Thermodynamics focuses on state functions: P, V, M, S, . . .
Nature often gives us response functions (derivatives):
1 ∂V
α≡
V ∂T P
�
�
1 ∂V
κT ≡ −
V ∂P T
�
�
1 ∂V
κS ≡ −
V ∂P
�
�
adiabatic
∂M
χT ≡
∂H T
�
�
8.044 L6B1
Example Non-ideal gas
Given
• Gas → ideal gas for large T & V
•
∂P
Nk
=
∂T V
V − Nb
•
∂P
N kT
2aN 2
=−
+
2
∂V T
(V − N b)
V 3
Find P
8.044 L6B2
�
dP =
P =
∂P
∂P
dV +
dT
∂V T
∂T V
�
� � ∂P �
∂T V
�
�
dT + f (V ) =
� �
Nk
V − Nb
�
dT + f (V )
N kT
=
+ f (V )
(V − N b)
8.044 L6B3
2
∂P
N kT
N
kT
2aN
I(V ) = −
= −
+
f
+
j
V
)
2
2
3
∂V T
(V − N b)
(V − N b)
j V V
)
f (V ) =
�
2aN 2
aN 2
dV = − 2
+ c
3
V
V
N kT
aN 2
P =
− 2
+ c
(V − N b)
V
but c = 0 since P → N kT /V as V → ∞
8.044 L6B4
Internal Energy U
Observational fact
initial
∆W
final
isolated
(adiabatic)
Final state is independent of how ΔW is applied.
Final state is independent of which adiabatic path
is followed.
8.044 L6B5
⇒ a state function U such that
ΔU = ΔWadiabatic
U = U (independent variables)
= U (T, V ) or U (T, P ) or U (P, V ) for a simple fluid
8.044 L6B6
Heat
If the path is not adiabatic, dU = d
/W
d
/Q ≡ dU − d
/W
d
/Q is the heat added to the system.
It has all the properties expected of heat.
8.044 L6B7
First Law of Thermodynamics
dU = d
/Q + d
/W
• U is a state function
• Heat is a flow of energy
• Energy is conserved
8.044 L6B8
Ordering of temperatures
dQ
T1
T2
When d
/W = 0, heat flows from high T to low T.
8.044 L6B9
Example Hydrostatic System: gas, liquid or simple
solid
Variables (with N fixed): P, V, T, U .
Only 2 are independent.
⎛
⎞
d
/Q ⎠
CV ≡
dT V
⎝
⎛
⎞
d
/Q ⎠
CP ≡
dT P
⎝
Examine these heat capacities.
8.044 L6B10
dU = d
/Q + d
/W = d
/Q − P dV
d
/Q = dU + P dV
d
We want
. We have dV .
dT
∂U
∂U
dU =
dT +
dV
∂T V
∂V T
�
�
�
�
8.044 L6B11
∂U
d
/Q =
dT +
∂T V
��
∂U
+ P dV
∂V T
d
/Q
∂U
⇒
=
+
dT
∂T V
��
∂U
dV
+P
∂V T
dT
�
�
�
⎛
�
⎞
�
�
�
�
d
/Q ⎠
∂U
CV ≡
=
dT V
∂T V
�
�
⎝
8.044 L6B12
⎛
⎞
d
/Q ⎠
∂U
⎝
CP ≡
=
+
dT P
" ∂T
T Vv
�
�
��
∂U
+P
∂V T
CV
��
CP − CV =
�
∂V
" ∂T
T Pv
��
�
αV
∂U
+ P αV
∂V T
�
�
The 2nd law will allow us to simplify this further.
∂U
Note that CP =
.
∂T P
�
�
8.044 L6B13
Paths Experimental conditions, not just math
fills container
∆V=0
floating piston
∆P=0
bath
insulation
~~~~~~~
∆T=0
∆Q=0
8.044
L6B14
8.044L12B1
∆Q = 0 could come from time considerations
Example Sound Wave
ρ(x)
v
x
too fast for heat to flow out of compressed regions
v=
1
ρκS
8.044 L6B15
8.044L12B2
Example Hydrostatic system: an ideal gas, PV=NkT
New information
!
