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Transcript of Physics 1501: Lecture 36, Pg 1 Physics 1501: Lecture 36 Today’s Agenda l Announcements çHomework...
Physics 1501: Lecture 36, Pg 1
Physics 1501: Lecture 36Physics 1501: Lecture 36TodayToday’’s Agendas Agenda
AnnouncementsHomework #12 (Dec. 9): 2 lowest droppedMidterm 2 … in class Wednesday
Honors’ students: Wednesdat at 3:30 in my office.
Today’s topicsChap. 17: ideal gas
» Kinetic theory
» Ideal gas law
» Diffusion Chap.18: Heat and Work
» Zeroth Law of thermodynamics» First Law of thermodynamics and applications» Work and heat engines
Physics 1501: Lecture 36, Pg 2
Kinetic Theory of an Ideal GasKinetic Theory of an Ideal Gas Pressure is
Physics 1501: Lecture 36, Pg 3
Does a Single Particle Have a Temperature?
Each particle in a gas has kinetic energy. On the previous page, we have established the relationship between the average kinetic energy per particle and the temperature of an ideal gas.
Is it valid, then, to conclude that a single particle has a temperature?
Concept of temperatureConcept of temperature
Physics 1501: Lecture 36, Pg 4
Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0u) and oxygen O2 molecules (molecular mass 32.0u). Assume that each behaves as an ideal gas and determine
the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K.
For nitrogen
Example: Example: Speed of Molecules in Air
Physics 1501: Lecture 36, Pg 5
Internal energy of a monoatomic ideal gasInternal energy of a monoatomic ideal gas The kinetic energy per atom is
Total internal energy of the gas with N atoms
Physics 1501: Lecture 36, Pg 6
Kinetic Theory of an Ideal Gas: summaryKinetic Theory of an Ideal Gas: summary
Assumptions for ideal gas:Number of molecules N is largeThey obey Newton’s laws (but move
randomly as a whole)Short-range interactions during elastic
collisionsElastic collisions with wallsPure substance: identical molecules
Temperature is a direct measure of average kinetic energy of a molecule
Microscopic model for a gas Goal: relate T and P to motion of the
molecules
Physics 1501: Lecture 36, Pg 7
Theorem of equipartition of energy
Each degree of freedom contributes kBT/2 to the energy of a system (e.g., translation, rotation, or vibration)
Kinetic Theory of an Ideal Gas: summaryKinetic Theory of an Ideal Gas: summary
Total translational kinetic energy of a system of N molecules
Internal energy of monoatomic gas: U = Kideal = Ktot trans
Root-mean-square speed:
Physics 1501: Lecture 36, Pg 8
Consider a fixed volume of ideal gas. When N or T is doubled the pressure increases by a factor of 2.
11) If T is doubled, what happens to the rate at which ) If T is doubled, what happens to the rate at which a single a single moleculemolecule in the gas has a wall bounce? in the gas has a wall bounce?
b) x2a) x1.4 c) x4
22) If N is doubled, what happens to the rate at which ) If N is doubled, what happens to the rate at which a single a single moleculemolecule in the gas has a wall bounce? in the gas has a wall bounce?
b) x1.4a) x1 c) x2
Lecture 36: Lecture 36: ACT 1ACT 1
Physics 1501: Lecture 36, Pg 9
DiffusionDiffusion The process in which molecules move from a region
of higher concentration to one of lower concentration is called diffusion.Ink droplet in water
Physics 1501: Lecture 36, Pg 10
Why is diffusion a slow process ?Why is diffusion a slow process ? A gas molecule has a translational rms speed of hundreds of
meters per second at room temperature. At such speed, a molecule could travel across an ordinary room in just a fraction of a second. Yet, it often takes several seconds, and sometimes minutes, for the fragrance of a perfume to reach the other side of the room. Why does it take so long?Many collisions !
Physics 1501: Lecture 36, Pg 11
Comparing heat and molecule diffusionComparing heat and molecule diffusion Both ends are maintained at constant concentration/temperature
Physics 1501: Lecture 36, Pg 12
concentration gradientbetween ends
diffusion constant
SI Units for the Diffusion Constant: m2/s
FickFick’’s law of diffusions law of diffusion For heat conduction between two side at constant T
L
Th Tc A
Energy flow
conductivity temperature gradientbetween ends
The mass m of solute that diffuses in a time t through a solvent contained in a channel of length L and cross sectional area A is
Physics 1501: Lecture 36, Pg 13
Chap. 18: Work & 1Chap. 18: Work & 1stst Law Law
The Laws of Thermodynamics
0) If two objects are in thermal equilibrium with a third, they are in equilibrium with each other.
