Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1st Law of Thermodynamics 4) 2nd Law of Thermodynamics 5) Gibbs Free Energy 6) Phase Diagrams and REAL Phenomena 7) Non-Electrolyte Solutions & Simple Mixtures 8) Chemical Equilibrium 9) Kinetics
Section 5.1. The Use Of Partial Differentials Goals 1) To understand the use of Partial Differentials in developing thermodynamic formulae.
Use Of Partial Differentials 8 Variables Of State:
U, H, S, G, A, P, V and T
Express V as a function of P and T: V = V(P,T) =
nRT P
If an ideal gas
How does V change with T (at constant P) for an ideal gas :
nR V = P T P
How does V change with P (at constant T) for an ideal gas :
nRT V = P2 P T It is generally true that if: V = V(P,T) then the total differential of V is:
V V dV = dP + dT P T T P nRT nR = dP + dT 2 P P
dV is an exact differential. For exact differentials, the order of differentiation doesn’t matter !!
V V = T P P T T P P T For an ideal gas:
nR V = P T P Therefore:
nR V = P T P2 P T
Likewise:
nRT V = P2 P T Therefore:
nR V = T P P2 T P Same Result !!!
Therefore:
V V = T P P T T P P T
For exact differentials, the order of differentiation doesn’t matter !!
Example: (a)
Show that:
G = V P T
(b)
V S = – T P P T
Solution: (a)
Express G as a function of T and P: G = H – TS = U + PV – TS dG = dU + d (P V) – d (T S)
= dU + P dV + V dP – T dS – S dT G is a State Function. Thus, if we derive: G = G(T,P)
for one path, it must
be true for ANY path. Therefore, choose a reversible path with P–V work only.
dU = dqrev + dwrev = dqrev – P dV But: Therefore:
dqrev = T dS dU = T dS – P dV
Substitute into the expression for dG above: dG = T dS – P dV + P dV + V dP – T dS – S dT = V dP – S dT
G G = dP + dT P T T P
Therefore we may conclude that:
G = V P T
G = –S T P
AND
Now, for an isothermal process dT = 0 Therefore we may write:
dG = V dP dG = V dP Integrate both sides of this equation: G2
P2
G1
P1
dG = V dP P2
ΔG = G2 – G1 = V dP P1
Likewise, for an isobaric process dP = 0 Therefore we may write:
dG = –S dT dG = – S dT Integrate both sides of this equation: G2
T2
G1
T1
dG = – S dT
T2
ΔG = G2 – G1 = – S dT T1
For biological processes at body temperature (37oC = 98.6oF) ΔS(37oC) ≈ ΔS(25oC) ≈ constant And thus: ΔG(37oC) – ΔG(25oC) ≈ – (T – 298) (ΔS(25oC))
(b)
dG is an exact differential Therefore:
G G = T P P T T P P T
From above we have:
G = V P T
AND
G = –S T P
Therefore:
V S = – T P P T Q.E.D.
(quod erat demonstrandum)
In the special case of an ideal gas:
nR V = P T P Therefore, we may now write:
S nR = – P T P
Now, for an isothermal process dT = 0 Therefore we may write:
dS nR = – dP P nR dS = – dP P Integrate both sides of this equation (at constant T) :
nR S dS = – P dP = – nR ln P
S2
P2
1
1
P2 P1
Therefore:
P ΔS = S2 – S1 = – nR ln 2 P1
Relations Among Partial Differentials Objective:
To obtain equations relating the state variables: T,P,V,U,H,G,S Closed system with no external fields and no reactions or phase changes
Usually think of: T,P,V
U,H,G,S,A
as Independent
as Dependent
We derived above: G = G(P,T) ie.
