Physics 2101
Section 3
May 3rd: Chap. 19
Announcements:
• Final Exam: May
y 11th
(Tuesday), 7:30 AM at Howe
HoweRussell 130
• Make up Final: May 15th
(Saturday) 7:30 AM at Nicholson
119
• Final Exam for those who need
extended
d d time
i
will
ill b
be at
Nicholson 109
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101--3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
Pressure, Temperature, & RMS speed
Assume the collision of the gas molecule with the wall is h
ll
f h
l l
h h
ll
elastic then:
Δpx = (−mvx ) − (mvx ) = −2mvx
The molecule travels to the back wall, collides and comes back. The time it takes is 2L/vx.
mvx2
Δp 2mvx
=
=
L
Δt 2 L / vx
But the pressure is F/A
n
2
xi
n
Fx
mv
1
m
p=
= 2∑
= 3 ∑ vxi2
A L i =1 L
L i =1
nM (vx2 )avg
nmN A 2
p=
(vx )avg =
3
L
V
2
nM (v 2 )avg nMv
M rms
=
=
3V
3V
If we calculated the average velocity and If
we calculated the average velocity
and
(vx2 )avg
(v
use the fact that the number in the sum is nNA then: v 2 = vx2 + vy2 + vz2
2
v
v =
3
2
x
(v )avg = vrms
2
RMS = Root‐Mean‐Square
M ‐‐‐Molar mass
RMS Speeds
We have
p=
For ideal gas
( )
nM v 2
3V
pV = nRT
1/ 2
⎛ 3 pV ⎞
vrms = ⎜
⎟
⎝ nM ⎠
vrms
ave
1/ 2
⎛ 3nRT ⎞
=⎜
⎟
⎝ nM ⎠
3RT
(19-22)
=
M
The RMS velocity depends on:
Molar mass & Temperature
Problem 19-3: Here are five numbers: 5, 11, 32, 67, and 89.
(a) What is the average value navg?
(b) Wh
Whatt iis th
the rms value
l nrms off the
th numbers?
b ?
(a)
navg
5 + 11 + 32 + 67 + 89
=
= 40.8
5
(b)
(n 2 )avg
1 n 2 5 2 + 112 + 32 2 + 67 2 + 89 2
= ∑ ni =
= 2714.41
n i =1
5
(n 2 )avg = 52.1
52 1
Problem 19-18: Calculate the rms speed of helium atoms at 1000K.
Helium has 2 protons and 2 neutrons, what is the molar mass?
Appendix F: M = 4.00 x10−3 kg / mol
U E
Use
Eqn 19-22
19 22
vrms
vrms
3RT
=
M
3 ( 8,31J / molK )(1000 K )
3RT
3
=
=
=
2.50
x
10
m/s
−3
M
4.00 ×10 kg / mol
Translational Kinetic Energy
The kinetic energy of a gas molecule K =
Its average kinetic energy K avg
Thus K avg =
⎛ mv 2 ⎞
=⎜
⎟
2
⎝
⎠avg
2
mv
.
2
2
mvrms
=
.
2
K avg
3kT
=
2
m 3RT 3RT
.
=
2 M
2NA
We finally get:
K avg
3kT
=
2
At a given temperature T all ideal gas molecules, no matter what their mass,
have the same average translational kinetic energy. When we measure the
temperature of a gas, we are also measuring the average translational
kinetic energy of its molecules.
Problem 19-26: What is the average translational kinetic
energy of nitrogen molecules at 1600 K?
Average Kinetic energy is
3
K avg = kT
2
K avg
(
)
3
3
= kT = 1.38x10 −23 J (1600K ) = 3.31x10 −20 J
2
2
Molar Specific Heat: Monatomic Ideal Gas
Molar Specific Heat at Constant Volume (Wby
Molar Specific Heat at Constant Volume (W
0)
b =0)
Q = nCV ΔT
&
Wby = 0
ΔE int = Q = nCV ΔT
Always true if
V = const.
= const
ΔE int = n(32 R)ΔT = n(CV )ΔT = Q
CV , monatomic
3
= R = 12.5J/mol ⋅ K
2
Molar Specific Heat: Monatomic ideal gas
Molar Specific Heat at Constant Volume (W
l
f
l
( by=0))
ΔE int = Q = nCV ΔT
CV ,monatomic
Always
y true for const. V
3
= R = 12.5 ⋅ J/mol ⋅ K
2
Molar Specific Heat at Constant Pressure (Wby=pΔV)
ΔE int = Q − W = nC p ΔT − pΔV
nCV ΔT = nC p ΔT − nRΔT
CV = C p − R
C p,monatomic =
or
C p = CV + R
5
R = 20.8 ⋅ J/mol ⋅ K
2
Molecular Specific heat Constant Volume CV
CV ,monatomic
3
= R = 12.5 ⋅ J/mol ⋅ K
2
ΔEint
i t = nCV ΔT
Work Done by Isothermal (ΔT = 0)
Expansion/Compression of Ideal Gas
On p‐V graph, the green lines are isotherms…
… each green line corresponds to a system at a constant temperature.
