Some terms that must be understood
Microscopic Variable
Macroscopic Variable
8.044 L5B1
Extensive (∝ N )
Intensive (=
f (N ))
V volume
P pressure
A area
S surface tension
L length
F tension
P polarization
E electric field
M magnetization
H magnetic field
·········
·········
T temperature
U internal energy
8.044 L5B2
Adiabatic Walls
Equilibrium
Steady State
Diathermic Walls
Complete Specification:
Independent and Dependent Variables
8.044 L5B3
Equation of State
P V = N kT
V = V0(1 + αT − KT P )
M = cH/(T − T0)
T > T0
In Equilibrium with Each Other
8.044 L5B4
OBSERVATIONAL FACTS
" 0 th Law"
if A
equilibrium
C
then A
and B
equilibrium
equilibrium
C
B
8.044 L5B5
" Law 0.5 ? "
YA
Many macroscopic states of B can be
in equilibrium with a given state of A
YB
A
1
B
1
XA
YB = f (XB)
also
XA , YA
XB
8.044 L5B6
THEOREM A "predictor" of equilibrium h(X, Y, ...) exists
h(X, Y)
• only in equilibrium
• state variable
• many states, same h
• different systems,
different functional forms
• value the same if
systems in equilibrium
Y
X
locus of
constant h
8.044 L5B7
XA , YA
XC , YC
XA , YA , XC , YC all free
[ P A , V A , PC , V C ]
equilibrium
XA , YA
XC , YC
keep same
adjust
XC = f 1 (YC , XA , YA )
[ P C = PA V A / V C ]
F1 (XC , YC , XA , YA ) = 0
[ PC V C - PA V A = 0 ] 8.044 L5B8
equilibrium
XB , YB
XC , YC
adjust
keep same
XB = g (YB , XC , YC )
[ P B = PC V C / V B ]
F2 (XC , YC , XB , YB ) = 0
[ PC V C - PB V B = 0 ] solve for X C
XC = f 2 (YC , XB , YB )
[ P C = PB V B / V C ]
same value as before
8.044 L5B9
f 1 (YC , XA , YA ) = XC = f 2 (YC , XB , YB ) 1
[ P A V A / V C = PB V B / V C ]
equilibrium
XA , YA
XB , YB
due to 0 th law
F3 (XA , YA , XB , YB ) = 0
1
+
2
2
F3 factors
YC drops out
8.044 L5B10
For this equilibrium condition
h ( X , Y ) = constant = h ( X , Y )
A
A
B
B
0,7B6SVV
[ PA VA = P B VB ]
8.044 L5B11
YA
YB
A
Isotherm at T 1
B
2
1
T2
1
T2
2
Isotherm at T 1
XA
XB
8.044 L5B12
Empirical Temperature: t
low density gas
Y
P
PV/N = constant'
2
PV/N = constant
2
1
1
X
V
8.044 L5B13
we could
• Define t
possible alternative
cg PV/N
t'
cg' (PV/N)α
• Use to find isotherms
in other systems
• Then in a simple paramagnet
t = cm (M/H)-1
t' = cm' (M/H)-α
Many possible choices for t
8.044 L• B
• • • •a •
PV = Nkt → t = PV/Nk
P
1
0.8
t
P
0.6
0.4
0.2
V
0
0
0.2
0.4
0.6
PV = constant
0.8
1
V
8.044 L5B14b
t' = (PV/Nk) 2
P
1
0.8
t'
P
0.6
0.4
0.2
V
0
0
0.2
0.4
0.6
PV = constant
0.8
1
V
8.044 L5B14c
t'' = PV/Nk
P
1
t''
0.8
P
0.6
0.4
0.2
V
0
0
0.2
0.4
0.6
PV = constant
0.8
1
V
8.044 L5B14d
Work
dW
differential of work done on the system
= - (work done by the system)
Hydrostatic system
dW = -PdV
dx
F
dW = Fdx = (PA)(-dV/A) = -PdV
8.044 L• •B• • • •
F
Wire
dW = FdL
F
P pushes, F pulls
dW = Fdx = (F)(dL) = FdL
Surface
dW = SdA
wire
2 SL
film
(2 surfaces)
F const raint
fram e
dW = Fdx = (SL)(dA/L) = SdA
8.044 L5B16
Chemical Cell (battery)
dW = EEMF dZCHARGE
Electric charges
dW = EdP
Magnetic systems
Field in absence of matter
as set up by external
sources. Does not include
energy stored in the field
itself in the absence of
the matter.
dW = HdM
8.044 L5B17
• All differentials are extensive
• Only -PdV has a negative sign
• Good only for quasistatic processes
• ∆W =
a
b
dW depends on the path
W is not a state function
8.044 L5B18
low density gas
Y
A
P
isotherm PV= NkT
P2
b
a
B
P1
X
c
b
a
V2
V1 V
dW = YdX
depends on Y(X)
8.044 L5B19
−P1(V2 − V1) = P1(V1 − V2)
(a) W1→2 =
(b) W1→2 =
−P2(V2 − V1) = P2(V1 − V2)
2
2
N kT
(c) W1→2 = − P (V ) dV = −
1
1 V
dV = −N kT
2
dV
1
V
2 = N kT ln V1 = P V ln V1
= −N kT ln V
1 1
V1
V2
V2
8.044 L5B20
MATH
I) 3 variables, only 2 are independent
F (x, y, z) = 0
⇒ x = x(y, z),


∂x 
1
⇒
=
�
�
,
∂y
∂y z

∂x z
y = y(x, z),
z = z(x, y) 
 �
�
�
�
∂x
∂y
∂z


∂y z ∂z x ∂x y
=
−1
8.044 L5B21
Given some W = W (x, y, z) where only 2 of the 3 variables
in the argument are independent,
then along a path where W is constrained to be constant
�
∂x
∂y
�
W
�
∂y
∂z
�
W
�
∂z
∂x
�
=1
W
8.044 L5B21a
then it follows that
∂x
∂z w
∂x
= ∂y w
∂y
∂z w
II) State function of 2 independent variables
S = S(x, y)


∂S 
∂S

dx +
dy
dS =
∂x y
∂y
| {z }
| {z x}
!
A(x,y )
B(x,y)
An exact differential
8.044 L5B22


∂ 2S
�
∂ 2S
�
∂A 
∂B
=
=
=
∂y x
∂y∂x
∂x∂y
∂x y

⇒ necessary condition, but it is also sufficient

Exact differential if and only
Then
� 2
1

�
�
∂A 
∂B
=
∂y x
∂x y
if

dS = S(x2, y2) − S(x1, y1) is independent of
the path.
8.044 L5B23
III) Integrating an exact differential
dS = A(x, y) dx + B(x, y) dy
1. Integrate a coefficient with respect to one variable
∂S
= A(x, y)
∂x y
!
S(x, y) =
Z
|
A(x, y) dx +f (y)
y
{z
fixed
}
8.044 L5B24
2. Differentiate result with respect to other variable


Z
d f (y)
∂S
∂

 =
A(x, y) dx +
= B(x, y)
∂y x
∂y
dy
3. Integrate again to find f (y)
d f (y)
=
dy
f (y) =


∂
B(x, y) −

∂y
Z
Z


A(x, y) dx 
{· · ·} dy
done
8.044 L5B25
MIT OpenCourseWare
http://ocw.mit.edu
8.044 Statistical Physics I
Spring 2013
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