Physics 2101
Section 3
April 26th: Chap. 1818-19
Announcements:
Ann n m nt :
• Exam #4, April 28th
(Ch. 13.613.6-18.8)
• Final Exam: May 11th
(Tuesday), 7:30 AM
• Make
M k up Final:
F l May
M
15th
(Saturday) 7:30 AM
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
The First Law of Thermodynamics
Eint = sum total of the energy of particles (molecules/atoms) in system
Internal Energy or Thermal Energy
• Eint ↑(increases) if work done to
↑
system or heat added to system
• Eint ↓(decreases) if work done by system or heat taken from system
• Although W and Q are path‐dependent, Eint is not.
ΔE int = E int, f − E int,i = Q − Wby
dE int = dQ − dWby
In thermodynamics Work is defined as done by system:
In thermodynamics, Work is defined as done by
ΔE int = Q + Won
Path Dependence of Work and ΔEint
Wby =
Vf
∫ dW = ∫ pdV
Vi
ΔE int = E int, f − E int,i = Q − W
isobaric: W = pΔV
isochoric: W = 0
isochoric: W = 0
Volume increases,
Pressure decreases:
area > 0 ‐> W > 0
(gas expands) Two step:
Volume increases then
l
h
Pressure decreases:
area > 0 ‐> W > 0
Work and Heat are NOT CONSERVATIVE: depends on path
Work
and Heat are NOT CONSERVATIVE: depends on path
ΔEint does NOT depend on path !!
NET WORK, Wnet, done by
system during a complete cycle is
system during a complete cycle is
shaded area. It can be pos., neg, or zero depending on path. Around a closed cycle
ΔEint is zero.
dE int = dQ − dWby
Figure shows four paths on p‐V
g
p
p
diagram along which a gas can be taken from state i to state f. Rank:
1) ΔEint ?
2) Work done by gas?
3) Heat transferred?
Special cases of First Law of Thermodynamics
ΔE int
= E int,
− E int,i
=Q −W
i t
i t f
i ti
1)
Adiabatic processes ‐ NO TRANSFER OF ENERGY AS HEAT Q = 0
a) rapid expansion of gasses in piston ‐ no time for heat to be transferred
b) if work is done by system (W>0), then ΔEint decreases
b) if work is done by system (W>0), then ΔE
c) NOTE: temperature changes!!
[ΔE int = −W ]adiabatic
2)
Constant‐volume processes
Constant‐volume
processes (isochoric)‐ NO WORK IS DONE W = 0
NO WORK IS DONE W = 0
a) if heat is absorbed, the internal energy increases
b) NOTE: temperature changes!!
3)
Cyclical process (closed cycle) ΔEint,closed cycle =0
a) net area in p‐V curve is Q
4)
Free Expansion : adiabatic process with no transfer of heat
a) happens suddenly
b) no work done against vacuum ΔE int = Q = W = 0
c) non‐thermal equilibrium process
5)
Isothermal: Temperature does not change
Wby =
V f =Vi
∫ pdV = 0
Vi
ΔE int = Q
ΔE int = 0 ⇒ Q = W
We’ll talk about this later…
Sample Problem 18‐5
Let 1.0 kg of liquid at 100°C be converted to steam at 100°C by boiling
at twice atmospheric pressure
tt i
t
h i
(2 t )
(2 atm) as shown. The volume of the h
Th
l
f th
‐3
3
water changes from an initial value of 1.0×10 m as a liquid to 1.671 m3
as a gas.
Here, energy is transferred from the thermal reservoir as heat until the liquid water is changed completely to steam. Work is done by the expanding
gas as it lifts the loaded piston against a constant atmospheric pressure.
a)
b)
c)
How much work is done by the system during the process?
How do we calculate work?
Wby =
How much energy is transferred as heat during the process?
What is the heat added?
What is the change in the system’s internal energy during the process?
Using 1st Law of Thermo:
Using 1st Law of Thermo:
Vf
∫ pdV = p(V
f
− Vi )
Vi
= (2 ⋅ atm)(1.01 ×10 5 ⋅ N/m 2 atm)(1.671⋅ m 3 − 0.001⋅ m 3 )
= 338 ⋅ kJ
no temperature change → only phase change
Q = mLV = (1.0 ⋅ kg)(2256 ⋅ kJ/kg) ≅ 2260 ⋅ kJ
ΔE int = Q − Wby
= 2260 ⋅ kJ − 338 ⋅ kJ ≅ 1920 ⋅ kJ
Positive! Energy mostly (85 %) goes into
separating H2O molecules
More Example…
18‐43:
18
43 Gas within a closed chamber undergoes the cycle shown in the p‐V
Gas within a closed chamber undergoes the cycle shown in the p V
diagram. Calculate the net energy added to the system as heat (Q) during on complete cycle.