From ideal gas law this means that for a given isotherm:
From ideal gas law, this means that for a given isotherm:
pV = constant
The work done by the gas is then: Wby =
Vf
∫
Vi
V
⎛ nRT ⎞
dV
pdV = ∫ ⎜
dV
=
nRT
⎟
∫
⎝
⎠
V
V
V V
Vf
f
i
i
⇒ p = (nRT )
1
V
Relates p and V
⎛Vf ⎞
⇒ Wby,isothermal = nRT ln⎜ ⎟
ΔT =0
⎝ Vi ⎠
⎛p ⎞
= nRT ln⎜ i ⎟
⎝ pf ⎠
Adiabatic Expansion of an ideal gas
Because a gas is thermally insulated, or expansion/compression happens suddenly ⇒ adiabatic
Remember “Adiabatic means Q = 0”
or, by 1st Law of Thermo ⇒ ΔEint = ‐Wby
In this case pVγ = constant where γ = Cp/CV = (R+ CV )/CV
example: monatomic gas γ = 5/3
p1V1 = p2V2
γ
γ
T1V1γ −1 = T2V2γ −1
Compare with Isothermal Expansion (ΔT = 0)
Compare with Isothermal Expansion (ΔT 0)
T1 = T2 ⇔
[p1V1 = p2V2 ]isothermal
Adiabatic Expansion of an ideal gas
Remember “Adiabatic means Q = 0”
or, by 1st Law of Thermo ⇒ ΔEint = ‐Wby
pVγ = constant
P f
Proof
dEint = Q − pdV
dEint = − pdV = nCV dt
⎛ p
ndT = − ⎜
⎝C
V
⎞
⎟⎠ dV
pV = nRT
or pdV+Vdp=nRdT
Remember C p − CV = R
giving
PdV + Vdp
ndT =
C p − CV
⎛ p
−⎜
⎝C
⎞
PdV + Vdp
dV
=
⎟⎠
C p − CV
V
dp ⎛ C p ⎞ dV
+⎜ ⎟
=0
p ⎝ CV ⎠ V
dp
dV
+γ
=0
p
V
Integrating gives
ln p + γ lnV = Constant
or pVγ = Constant
γ
piVi = p f V f
TV
i i
γ −1
γ
= Tf V f
γ −1
Adiabatic Expansion of an Ideal Gas
Consider the ideal gas in fig. a. The container is well
i l t d When
insulated.
Wh the
th gas expands,
d no heat
h t is
i transferred
t
f
d
to or from the gas. This process is called adiabatic.
Such a process is indicated on the p - V diagram of fig. b
by the red line. The gas starts at an initial pressure pi and
initial volume Vi . The corresponding final parameters
are p f and V f . The process is described by the equation
piVi = p f V f . Here the constant γ =
γ
γ
Cp
CV
.
Using the ideal gas law we can get the equation
TV
i i
γ −1
= Tf V f
γ −1
Vi γ −1
→ T f = Ti γ −1
Vf
If V f > Vi we have adiabatic expansion and T f < Ti .
If V f < Vi we have adiabatic compression and T f > Ti .
Ti = T f
piVi = p f V f
Free Expansion
In a free expansion, a gas of initial volume Vi and initial
pressure pi is
i allowed
ll
d to
t expand
d in
i an empty
t container
t i
so that the final volume is V f and the final pressure p f .
In a free expansion Q = 0 because the gas container is insulated. Furthermore,
since the expansion takes place in vacuum the net work W = 0.
The first law of thermodynamics predicts that ΔEint = 0.
Since the gas
gas is assumed to be ideal there is no change in temperature: Ti = T f .
Using the law of ideal gases we get the following equation,
which connects the initial with the final state of the gas:
piVi = p f V f .
3nRT
Eint =
2
Internal Energy of an Ideal Gas
Consider a monatomic gas such as He, Ar, or Kr. In this case the internal energy
Eint of the gas is the sum of the translational kinetic energies of the constituent
atoms. The average translational kinetic energy of a single atom is given by the
3kT
. A gas sample
l off n moles
l contains
i N = nN
N A atoms.
2
nN A 3kT 3nRT
=
.
The internal energy of the gas Eint = NK avgg =
2
2
equation
i K avg =
The equation above expresses the following important result:
The internal energy Eint of an ideal gas is a function of gas temperature
only; it does not depend on any other parameter.
Work Done by Isobaric (ΔP = 0) Expansion of an Ideal Gas
p
⇒ Wby,isobaric = pΔV
ΔP=0
= nRΔT
Problem 19-6: A quantity of ideal gas at 10.0 0C and 100 kPa
occupies a volume of 2.50 m3. (a) How many moles of the gas
are present?
t? (b) If th
the pressure iis now raised
i d tto 300 kPa
kP and
d
the temperature to 30.0 0C, how much volume does the gas
occupy?