dG = V dP – S dT
If you need to calculate ΔG resulting from ΔV first find ΔP and ΔT corresponding to ΔV. Thus: G2 – G1
for
P1 T1 → P2 T2
P2
T2
P1
T1
G2 – G1 = V dP – S dT
For a chemical reaction or a phase change: P2
T2
P1
T1
ΔG2 – ΔG1 = V dP – S dT
G = H – TS dG = V dP – S dT
G G = dP + dT P T T P
For Isobaric changes:
dP = 0
Only 2nd term is important !!
dG = –S dT dG = – S dT
ΔG = G2 – G1 = T2
G2
T2
G1
T1
dG = – S dT
≈ – S dT = – S (T2 – T1) T1
A (Helmholtz Free Energy): A = U – TS dA = dU – d (T S) = dU – T dS – S dT = – P dV + T dS – T dS – S dT = – P dV – S dT
Therefore:
A = –P V T
AND
A = –S T V
A = U – TS dA = – P dV – S dT
A A = dV + dT V T T V dV = 0
For Isochoric changes:
Only 2nd term is important !!
dA = –S dT dA = – S dT
ΔA = A2 – A1 = T2
A2
T2
A1
T1
dA = – S dT
≈ – S dT = – S (T2 – T1) T1
dA is an exact differential. Therefore:
A A = V T T V V T T V
We obtain:
S P = V T T V
Will be useful for dU later !!
Corresponding equations for: H = H(T,P)
H H dH = dT + dP T P P T But:
Now:
H CP ≡ T P H = G + TS
dH = dG + T dS + S dT = V dP – S dT + T dS + S dT
dH = V dP + T dS
(A)
H Obtain the correct expression for by writing P T the derivative form of (A) and specifying that T is constant:
H S = V + T P T P T
S We already have an expression for in terms of P , V , T: P T H V = V – T P T T P
Therefore:
H H dH = dT + dP T P P T V = CP dT + V T dP T P
(B)
Can integrate (B) to give: H2 – H1
for a finite change:
For an isobaric process:
ΔH = H2 – H1 =
H2
T2
H1
T1
P1 T1 → P2 T2
dP = 0
Then:
T2
dH = C dT ≈ CP dT = CP (T2 – T1) P
T1
For S we have:
S S dS = dT + dP T P P T Use the integrated form:
S2 – S1 =
dq rev T
C p dT T T
T2
=
Therefore:
Cp S = T T P
1
And thus:
dS =
CP V dT – dP T T P nR V = P T P
For an Ideal Gas:
dS = Can integrate to give:
ΔS = S2 – S1
CP nR dT – dP T P
S2 – S1
for a finite change:
P1 T1 → P2 T2
P CP nR = dS = dT – dP ≈ CP ln S T T P P S2
T2
2
1
1
1
T = CP ln 2 + nR ln T1
P1 = CP ln P2
T2 – nR ln T1
T2 + nR ln T1
V2 V1
P2 P1
For U the most useful variables to use are T and V instead of T and P:
U U dU = dT + dV T V V T Recall: Since:
U CV ≡ T V dU = T dS – P dV We have:
U S = T – P V V T T S Need to write in terms of P , V , T V T
From the discussion of A (Helmholtz Free Energy) we derived:
S P = V T T V Therefore:
P dU = CV dT + T P dV T V
For an isochoric process: Can integrate to give:
ΔU = U2 – U1 =
U2 – U1
dV = 0
for a finite change:
U2
T2
U1
T1
T2
Then: P1 T1 →
P2 T2
dU = C dT ≈ CV dT = CV (T2 – T1) V
T1
We can now calculate the change in: U,H,S,G,A for a system in which P1 V1 T1
→
P2 V2 T2
Following 4 equations yielded many useful results: dU = – P dV + T dS dH = + V dP + T dS dG = + V dP – S dT dA = – P dV – S dT
Section 5.1. The Use Of Partial Differentials Goals 1) To understand the use of Partial Differentials in developing thermodynamic formulae. Progress 1) We can now calculate the change in: U,H,S,G,A for a system in which: P1 V1 T1 → P2 V2 T2
