure.
-F: Helmholtz free energy
To find ΔF, you have to use the following equation:
ΔF = ΔU – T ΔS With T constant, (Eq 4)
Where:
ΔF = change in Helmholtz free energy
ΔU= change in internal energy
T = temperature
ΔS = change in entropy
And also, you have to use the equation for internal energy:
ΔU = q + w (Eq 5)
The complete exercise is on the document attached.
b. For this problem we have to establish two states, A and B, based on the data given from the problem:
State A:
V1 = 2 dm³
T1 = 300K
State B:
V2 = 6 dm³
T2 = ?
Due to the adiabatic properties of the process, this expansion makes that change on heat “q” equals to 0:
Δq= 0
SO we have to ask ourselves what is the value of the change in entropy. But we don’t know if the process is reversible or not. Also, we don’t know if the process is static or not, and the volume could be hard to define.
The complete exercise is on the document attached.
