PHY 113 C General Physics I
11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 22:
Chapter 21: Ideal gas equations
1. Molecular view of ideal gas
2. Internal energy of ideal gas
3. Distribution of molecular speeds in ideal
gas
4. Adiabatic processes
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
1
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
2
From Webassign (Assignment #19)
A combination of 0.250 kg of water at 20.0°C, 0.400 kg
of aluminum at 26.0°C, and 0.100 kg of copper at
100°C is mixed in an insulated container and allowed
to come to thermal equilibrium. Ignore any energy
transfer to or from the container and determine the
final temperature of the mixture.
Thermally insulated container  Q  0
Q  0   mi ci TF  TIi 
i
0  0.25  cwater  TF  20   0.4  c Al  TF  26   0.1  cCu  TF  100 
4186 J/(kg*oC)
900 J/(kg*oC)
387 J/(kg*oC)
(From Table 20.1)
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
3
From Webassign (Assignment #19)
A thermodynamic system undergoes a process in
which its internal energy decreases by 465 J. Over the
same time interval, 236 J of work is done on the
system. Find the energy transferred from it by heat.
Eint  Q  W
Q  Eint  W  465 J  236 J  701J
Note: Sign convention for Q :
Q>0  system gains heat from environment
iclicker question:
Assuming the system does not change phase, what can
you say about TF versus TI for the system?
A. TF>TI
B. TF<TI
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
4
From Webassign (Assignment #19)
A 2.20-mol sample of helium gas initially at 300 K, and
0.400 atm is compressed isothermally to 1.80 atm.
Note that the helium behaves as an ideal gas.
(a) Find the final volume of the gas.
(b) Find the work done on the gas.
(c) Find the energy transferred by heat.
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
5
From Webassign (Assignment #19)
A 2.20-mol sample of helium gas initially at 300 K, and
0.400 atm is compressed isothermally to 1.80 atm.
Note that the helium behaves as an ideal gas.
(a) Find the final volume of the gas.
P1V1  n1 RT1
P2V2  n2 RT2
P2V2 n2 RT2

P1V1 n1 RT1
V2 P1

V1 P2

P1  n1 RT1  P1

V2  V1  
P2  P1  P2
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
6
From Webassign (Assignment #19)
A 2.20-mol sample of helium gas initially at 300 K, and
0.400 atm is compressed isothermally to 1.80 atm.
Note that the helium behaves as an ideal gas.
(b) Find the work done on the gas.
(c) Find the energy transferred by heat.
Vf
 nRT 
W    PdV    
dV   nRT ln
V 
 Vi
Vi
Vi 
Vf
11/14/2013
Vf
PHY 113 C Fall 2013 -- Lecture 22

  Q

7
From Webassign (Assignment #19)
One mole of an ideal gas does 2 900 J of work on its
surroundings as it expands isothermally to a final pressure of
1.00 atm and volume of 28.0 L.
(a) Determine the initial volume of the gas.
(b) Determine the temperature of the gas.
1.013 105  0.028
Pf V f  nRT T 

 341.14o K
nR
1  8.314472
Pf V f
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
8
From Webassign (Assignment #19)
One mole of an ideal gas does 2 900 J of work on its
surroundings as it expands isothermally to a final pressure of
1.00 atm and volume of 28.0 L.
(a) Determine the initial volume of the gas.
For isothermal process :
Vf
 nRT 
W    PdV    
dV   nRT ln
V 
 Vi
Vi
Vi 
Vf
Vf

  2900 J

(b) Determine the temperature of the gas.
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
9
From Webassign (Assignment #19)
In the figure, the change in internal energy of a gas
that is taken from A to C along the blue path is
+795 J. The work done on the gas along the red
path ABC is -530 J.
(a) How much energy must be added to the system by heat
as it goes from A through B to C?
(b) If the pressure at point A is five times that of point C, what
is the work done on the system in going from C to D?
(c) What is the energy exchanged with the surroundings by
heat as the gas goes from C to A along the green path?
(d) If the change in internal energy in going from point D to
point A is +495 J, how much energy must be added to the
system by heat as it goes from point C to point D?
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
10
Review:
Consider the process described by ABCA
iclicker exercise:
What is the net work done
on the system in this
cycle?
A. -12000 J
B. 12000 J
C. 0
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
11
Equation of “state” for ideal gas
(from experiment)
8.314 J/(mol K)
PV  nRT
temperature in K
volume in m3 # of moles
pressure in Pascals
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
12
Ideal gas -- continued
Equation of state : PV  nRT
1
1
Internal energy :
Eint 
nRT 
PV
 1
 1
  parameter depending on type of ideal gas
 53 for monoatomic
7
 5
for diatomic
..............................