In one complete cycle, ΔEint,cycle = 0 so Q=W. To find Q, calculate W!
Wby = WA →B + WB →C + WC →A
=
VB
∫p
A →B
dV +
VA
=
VC
∫p
B →C
VB
4
∫(
20
3
V+
1
10
3
1
2
C →A
dV
1
)dV + ∫ (30)dV + ∫ p
C →A
4
(( )
20
3
∫p
VC
1
=
dV +
⎛ 20
10 ⎞
p A →B (V ) = ⎜ V ⋅ m -3 + ⎟ ⋅ Pa
⎝ 3
3⎠
p B →C (V ) = 30 ⋅ Pa
ΔVC →A = 0
VA
1
4 ⋅ m3
)1⋅ m
V 2 + 103 (V )
dV
1⋅ m 3
+ 30(V ))
+ 0 = −30 ⋅ J
3
3 (
4⋅m
W = Q = ‐ 30 J
Or just add up area enclosed!
CCW ‐ negative CW ‐ positive
Chapt. 19: Kinetic Theory of Gases
Kinetic Theory of Gases
Thermodynamics = macroscopic picture
Gases = micro Gases micro ‐> macro picture
macro picture
IDEAL GAS LAW
n = number of moles
p = nRT p
pV
pV
= NkT
N = number of particles
k = 1.38×10‐23 J/K
R = kNA
R = 8.315 J/(mol⋅K) = 0.0821 (L⋅atm)/(mol⋅K) 0 0821 (L atm)/(mol K)
= 1.99 calories/(mol⋅K) Monoatomic ideal gas : He, Ar, Ne, Kr… (no potential energies)
3
3
E int,monotonic
=
N
kT
=
(2 ) 2 nRT
int monotonic
ΔE int,monotonic = 32 nR(ΔT )
The internal energy
of an ideal gas depends f id l
d
d
only on the temperature
Avogadro’ss Number
Avogadro
Number
One mole is the number of atoms in 12 g sample of carbon‐12
C(12)—6 protrons, 6 neutrons and 6 electrons
12 atomic units of mass assuming mP=mn
NA=6.02 x 1023 mol‐1
So the number of moles n is given by n=N/NA
Another way to do this is to know the mass of one molecule: then
N=M(sample)/mNA
19-1: Gold has a molar mass of 197 g/mol. (a) How many
moles of gold are in 2.50g sample of pure gold? (b) How
many atoms are in the sample? Au has 79 protons and ~118
neutrons.
n is number of moles (19-3)
n=
M sample
M (molar
( l mass))
≡
M sample
m(one
(
atom)
t ) NA
M sample
2.5g
n
n=
=
= 0.0127 mol.
M (molar mass) 197g / mol.
(b) Number of atoms Eqn.
Eqn 19.2
19 2
n=
N
or N=n • N A
NA
N = ( 0.0127 ) ( 6.02 x1023 ) = 7.64 x1021
19.2: Find the mass in kilograms of 7.5 x 1024 atoms of arsenic, which
has
as a molar
o a mass
ass o
of 74.9
9 g/
g/mole.
o e As
s has
as 33 p
protons
oto s a
and
d ~42 neutrons
eut o s
M
NA
Where M is the molar mass
Each atom has a mass m=
The molar mass of As is 74.9 g/mol.
7 5 10 )(74.9x10
7.5x10
74 9 10 kg
k / moll )
(
Total M=
(6.02x10 mol )
−3
3
24
24
M=0.933
M
0.933 kg
−1
19-3: The best laboratory vacuum has a pressure of about 1.00 x
10-13 Pa or 10-18 atm. How manyy g
gas molecules are there p
per
cubic centimeter in such a vacuum at 293 K?
Get units straight
−6
V=1.0x10 m
Ideal gas Law
3
pV = nRT
p
p = 10 −13 Pa
R gas constant
T=293 K
R=8.31 J/mol. • K
(10
−13
Pa )(10−6 m3 )
p
pV
n=
=
= 4.1
4 1x10−233 mole
l
RT ( 8.31J / mol.K )( 293K )
N = nN A = ( 4.1x10−23 mole )( 6.02 x1023 mole −1 ) = 25 molecules