Transcript Equation sheet and constants for Final Exam: Thermodynamics (Physics 160) k  1.381 10 23 J / K 1 atm  10

Equation sheet and constants for Final Exam: Thermodynamics (Physics 160)
k  1.381 10 23 J / K
1 atm  10 5 Nm -2
N A  6.0221023
R  8.315 J mol-1 K -1
1 liter  10 3 m3
mp 1.6731027 kg
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1 liter H 2O = 1 kg
1 atm  1.013 10 5 N  m-2
T in o C = T in K - 273.15
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Specific
 heat of water = 4190 J/kg-K
Specific heat of ice = 2000. J/kg-K
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Heat 
of fusion of water = 334 x 103 J/kg
Heat of vaporization of water = 2256 x 103 J/kg
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U 
CV   
T V
PV  NkT
PV  nRT


3kT
m
v rms 
Uthermal 


f
NkT
2

U  QW




Vf
W    P(V )dV
Vi

dP
L

dT TV
H 
CP   
T P
 m 3/2
2
D(v)  
4v 2emv /2kT


2kT 
CP  CV  nR
(N,n) 

N!
n!(N  n)!
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PV  constant
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


VT
f /2
 constant
f 2

f
Q
C
T
dU TdS  PdV  dN
(q  N 1)!
(N,q) 
q!(N 1)!


 V 4m U3/2  5 
S  Nkln
 
2  
N 


3Nh

 2 
 

H U  PV

N! N N eN 2N
ln(N!)  N ln(N)  N




S  k ln
S 
Q
T

dF  SdT  PdV  dN
dG  SdT  VdP  dN
dH  TdS  VdP  dN
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O
C
O