∂U
=0,
∂V T
3 possible sources Experiment
free
expansion
bath initially at Ti
observe Tf = Ti
8.044 L6B16
8.044L12B3
No work done so ΔW = 0
Tf = Ti ⇒ ΔQ = 0
together
⇒ ,
ΔUU== 0Q
here
→
(∂U/∂VU=)T = 0Q
quasi-static changes
,
• Physics: no interactions, single particle energies
only ⇒ (∂U/∂V )T = 0
• Thermo: 2nd law + (P V = N kT ) ⇒ (∂U/∂V )T = 0
8.044 L6B17
Consequences
∂U
∂U
dU =
dT +
dV
, ∂T
vt Vv
, ∂V
vt Tv
�
�
CV
U =
�
�
0
� T
CV (T ') dT ' + constant
,
vt
v
0
set=0
In a monatomic gas one observes CV = 3
2 N k.
Then the above result gives U = CV T = 3
2 N kT .
8.044 L6B18
∂U
CP − CV = (
+P )
, ∂V Tp
�
�
0
= Nk
∂V
, ∂T Pp
�
�
∂ (N kT /P ) =N k/P
P
∂T
for any ideal gas
Applying this to the monatomic gas one finds
3
5
CP = N k + N k = N k
2
2
5
γ ≡ CP /CV =
3
8.044 L6B19
Adiabatic Changes
d
/Q = 0
Find the equation for the path.
Consider a hydrostatic example.
∂U
∂U
d
/Q =
dT +
+ P dV = 0
T
� ∂T
�� V�
� ∂V ��
�
�
�
��
CV
�
�
�
(CP −CV )/αV
⎛
⎞
(γ − 1)
∂T
CP − CV ) ⎠ 1
⎝
=−
=−
∂V ΔQ=0
CV
αV
αV
�
This constraint defines the path.
8.044 L6B20
Apply this relation to an ideal gas.
1 ∂ N kT
1 Nk
1V
1
1 ∂V
α≡
=
=
=
=
V ∂T P
V ∂T
P
V P
VT
T
P
�
�
�
�
�
�
Path
dT
T
= −(γ − 1)
dV
V
⎛
⎞
dT
dV
T⎠
V
⎝
= −(γ − 1)
→ ln
= −(γ − 1) ln
T
V
T0
V0
⎛
⎝
⎞
T⎠
=
T0
⎛
⎝
⎞−(γ−1)
V
⎠
V0
8.044 L6B21
Adiabatic
Isothermal
Adiabatic
1=c
γ γ−
T
V
1
=c
TV
Isothermal
P
""
P
V
=
c
′′
PV = c
adiabat
−1
γ γ ′ c"
PP
V V==
c
γ = 5/3 (monatomic)
PP∝∝
V V1
isotherm
5/3
P γ∝=
V 5/3
(monatomic)
dP
=
dV
5P
3V
P ∝
V −5/3
dP
5P
=−
dV
3V
dP
P
=−
dV
P V
dP
dV
=
V
V
8.044L12B9
8.044 L6B22
Expansion of an ideal gas
insulation
F
rupture diaphragm
adiabatic ∆Q = 0
not quasistatic
∆W = 0
∆U = 0
T
slowly move piston
adiabatic ∆Q = 0
quasistatic
∆W is negative
∆U = is negative
T
adiabat
constant U
V
V
8.044
L6B23
8.044L12B10
Starting with a few known facts,
/W , and state function math,
1st law, d
one can find
relations between some thermodynamic quantities,
a general expression for dU ,
and the adiabatic constraint.
Adding models for the equation of state and the heat
capacity allows one to find
the internal energy U
and the adiabatic path.
8.044 L6B24
MIT OpenCourseWare
http://ocw.mit.edu
8.044 Statistical Physics I
Spring 2013
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