Application of the First Law to Ideal Gases
Calculate q,w, U, and H for ideal gas processes:
dU = q + w
dU = Cv dT dH = Cp dT
U
U = q + w
for any process since = 0
VT
Isothermal Reversible Expansion
dT = 0 so dU =
and dH =
q = w
V2
V1
w = nRT ln
P2V2 = P
Efficiency of Cyclic Processes- Carnot Cycle
w1
P
Reservoir, TH
qH >0
gas
q
w
4
H
TH
w
qL <0
w2
w3
q
L
Reservoir, T L
T
L
V
U = q + w
U = q H + qL + w1 + w2 + w3 + w4
w = w1 + w2 + w3 + w4
w = (qH + qL)
-w
= qH
=
qH + q L
qH
Comments:
for reversible or
G Solvation
solG = solGVdW + solGcav + solGelec
H
VdWG = solute-solvent Van der Waals
N
H
r = 1-5
H
cavG = work to create cavity in solvent
r = 78.54
= surface tension x surface area
= 10-37 J/2 includes entropy penalty for rearrangement of water molecule
Work and G
dU = T dS P dV
dw = F dx
dU = T dS P dV + F dx
dG = -S dT + V dP + F dx
(mN/m)
72.8
22.3
18.4
51.1
P, T
dq
Surface Tension: dw = d
Interface
water/air
ethanol/air
hexane/air
hexane/water
dV
Tsurr
J/m2 or N/m
d = 8r dr
dG = -S dT + V dP + d
Ex
Thermodynamic Potential Functions
From Mechanics: dU = - F dr
U
dU
force = dr
r
U(S,V)
dU = dq + dw
H(S,P)
A(V,T)
G(P,T)
dU = TdS - PdV
dU = (
H = U + PV
A = U - TS
G = H - TS
dH = ? from dU = TdS - PdV
H = U + PV
dH = dU + PdV + VdP
dH = TdS + VdP
dG = ?
Partial Derivative Conversion
= CV +
V
=
U
P
+ V
T V TV
=
1
1
=
Cp
H
T P
H = U + PV
definition
invert
misplaced numerator
misplaced numerator
H
T V
T
HP
H
VT
misplaced denominator
H
T V
T
PH
misplaced constant variable
chain rule
misplaced consta
Chemical Potential of Solutions
(g) = (g) + RT ln P/P
Pure gas
A(g) = (g) + RT ln PA/P
A
*(l) = (g) + RT ln P*/P
A
A
A
A(l) = (g) + RT ln PA/P
A
A(l) = *(l) + RT ln xA
A
B(l) = B(l) + RT ln xB
Colby College
PB
xA
A(l) = *(l) + RT ln PA/P*
A
A
PA
xB
Henry's Law constants and Free Energies of solvation.
solG = - RT ln K
The units are indicated as subscripts: p=pressure, x=mole fraction, and c=molarity.
substance
Kcc
Kpc
solGpx
unitless atm /M
kJ/mol
294.
0.216
5.32
-14.1
358.
0.263
6.47
-14.6
433.
0.3
Thermodynamic Equations of State
Put dU and dH soley in terms of P, V, T
U
U
dU = (V) dV + (T ) dT
T
V
dH = (
H
) dP + (H) dT
P T
T P
(U)
V
T
=(
(U)
V
P
= -P + T( ) = -P + T
T V
T
(A+TS)
S
) = (A) + T (V)
V
V T
T
T
nRT
ideal gas: P = V
(H)
P
(H)
P
T
T
=(
Foundations of Thermodynamics
PV=nRT
2
(P + a n 2 )( V - nb ) = nRT
1 V
= V( T )P
-1 V
= V ( P )T
U
Cv = ( T )V
H
CP = (T )P
P
= (T)V
V
_
T
dU = dq + dw
dU = TdS - PdV
(V)S = -(P)V
S
T
V
H = U + PV
dH = TdS + VdP
(P)S = ( S )P
S
A = U - TS
dA = -SdT - P
Integrating the Basic Derivatives
dV = V dT
cst. P
Good: V Vo
V
V
T
dV =
Vo
V dT
To
V
o
V-Vo = Vo (T - To)
T
V = Vo T
To
Better: V Vo + Vo (T - To)
V
V
T
dV =
Vo
(Vo + Vo (T - To) dT
To
V
o
V-Vo = V o (T - To) + Vo 2
(T - To)2
2
T
To
Best: dV = dT
V
V
1
Differential Scanning Calorimetry
sample
reference
Linear Temperature Scan
Tsample Tref
T
dT
= 20C min 1
dt
+
Heater
sample
power
monitor
Heat flow =
Scan
Control
Heater
T
reference
power
monitor
qp
dt
Jq =
time
endotherm
+
dq
dt
trCp
heat flux
mJ s-1
e
Entropy and the Ideal Gas
Ideal gas:
( U ) T = 0
V
dU = CV dT
S
for all processes
dU = dq - P dV
T or V
dq = dU + P dV
dqrev
dT nR
dS = T = CV T + V d V
at constant V
dT
S = CV T
T2
S = CV ln T
1
assume Cv = cst
at constant T
V2
nR
S = V dV = nR ln V
Van der Waals Liquifaction
90
(atm)
80
70
vapor
liquid
2-phase
60
P
P
(atm)
supercritical fluid
40
30
50
20
V (L)
0
0.2
0.4
0.6
an2
P + V2 (V - nb) = nRT
2P
I. ( 2)T = 0
V
II. (
RTc a
Pc= V -b - 2
c
Vc
R
P
)T = (VcTc 2 + 2a =
-b) V3
V
c
III. (
2
2P
)T = (