•  Main points :

•  Temperature and thermometers •  Thermal expansion

•  Ideal gas Law:

Some slides relevant to Wed. Lecture. The notes for the lecture given by Dr. Nagy can be found at

TVVTAATLL

Δ=ΔΔ=ΔΔ=Δ

0

0 ; ;β

γα

PV = nRT = NkBT

Physics 231 Lecture 29

Heating something increases its thermal energy •  Every body of matter (solid, liquid or gas) consists of atoms in rapid motion •  Heating increases the random motion of these particles. This additional random

motion adds to the thermal energy part of the kinetic and potential energies of the individual molecules/atoms making up the matter

•  Solids –  Individual particles locked in place/connected to neighbors via spring like

bonds –  Vibrate about equilibrium position (kinetic + potential energy)

Temperature is the observable quantity associated with thermal motion

Zeroth Law of Thermodynamics

•  If objects A and B are in thermal equilibrium with a third object, C, then A and B are in thermal equilibrium with each other.

•  This will mean that these objects are at the same temperature, an abstract concept that we will now make more concrete with examples of thermometers that change their properties with temperature.

Different Temperature Scales

•  Three different temperature scales are commonly in use: –  Fahrenheit –  Celsius –  Kelvin

•  Celsius define such that the freezing point of water is at a temperature of about 0o C and boiling temperature is at about 100o C.

•  The conversions from Celcius to Farenheit and to Kelvin are as follows:

15.273

3259

+=

+=

CK

CF

TT

TT

•  At TK=0, molecules in gas are nearly at rest and many material have unusual properties. What is this temperature in Fahrenheit? a) 00 F b) -2730 F c) – 4600 F d) -9230 F

TC = TK − 273.15= −273.150C

TF =95

TC + 32

= 95

−273.15( ) + 32 = −459.670 F

Thermometers

•  While we know what is hot or cold. Much of what we think is simply perception: –  Wind makes day “feel” colder because moving air strips away the

insulating layer of warm air next to our skin. –  Cold metal feels colder than equally cold wood. –  Humid winter days feel colder than dry winter days. How do we

construct a reliable thermometer? •  Need a quantity that has a unique dependence on

temperature. Example the length of a column of mercury in a thermometer grows linearly with temperature.

•  Many solids and liquids expand linearly with T: •  Example: A aluminum rod which is 1.0000 m long at 0°C.

What is its length at 100 °C? (αAl=23x10-6 / °C) a) 0.9989 m b) 1.0011 m c) 1.0023 m d) 1.0045 m

TLL Δ=Δ 0α

0L L TαΔ = Δ ( )( )( )6 o o23x10 / C 1.0000m 100 C−=L 0.0023mΔ =

0L L L= + Δ 1.0023m=

Temperature and Thermal Expansion

Slide 12-42

Area and Volume expansion •  The areas and volumes of many objects also expand with temperature. •  Example: Calculation of the area expansion coefficient ! for a

rectangular aluminum plate.

x

y

( )( ) TyyTxxyyxxxyAf Δ=ΔΔ=ΔΔ+Δ+== 0000 ; ; αα( )( ) ( )( )TTyxTyyTxxAf Δ+Δ+=Δ+Δ+=⇒ αααα 11000000

⇒ Af = A0 1+ 2α ⋅ΔT +α 2ΔT 2( )⇒ΔA = Af − A0 = 2α ⋅ΔTA0

ΔA = γ ⋅ΔTA0 where γ = 2α•  By analogy, the volume expansion coefficient β=3α for a uniform

material, where: •  Quiz: A swimming pool contains 110 m3 ( about 30,000 gal of water). The

sun heats the water from 17 to 27°C. What is the change in the volume of the water? β=2.07x10-4 / °C. –  a) 0.08 m3

–  b) 0.13 m3

–  c) 0.17 m3

–  d) 0.23 m3

0TVV Δ=Δ β

30V 110m = 0V V T βΔ = Δ -4 o 2.07x10 / Cβ =

( )( )3 -4 o o 3V 110m 2.07x10 / C 10 C 0.228mΔ = =

Example

•  A brass ring of inner diameter 10.00 cm at 20.0°C is heated and slipped over an aluminum rod of outer diameter 10.01 cm at 20.0°C. Assuming the average coefficients of linear expansion are constant, to what temperature must this combination be cooled to separate them? (αAl=24x10-6 /oC, αBr=19x10-6 /oC)

Br AlWant D D>

Br Br,0 BrD D D ; = + Δ Al Al,0 Al D D D= + Δ

Br,0 Br Al,0 AlD D D D+ Δ > + Δ

Br,0 Al,0 Al BrD D D D⇒ − > Δ − Δ

Al Al Br Br10.00 10.01 0.01 D T D Tα α− = − > Δ − Δ ( )Al Al Br BrD D Tα α= − Δ

( )6 6 50.01 (24x10 10.01 19x10 10.00) T 5.0x10 T− − −⇒ − > ⋅ − ⋅ Δ = Δ

05

0.01 C T5.0x10−−⇒ > Δ 0199 C T⇒ − > Δ

0fT 20 C T⇒ = + Δ 0179 C= −

The special case of Ice •  Many material contract when changing from liquid to solid. Ice is

an exception. The lowest energy configuration of the solid occupies a larger volume than the same mass of liquid.

•  The fact is key to spread of life across the planet because it means that ice is on the surface of large bodies of water where it will melt during summer rather than on the bottom.

Phases of Matter

Slide 12-16

Reading Quiz 2.  A sample of nitrogen gas is in a sealed container with a constant

volume. Heat is added to the gas. The pressure

A.  increases B.  stays the same C.  decreases D.  can’t be determined with the information given

Slide 12-8

•  Ideal gas pressure depends linearly on temperature. •  Here n=number of moles of the gas. There are NA = 6.02x1023

molecules or atoms (if the atoms don’t combine into molecules) per mole. One mole = .0224 m3 of gas at T=0oC and P=1 atm.

•  R=8.31 J/(mole�K) •  Example: A molecular gas is contained in an 8.0-L vessel at a

temperature of 20°C and a pressure of 9.0 atm. (a) Determine the number of moles of gas in the vessel. (b) How many molecules are in the vessel?

Ideal gas

•  Ideal gas pressure depends linearly on temperature.

Kelvinin measured is T If

KTVnRbaTP =+=

PVa) nRT

= ( ) ( )( ) ( )

5 39 1.01x10 Pa .008m8.31J / K 293K

= 3.0 moles=

Ab) N nN= 233 6.02x10= ⋅ 241.8x10 molecules=

Example

•  Gas is confined in a tank at a pressure of 10.0 atm and a temperature of 15.0°C. If half of the gas is withdrawn and the temperature is raised to 65.0°C, what is the ratio of the final density over the initial density?

•  What is the new pressure in the tank?

00

N m V

ρ = ff

N m V

ρ =0

1 N m2

V= 0

12ρ=

00 B 0

NP k T V

= ff B f

N P k T V

=

fB f

f

00B 0

N k TP VNP k TV

= f f

0 0

N TN T

= 1 338 0.592 288

= =

fP 0.59 10atm= ⋅ 5.9atm=