Note that at this point, the above equation for Eint
is completely unjustified…
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
13
From The New Yorker Magazine, November 2003
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
14
Microscopic model of ideal gas:
Each atom is represented as a tiny hard sphere of
mass m with velocity v. Collisions and forces between
atoms are neglected. Collisions with the walls of the
container are assumed to be elastic.
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
15
What we can show is the pressure exerted by the
atoms by their collisions with the walls of the
container is given by:
2N1
2N
2
P
3V
2
mv
avg

3V
K
avg
Proof:
Force exerted on wall perpendicular to x-axis by an atom which
collides with it:
pix 2mi vix
Fix 

t
t
t  2d / vix
2mi vix mi vix2
 Fix 

2d / vix
d
-vix
vix
number of atoms
x
Fix
mi vix2 N
d
P 

mi vx2
A
dA V
i
i
average over atoms
volume
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
16
2mi vix mi vix2
Fix 

2d / vix
d
Fix
mi vix2 N
P


mi vix2
A
dA
V
i
i
Since molecules are equally likely to move along x, y, z :
mi vix2  mi viy2  mi viz2
Also note that vi2  vix2  viy2  viz2
vi2  vix2  viy2  viz2  3 vix2
N
N
2N
2
2
P
mi vix 
mi vi 
V
3V
3V
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
1
2
mi vi2
17
iclicker question:
What should we call
1
2
2
i i
mv
?
A. Average kinetic energy of atom.
B. We cannot use our macroscopic equations
at the atomic scale -- so this quantity will go
unnamed.
C. We made too many approximations, so it is
not worth naming/discussion.
D. Very boring.
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
18
2N 1
2
P
m
v
i i
3V 2
Note : N  number of atoms
mi  mass of atom (6.6 10-27 kg for He atom)
Nmi  nM where M denotes the molar mass (0.004 kg for He atom)
n  number of moles of atoms
Connection to ideal gas law :
2 1
2 
PV  n 2 Mvi   nRT
3

2 1

  2 Mvi2   RT
3

1
2
or
1
2
2
i
Mv
3
 RT
2
Mvi2  average kinetic energy of mole of ideal gas atoms
3
Eint  nRT
2
11/14/2013
for mono atomic ideal gas
PHY 113 C Fall 2013 -- Lecture 22
19
Average atomic velocities:
(note <vi>=0)
3
1
Mv

RT
2
2
3RT
2
vi 
M
2
i
Relationship between
average atomic velocities
with T
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
20
Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
21
Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg
Molecular mass
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
22
Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg
Molecular mass
M in units of 0.001 kg/mole
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
23
For monoatomic ideal gas:
3
Eint  nRT
2
General form for ideal gas (including mono-, di-, polyatomic ideal gases):
1
Eint 
nRT
 1
 53 for monoatomic
7
   5 for diatomic
 ..............................

11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
24
Macroscopic Microscopic
nR  Nk B
8.314 J/mole oK
1.38 x 10-23 J/molecule oK
1 mole  6.022  10 molecules
23
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
25
Big leap!
Internal energy of an ideal gas:
1
3
N
n
Eint  N  m v 2   N k BT 
k BT 
RT
2
γ-1
γ-1
2

derived for
monoatomic ideal gas
Gas
He
N2
H2O
11/14/2013
more general relation
for polyatomic ideal gas
 (theory)
5/3
7/5
4/3
PHY 113 C Fall 2013 -- Lecture 22
 exp)
1.67
1.41
1.30
26
Comment on “big leap” – case of diatomic molecule
w
vCM
Eint  ECM  Erot
1
1 2
2
 MvCM  Iw
2
2
Previously, we have shown :
Note: We are assuming
3
2
1
 RT
2 MvCM
that molecular vibrations
2
are not taking much
Educated guess :
energy
2
2
1
 RT
2 Iw
2
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
27
Comment on “big leap” – continued
Big leap!
Internal energy of an ideal gas:
1
3
N
n
Eint  N  m v 2   N k BT 
k BT 
RT
2
γ-1
γ-1
2

derived for
monoatomic ideal gas
more general relation
for polyatomic ideal gas
 can be measured for each gaseous system
Note:  = CP/CV
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
28
Determination of Q for various processes in an ideal gas:
n
Eint 
RT
γ-1
n
Eint 
RT  Q  W
γ-1
Example: Isovolumetric process – (V=constant  W=0)
n
Eint i  f 
RTi  f  Qi  f
γ-1
n
RTi  f  nCV Ti  f
In terms of “heat capacity”:Qi  f 
γ-1
R
CV 
γ-1
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
29
Example: Isobaric process (P=constant):
Eint i  f
n

