Physics 231 Lecture 29 - NSCLlynch/phy231_2011/lecture29.pdf=−179 C0 The special case of Ice •...

12
Main points of todays lecture: Physics 231 Lecture 29 Main points of today s lecture: Temperature 0 celcius farenheit 5 T (T 32 ) 9 = Thermal expansion celcius farenheit celcius kelvin 9 T T 273.15 = Ideal gas properties: T V V T A A TL L Δ = Δ Δ = Δ Δ = Δ 0 0 ; ; β γ α Ideal gas properties: KE N v m N P T Nk nRT PV B 2 1 2 2 = = = = T k E K KE KE V v m V P B 2 3 3 2 3 = = = 2

Transcript of Physics 231 Lecture 29 - NSCLlynch/phy231_2011/lecture29.pdf=−179 C0 The special case of Ice •...

• Main points of today’s lecture:

Physics 231 Lecture 29• Main points of today s lecture:• Temperature

0celcius farenheit

5T (T 32 )9

= −

• Thermal expansion

celcius farenheit

celcius kelvin

( )9

T T 273.15= −p

• Ideal gas properties:TVV

TAATLLΔ=Δ

Δ=ΔΔ=Δ

0

0 ; ;β

γα

• Ideal gas properties:

KENvmNP

TNknRTPV B

212 2

=

=

==

TkEKKE

KEV

vmV

P

B23

323

=≡

=

=

2

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.

• Allows a definition of temperature• Allows a definition of temperature

Different Temperature Scales

Th diff t t t l• Three different temperature scales are commonly in use:– Fahrenheit– Celsius– Kelvin

• Celsius define such that the freezing gpoint 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:

• At TK=0, molecules in gas are nearly at rest and many material have unusual properties. What is this temperature in Fahrenheit?

T T 273 15 0273 15 C

15.273

3259

+=

+=

CK

CF

TT

TTa) 00 Fb) -2730 Fc) – 4600 F

C KT T 273.15= − 0273.15 C= −

F C9T T 325

= +15.273+CK TT )

d) -9230 F ( ) 09 273.15 32 459.67 F5

= − + = −

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 p p g ya 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.

TLL Δ=Δ 0αExample: 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 mb) 1 0011 m

0L L TαΔ = Δ ( ) ( ) ( )6 o o23x10 / C 1.0000m 100 C−=L 0 0023mΔ =b) 1.0011 m

c) 1.0023 md) 1.0045 m

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 γ for a

rectangular aluminum plate. ( )( ) TyyTxxyyxxxyAf Δ=ΔΔ=ΔΔ+Δ+== 0000 ; ; αα

y

f

( )( ) ( )( )TTyxTyyTxxAf Δ+Δ+=Δ+Δ+= αααα 11000000

( ) ααα 221 0022

0 Δ=−=ΔΔ+Δ+= TAAAATTAA ff

x αγγ 2where0 =Δ=Δ TAA• By analogy, the volume expansion coefficient β=3α for a uniform

material, where:0TVV Δ=Δ β

• 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.

β– a) 0.08 m3

– b) 0.13 m3

c) 0 17 m3

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 = =– c) 0.17 m3

– d) 0.23 m3

( ) ( )

Example

A b i f di t 10 00 t 20 0°C i h t d d• A brass ring of diameter 10.00 cm at 20.0°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0°C. Assuming the average coefficients of linear expansion are constant to what temperature must this combination be cooledconstant, to what temperature must this combination be cooled to separate them? (αAl=24x1010--66 //ooCC, , αBr=19x1010--66 //ooCC))

Br AlWant D D>

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

Br,0 Br Al,0 AlD D D D+ Δ > + Δ

Br,0 Al,0 Al BrD D D D − > Δ − Δ10.00 10.01−

( )Al Al Br BrD T D Tα α> Δ − Δ

( )A Al Br Br0.01 D D Tα α − > − Δ( )6 6 50.01 (24x10 10.01 19x10 10.00) T 5.0x10 T− − − − > ⋅ − ⋅ Δ = Δ

0 01− 0199 C Δ05

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

i Th l fi i f h lidan 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

In the solid form, ice, the molecules are farther apart than theyare in the liquid form, water.

Slide 12-16

Reading Quiz 2. A sample of nitrogen gas is in a sealed container with a constant2. A sample of nitrogen gas is in a sealed container with a constant

volume. Heat is added to the gas. The pressure

A. increasesB. stays the sameC. decreasesC. decreasesD. can’t be determined with the information given

Slide 12-8

Ideal gas

Id l d d li l t t• Ideal gas pressure depends linearly on temperature.

KTVnRbaTP =+=

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

Kelvinin measured is T If

g A 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 g ( ) ymolecules are in the vessel?

PVa) nRT

= ( ) ( )( ) ( )

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

= 3.0 moles=( ) ( )8.31J / K 293K

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

Example

G i fi d i t k t f 10 0 t d t t f• 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?

1 N0

0N m

V

ρ = ff

N m V

ρ =0N m

2 V

= 012

ρ=

• What is the new pressure in the tank?

0NP k T fNP k T00 B 0P k T

V= f

f B f P k T V

=

fB f

N k TP V f fN T 1 273 65 1 338+f

00B 0

P VNP k TV

= f f

0 0

N TN T

= 1 273 65 1 338 0.592 273 15 2 288

+= = =+

fP 0.59 10atm= ⋅ 5.9atm=