(100x10 Pa )(
)(2.50m )
3
3
p
pV
n=
=
= 106moles
106 l
RT (8.31J / mol • K )(283K )
(b)
p f Vf
piVi
=
Tf
Ti
⎛ pi ⎞ ⎛ T f ⎞
V f = Vi ⎜ ⎟ ⎜ ⎟ = 0.892m 3
⎝ p f ⎠ ⎝ Ti ⎠
Remember special cases…
Ideal Gas:
ΔEint = 32 nR ( ΔT )
PV = nRT
Adiabatic expansion/contraction ‐ NO TRANSFER OF ENERGY AS HEAT Q = 0
[ΔE int = −W ]adiabatic
Constant‐volume processes (isochoric)‐
NO WORK IS DONE W = 0
NO WORK IS DONE W=0
ΔE int = Q
Constant‐pressure processes
ΔE int = Q − pΔV
CV = C P − R
3) Cyclical process (closed cycle)
a) net area in p‐V curve is Q
Wby =
V f =Vi
∫ pdV = 0
Wby =
∫ pdV
= area under p − V graph
Vi
QΔV = 0 = nCV ΔT
Wby =
V f =Vi
∫ pdV = pΔV
Vi
QΔP = 0 = nC
C P ΔT
ΔEint,closed cycle =0
ΔE int = 0 ⇒ Q = W
19#46: One mole of an ideal atomic gas goes from a to c along the diagonal pate. The scale of the g
g
p
vertical axis is set by pab=5.0 kPa and pc=2.0 kPa, and the scale of the V axis is Vbc4.0 m3 and Va=2.0 m3. (a) what is the change in the internal Energy? (b) How much Energy is added to the gas? (c) How much heat is required if the gas goes from a to c via abc?
For any straight line on pv plot
it is easy to prove
Wstraight
⎛ 3⎞
Eint = n ⎜ ⎟ RT
⎝ 2⎠
⎛ p f + pi ⎞
=⎜
ΔV
⎟
⎝ 2 ⎠
⎛ 3⎞
⎛ 3⎞
(a) Eint c − Eint a = ⎜ ⎟ (Tc − Ta ) = ⎜ ⎟ (pcVc − paVa ) = −3000J
⎝ 2⎠
⎝ 2⎠
⎛ p + pc ⎞
(c) ΔEint found in (a)
(b) W = ⎜ a
V
−
V
=
7000J
(
)
a
⎝ 2 ⎟⎠ c
W = 5.0x10 3 Pa 2m 3 = 10000J
Q = ΔEint + W = 2000J
Q = 5000J
(
)( )
19 #63: 1.00 mol of an ideal monatomic gas goes through the cycle shown in the Figure. The temperatures are T1
the cycle shown in the Figure. The temperatures are T
=300 K, T2=600 K, and T3= 455K. For 1 to 2, what are (a) heat Q, (b) the change in internal energy, and (c) the work done W? For 2 to 3, what are (d) Q, (e) change in Eint, and (i) W? F th f ll
(i) W? For the full cycle, what are (j) Q, (k) change in E
l
h t
(j) Q (k) h
i Eint, and (l) W. The Initial pressure at point 1 is 1.00 atm. What are the (m) volume and (n) pressure at point 2 and the (o) volume and (p) pressure at point 3.
(p) p
p
(a, b, c) For 1 ⇒ 2, ΔV = 0, W = 0, ΔEint = n
(d, e, f) For 2 ⇒ 3, Q = 0, ΔEint = n
For 3 ⇒ 11, ΔEint = n
Q = ΔEint + W =
( R ) ΔT < 0;
3
2
5
nRΔT ;
2
( R ) ΔT = n ( C ) Δ T = Q ;
3
2
( R ) ΔT = −W ;
3
2
V
Here ΔT = T3 − T2
W = pΔV = nRΔT < 0;
Here ΔT = T1 − T3
Here ΔT = T2 − T1
Problem 19-21: (a) Compute the rms speed of a nitrogen molecule at
20.0 0C. Each N atom has 7 protons and 7 neutrons? (a) what is the rms
speed at 300K and 20
20.0
0 0C.
C At what temperatures will the rms speed be
(b) half that value and (c) twice that value?
Table 19-1: M = 28 g / mol
(a) Use Eqn 19-22
3RT
M
= 517 m / s (for T = 300 K )
vrms =
vrms
v1 = 517 m / s for T = 300K
v2 = ( 517 m / s )
293K
300 K
Set up ratios
v2
3RT2 / M
T
=
= 2
v1
3RT1 / M
T1
(b) Set v3 =
v2
and solve
2
T3 = 73K
v2 = 511m / s
(c) Set v 4 = 2v2 and solve
T4 = 1170 K