1) There is a quantity known as internal energy that in an isolated system always remains the same.
2) There is a quantity known as entropy that in a closed system always remains the same (reversible) or increases (irreversible).
Physics 1501: Lecture 36, Pg 14
Zeroth Law of ThermodynamicsZeroth Law of Thermodynamics
Thermal equilibrium: when objects in thermal contact cease heat transfersame temperature
T1T2
U1 U2
=
If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in
thermal equilibrium with each other.
AC
B
Physics 1501: Lecture 36, Pg 15
11st st Law: Work & HeatLaw: Work & Heat Two types of variables
State variables: describe the system
(e.g. T, P, V, U). Transfer variables: describe the process
(e.g. Q, W).
=0 unless a process occurs
change in state variables. Work done on gas
W = F d cos = -F y
= - PA y = - P Vvalid only for isobaric processes
(P constant)If not, use average force or calculus:
W = area under PV curve
PV diagram
Physics 1501: Lecture 36, Pg 16
11st st Law: Work & HeatLaw: Work & Heat Work:
Depends on the path taken in the PV-diagramSame for Q (heat)
Physics 1501: Lecture 36, Pg 17
b) = |W1|a) > |W1| c) < |W1|
i
f
p
V
21
Consider the two paths, 1 and 2, connecting points i and f on the pV diagram. The magnitude of the work, |W2|,
done by the system in going from i to f along path 2 is
Lecture 36: Act 2Lecture 36: Act 2WorkWork
Physics 1501: Lecture 36, Pg 18
First Law of ThermodynamicsFirst Law of Thermodynamics
Isolated systemNo interaction with surroundingsQ = W = 0 U = 0.Uf = Ui : internal energy remains constant.
First Law of Thermodynamics
U = Q + W
variation of internal energyheat flow “in” (+) or “out” (-)
work done “on” the system
Independent of path in PV-diagramDepends only on state of the system (P,V,T, …)Energy conservation statement only U changes
Physics 1501: Lecture 36, Pg 19
Other ApplicationsOther Applications
Cyclic process:Process that starts and ends at the stateMust have U = 0 Q = -W .
Adiabatic process:No energy transferred through heat Q = 0.So, U = W .Important for
» expansion of gas in combustion engines
» Liquifaction of gases in cooling systems, etc. Isobaric process: (P is constant)
Work is simply
Physics 1501: Lecture 36, Pg 20
Other Applications (continued)Other Applications (continued) Isovolumetric process:
Constant volume W =0.So U = Q all heat is transferred into internal energy
» e.g. heating a “can” (no work done).
Isothermal process:T is constantUsing PV=nRT, we find P= nRT/V.Work becomes :
PV is constant.PV-diagram: isotherm.
Physics 1501: Lecture 36, Pg 21
p
V
Identify the nature of paths A, B, C, and D Isobaric, isothermal, isovolumetric, and adiabatic
Lecture 36: Act 3Lecture 36: Act 3ProcessesProcesses
A
B
C
D
T1
T2
T3T4
Physics 1501: Lecture 36, Pg 22
Heat EnginesHeat Engines We now try to do more than just raise the temperature of an
object by adding heat. We want to add heat to get some work done!
Heat engines: Purpose: Convert heat into work using a cyclic process Example: Cycle a piston of gas between hot and cold
reservoirs* (Stirling cycle)
1) hold volume fixed, raise temperature by adding heat
2) hold temperature fixed, do work by expansion
3) hold volume fixed, lower temperature by draining heat
4) hold temperature fixed, compress back to original V
Physics 1501: Lecture 36, Pg 23
Heat EnginesHeat Engines Example: the Stirling cycle
GasT=TH
GasT=TH
GasT=TC
GasT=TC
V
P
TC
TH
Va Vb
1 2
34
We can represent this cycle on a P-V diagram:
11 2
34*reservoir: large body whose temperature does not change when it absorbs or gives up heat
Physics 1501: Lecture 36, Pg 24
Identify whether Heat is ADDED or REMOVED from the
gas Work is done BY or ON the gas for each
step of the Stirling cycle:
V
P
TC
TH
Va Vb
1 2
34
ADDEDREMOVED
BYON
1
HEAT
WORK
step
ADDEDREMOVED
BYON
2
ADDEDREMOVED
BYON
3
ADDEDREMOVED
BYON
4
Heat EnginesHeat Engines