RTi  f  Qi  f  Wi  f
γ-1
In terms of “heat capacity”:
Qi  f
n
n

RTi  f  Pi V f  Vi  
RTi  f  nRTi  f  nC P Ti  f
γ-1
γ-1
R
γR
γR
 CP 
R
CP 
γ-1
γ-1
γ-1
Note:  = CP/CV
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
30
Summary
Suppose : Eint  XnRT
where X is a constant
 CV  XR
 C P  CV  R
Define :
CP

CV
From algebra :
CP
R

 1
CV
CV
R
 CV 
 XR
 1
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
R
X
 1
nRT
Eint 
 1
31
iclicker question:
The previous discussion
A. Made me appreciate the  factor in thermo
analyses
B. Made me want to scream
C. Put me to sleep
D. No problem – as long as this is not on the test
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
32
More examples:
Isothermal process (T=0)
n
Eint 
RT
γ-1
n
Eint 
RT  Q  W
γ-1
T=0  Eint = 0  Q=-W
Vf
dV
 W   PdV  nRT 
nRT ln
V
 Vi
Vi
Vi
Vf
11/14/2013
Vf
PHY 113 C Fall 2013 -- Lecture 22



33
Even more examples:
Adiabatic process (Q=0)
Eint  W
n
RT   PV
γ-1
PV  nRT
PV  PV  nRT
nRT  γ-1PV  PV  PV
V P
γ

V
P
 V fγ 
 Pf 


  ln γ  ln   PiVi γ  Pf V fγ
V 
 Pi 
 i 
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
34
PiVi 
Adiabat : P  
V
PiVi
Isotherm : P 
V
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
35
iclicker question:
Suppose that an ideal gas expands adiabatically. Does
the temperature
(A) Increase
(B) Decrease
(C) Remain the same
PiVi γ  Pf V fγ
Ti
PiVi  nRTi  Pi  nR
Vi
TiVi γ-1  T f V fγ-1
 Vi
T f  Ti 
V f
11/14/2013




γ -1
PHY 113 C Fall 2013 -- Lecture 22
36
Review of results from ideal gas analysis in terms
of the specific heat ratio   CP/CV:
n
R
Eint 
RT  nCV T ; CV 
γ-1
γ-1
γR
CP 
γ-1
For an isothermal process, Eint = 0  Q=-W
Vf
 W   PdV nRT ln
 Vi
Vi
Vf

V
  PiVi ln f

 Vi



For an adiabatic process, Q = 0
PiVi γ  Pf V fγ
TiVi γ-1  T f V fγ-1
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
37
Note:
It can be shown that the work done by an
ideal gas which has an initial pressure Pi and
initial volume Vi when it expands adiabatically
to a volume Vf is given by:

PiVi   Vi
W    PdV  
1

 V
γ

1
Vi
  f
Vf
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22




γ 1




38
Examples process by an ideal gas:
P (1.013 x 105) Pa
Pf
B
C
AB
Q
W
A
Pi
Vi
11/14/2013
CD
DA
Vi ( Pf  Pi )
γPf (V f  Vi )
 V f ( Pf  Pi )
-γPi (V f  Vi )
γ -1
γ -1
γ -1
γ -1
0
-Pf(Vf-Vi)
D
Eint
BC
0
Pi(Vf-Vi)
Vi ( Pf  Pi )
Pf (V f  Vi )
 V f ( Pf  Pi )
-Pi (V f  Vi )
γ -1
γ -1
γ -1
γ -1
Efficiency as an engine:
Vf
e = |Wnet/ |/Qinput
PHY 113 C Fall 2013 -- Lecture 22
39
From Webassign (#19)
An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as
shown below. (Let the factor n = 2.6.)
(a) Find the net work done on
the gas per cycle for 2.60 mol
of gas initially at 0°C.
(b) What is the net energy
added by heat to the system
per cycle?
Wnet  Pf  Pi V f  Vi   n  1 PiVi  Qnet
2
11/14/2013
PHY 113 C Fall 2013 -- Lecture 22
40