Download Specific Heat of p Diatomic Gases and Solids The Adiabatic Process

Specific
p
Heat of
Diatomic Gases and
Solids
The Adiabatic Process
Ron Reifenberger
Birck Nanotechnology Center
Purdue University
February 22
22, 2012
Lecture 7
1
Specific Heat for Solids and
Di t i Gasses
Diatomic
G
In this lecture, you will learn what determines
the specific heat of diatomic gasses and
elemental solids.
solids
The physics behind an adiabatic gas process
will also be analyzed.
analyzed
2
From last lecture, you learned how to
evaluate the heat capacity for an ideal
monatomic
t i gas
1
1 
2
KEav  K av  m  vav   3  kT  ( for one molecule)
2
2 
The internal energy of N atoms in an ideal monatomic
gas is an intrinsic,
intrinsic distributed property of the system
For ideal gas
of N atoms:
CV 
Eint  KE = 3(½ NkT) = 3(½ nRT)
dEint
dT
= 3/2 nR
CP = CV + nR = 5/2 nR
equipartition
theorem
monatomic
gas
3
What makes diatomic
molecules different?
Rotational motion - another way to
distribute energy?
gy
 (rad/s)
Velocity v
r
KE = ½ mv2
 = 2f
Mass m
For now, no linear translation
v = r
KE = ½ mr2 2 = ½ I 2
Typical value: CO, I  1.45 x 10-46 kg•m2
4
By way of review, . . .
Linear Motion Rotational Motion
Displacement
d = vot+½ at2
 = ot + ½ t2
V l i
Velocity
v = vo + att
 = o + t
Inertia
m
I
Newton’s 2nd Law
F = m a
 = I 
Momentum
p = mv
L = I 
Conservation of
momentum
Kinetic Energy
If F=0, then
p = constant
½ mv2
If =0, then
L = constant
½ I2
5
For diatomic molecules, additional energy can
be disbursed in rotational motion
5 degrees
of
f f
freedom
translation
+
rotation
KE = ½mvx2 + ½mvy2 + ½mvz2 + ½Ixx’ xx’2 + ½Iyy’ yy’2
Eint = 5 (½ nRT)  Cv = 5/2 nR  CP = 7/2 nR 6
7
Any other way to distribute energy?
vibrational motion
Work Done = Change in Elastic Energy
W = F • d
= ½ • kx • x
spring
constant
+Force
F
k (N/m)
x=0
k
xfinal
Frestore= -kxfinal
Work done is
equal to the area 8
It is possible that diatomic molecules may
vibrate! Usually,
Usually this occurs at temperatures
much higher than room temperature. If
vibration becomes important, then additional
energies must be added to heat capacity to
include this possibility.
1 mole of H2 gas
7
R
2
5
R
2
3
R
2
9
To calculate the specific heat for ELEMENTAL
solids we need a model that describes how
solids,
atoms interact with each other
KE = ½mvx2 + ½mvy2 + ½mvz2
+ ½kx2 + ½ky2 + ½kz2
6 degrees of freedom
Eint = 6 (½ nRT)
for one mole, n=1
c (molar specific heat) = 3R
k = effective spring constant [N/m]
c  24.9 J/mole K
Law of Dulong-Petit
10
Elemental Solid
Aluminum
Bismuth
Copper
Coppe
Gold
Platinum
Silver
Tungsten
CV (J/mole K)
23.4
25.3
23.8
3.8
24.5
25 4
25.4
24.4
24 4
24.4
Dulong-Petit
Dulong
Petit value
c  24.9 J/mole K
11
A Thermodynamic Process: How you go
f
from
iinitial
iti l tto fi
finall state
t t
P
(P2, V2, T2 )
FINAL
STATE
2
1
4
3
(P1, V1, T1)
INITIAL
STATE
T TE
There are many ways to go
from (P1,V1,T1) to (P2,V2,T2)
V
12
How to
measure P-V
f a gas?
for
?
13
The Adiabatic Process for an Ideal Gas
“”
 is the Greek
word for impassable
Adiabatic
Thermodynamic Process
Quasi-static
Adiabatic Process
insulation
fast!
no time for heat to
flow in or out
slow!
no heat flows in or out
Eint = Qin + Won
14
Using the 1st Law for Ideal Monatomic Gas
Process
Eint
Qin
Won=
Isobaric
nc'VT
nCPT
-P V
Adi b ti
Adiabatic
nc''VT
0
Eint
Isochoric
nc'VT
Eint
0
Isothermal
0
-W
-nRT ln(Vf /Vi)
nc'VT
E
Eint-W
W
(PV area)
General
15
Adiabatic Process ONLY
Eint = Qin + Won
nc’VdT= CV dT = 0 + (-PdV)
CV dT = -nRT dV/V
dT
nR
= T
CV


