Physics 231
Lecture 30
•
•
Main points of today’s lecture:
Ideal gas law:
PV = nRT = Nk BT
2  N  1
2  N
3
2 
m
v
KE
;
KE
KE
k BT
=
≡
=
  

 
3 V 2
3 V
2
• Heat and heat capacity: Q = cmΔT
• Work in thermodynamic processes: Wsystem = PΔV
• First Law of Thermodynamics: ΔU = ΔQ − PΔV
P=
Checking Understanding: Pressure and Forces
The two identical cylinders contain
samples of gas. Each cylinder has
a lightweight piston on top that is
free to move,
move so the pressure inside
each cylinder is equal to atmospheric
pressure. One cylinder contains
hydrogen the other nitrogen
hydrogen,
nitrogen. Both
gases are at the same temperature.
The number of moles of hydrogen is
A. greater than the number of moles of
nitrogen.
B equall to
B.
t the
th number
b off moles
l off nitrogen.
it
C. less than the number of moles of nitrogen.
PV = nRT

n=
TH =TN
VH =VN
PH =PN
nH/n
/ N=??
PV
RT
Slide 12-24
Checking Understanding: Pressure and Forces
The two identical cylinders contain
samples of gas. Each cylinder has
a lightweight piston on top that is
free to move,
move so the pressure inside
each cylinder is equal to atmospheric
pressure. One cylinder contains
hydrogen the other nitrogen.
hydrogen,
nitrogen The
mass of gas in each cylinder is the same.
The temperature of the hydrogen gas is
A. greater than the temperature of the
nitrogen.
B equall to
B.
t the
th temperature
t
t
off the
th nitrogen.
it
C. less than the temperature of the nitrogen.
TH /TN=?
VH =VN=V
PH =PN=P
mH=mN=m
PV
1
PV

PV = nRT
T=
n
T
n R n
1
nR
 H = H = H = N =
m
PV
1
TN
n H 14
m = n H 2g = n N 28g  n H = m & n N =
28g
n NR nN
Slide 12-26
2g
Reading Quiz
2. When the temperature of an ideal gas is increased, which of the
following also increases? (1) The thermal energy of the gas; (2) the
average kinetic energy of the gas; (3) the average potential energy
of the gas; (4) the mass of the gas atoms; (5) the number of gas
atoms.
A.
A
B.
C.
D
D.
E.
11, 2,
2 and 3
1 and 2
4 and 5
2 and 3
All of 1–5
Slide 11-8
Reading Quiz
What is the mass, in u, of a molecule of carbon dioxide, CO2?
A.
B
B.
C.
D.
E
E.
12
24
32
36
44
Each nucleon has a mass of about 1 u.
Most carbons has 6 protons and 6 neutrons
for a total of 12 nucleons or m=12u.
Most Oxygens have 8 protons and 8 neutrons
for a total of 16 nucleons or m=16u.
The total mass is 12+2*16 u = 44 u
Slide 12-12
Kinetic properties of an ideal gas
•
Consider the gas volume at shown on the right. A
molecule with mass m is elastically scattered by the
wall which supplies an impulse force F1.
F1Δt = mvx − m( − v x ) = 2mvx  F1 =
•
•
2mvx
Δt
wall
After a time Δt=2d/(vx), the mass bounce off the far
wall and return to bounce of this wall again. If we
use this value for Δt in the expression above, we
can get the average force for much large time
intervals.
2mv
F1 =
•
wall
x
(2d / v x )
d
A
= mvx2 / d
Multiplying by the number of molecules in the gas and dividing by the area A,
we get the pressure on the wall. P = N mv 2 = N mv 2
Ad
x
V
x
This would be correct if each molecule was moving the x direction with
exactly velocity vx. To correct for the differences in velocities, we use the
g <vx2> instead. Also <v2>= <vx2>+ <vy2>+ <vz2>=3 <vx2>. Thus”
average
P=
N
2 N 1
2N
2 
=
< KE >
m v2 / 3 =
m
v


V
3 V 2
 3V
Average kinetic energy of gas molecules
•
If we combine the last expression with the ideal gas law equation of
state, we get a useful expression for the means kinetic energy of gas
molecules:
R
2N
n
N
P=
< KE > P = RT N = nN A k B ≡
 P = k BT
NA
3V
V
V
2N
N
1
1
3

< KE >= k B T  < KE >= m v 2 ≡ mv 2rms = k B T v rms =
3V
V
2
2
2
•
3k BT
m
note: increasing temperature increases thermal (KE) energy
(adding heat energy Q)
Example: If the translational rms speed of water molecules (H2O) in a
volume
l
off air
i is
i 648 m/s,
/ what
h is
i the
h translational
l i l rms speedd off carbon
b
dioxide (CO2)? Both gases are at the same temperature.
(vrms≡[<v2>]1/2)
v rms,H 2O
 v rms,CO2 =
3k B T
=
m H2O
m H2O
m CO2
v rms,H2O =
v rms,CO2
3k BT
=
=
m CO2
N A m H2O
N A mCO2
v rms,H2O =
m H2O 3k BT
=
m CO2 m H2O
m H2O
m CO2
3k B T
m H 2O
2 + 16
( 648m / s) = 414m / s
12 + 2 ⋅16
quiz
•
Consider
C
id a gas mixture
i t
off the
th two
t monoatomic
t i gases: 50% Helium
H li
(Mmole=4.0 g; Mmole is the mass of one mole of the gas.) and 50%
Argon (Mmole=36.0 g). The ratio vrms,He / vrms,Ar of the rms speed for
helium atoms divided by the rms speed for Argon atoms is:
– a) .06
3k BT
3k BT
– b) .17
v rms,Ar
=
v
=
rms Ar
rms He
rms,He
m Ar
m He
– c) .67
– d) 1.3
3k BT
– e)) 3.0
1
v
m He
m He m He m Ar
 rms,He =
=
1 m He m Ar
v rms,Ar
3k BT
m Ar
m Ar
=
m Ar
36
=
=3
m He
4
Speed and Kinetic Energy of Gas Molecules
1 2
Kavg = mv rms
<KE>
2
<KE>
2K
avg
T=
3 kB
vrms =
3kBT
m
kB = 1.38 × 10 −23 J/K
 m 
F(v) = 4π 
 2π k BT 
3/2
 mv 2 
v exp  −
 2k T 
2
B
Slide 12-17
The Ideal Gas Model
2 <KE>
Kavg
T=
3 kB
3
E th = N < KE >= Nk BT
2
Transfer of heat at constant
volume:
Slide 11-18
Checking Understanding
Two containers of the
same gas (ideal) have
these masses and
temperatures:
• Which gas has
atoms with the
l
largest
t average
thermal
th
l
energy?
• Which container of gas has the largest thermal energy?
A. P, Q
1
3
2
< mv >= k BT, which is independent of mass
B. P, P
2
2
C. Q, P
Q hhas the
h hi
highest
h T  highest
hi h thermal
h
l
D. Q, Q
energy per atom
3
th P
E th = Nk BT  E th,P
2
E th,Q
3
N P k BTP
= 2
3
N Q k BTQ
2
=
N P TP 100 273 + 0
=
>1
N Q TQ
20 273 + 50
Slide 11-21