dV
   ln ( V )  C  ln ( V
V
ln ( T )  ln (
1
V

 T
)  ln 
1 V
nR/CV
 ln (TV

dV
V

 1 
)  C  ln     C
V


  ln  T V


  C
NOTE:
the final result
follows directly
from the 1st Law
) = constant
Key Idea: the quantity on the left is
constant during an adibatic process
involving an ideal gas!
16
Rewrite:
nR
R
CP - CV
=
= -1
CV
CV
TV
-1
= constant
  CP/CV
if T,V are known
rewrite, using the ideal gas law:
PV V
nR
R
-1
= constant
if the number of moles is constant during
the p
process,, then
PV

= some other constant
if P,V are known
17
Adiabatic Compression
PV
γ

= constant
Monatomic gas
Diatomic gas
5/3≈1.667
7/5=1.400
18
Calculating the Work in an
Adiabatic Compression
dEint  dQin  dWon
dWon  dEint  dQin
0
dWadiabatic  dEint  Cv dT
Tf
Wadiabatic  Cv  dT  Cv T f  Ti  only depends on ΔT !
Ti
PV
nR
 Pf V f PV

i i
 Cv 


nR
nR


using
g PV  nRT  T 
Wadiabatic
PV


f
f
 PV
i i
 1
since
nR
CV
= -1
19
Example:
If 1 mole
l of
f an ideal
id l di
diatomic
t i gas, initially
i iti ll att 310 K
K,
expands adiabatically from 12 L to 19 L, what is the
final temperature?
p
TV
-1
= constant
T,V
For diatomic gas, CV = 5/2 R
For diatomic gas, CP = 7/2 R
  CP/CV = 1.4000
TiVi
-1
= TfVf
for diatomic gas
-1
Solving for Tf gives 258 K
20
Example:
30 moles of a monatomic ideal gas
att 1 atmosphere
t
h
pressure expands
d
adiabatically from an initial volume
of 1.5 m3 to 3.0 m3.
PV

= constant
Initial
state
P
Tf
Final
Ti
state
V
P,V
a. What
Wh iis the
h final
fi l pressure?
?
CP
pp
to [5/3])
[
])
 =
= 1.667 ((Note this is an approximation
CV

constant = Pi Vi = 1.99 x 105 Pa m5

Pf Vf = 1.99 x 105 Pa m5
Pf = 3
3.19
19 x
104
Pa → Tf=
PfVf
nR
Pi = 1.01 x 105 Pa → Ti=607 K
=383 K
21
b. What is the work done on the gas?
Won    PdV
Note that P must be inside the integral!
PV   constant  C
= 1.99 x 105
  1
dV
V
|
Won  C    C
  1
V
  1.667
Vf
Vi
1    0.667 
1
3

2
  1
 3
Won  C    V f0.667  Vi 0.667 
 2
3C
3C

0.4807  0.7631    0.2824 
2
2
 84,350 J  84.4 kJ
(also equals change in internal energy)
22
The work can also be calculated from
Wadiabatic 
Pf V f  PV
i i
 1
3.19 104  3  1.013 105 1.5

1.667  1
 84,375
84 375 J
γ-1
C. Is TV
really
y a constant?
 1
 1
TV

T
V
i i
f f
6071.5
1 5
0.667
 383 3
0.667
6071.310  3832.080
795.2  796.7
23
To do better, use 2/3=0.666666667 instead of 0.667 throughout the calculation.
In summary, for the process
under consideration,
consideration we have:
PV
1 atm
t

= constant
607 K
383 K
0.32 atm
1.5 m3
3.0 m3
Area under curve ≈ 84,375 J
24