msu physics 231 fall 2015 1 physics 231 topic 12: temperature, thermal expansion, and ideal gases...

MSU Physics 231 Fall 2015 1

Physics 231Topic 12: Temperature,

Thermal Expansion, and Ideal Gases

Alex BrownNov 18-23 2015


3rd midterm

final Thursday 8-10 pm

homework

makeup Friday final 9-11 am


Key Concepts: Temperature, Thermal Expansion, and Ideal

GasesTemperature and Thermometers

Thermal Energy & Temperature

Thermal ExpansionCoefficient of thermal expansion

Ideal GasesState Variables

Ideal gas law

Kinetic Theory of Gases

Kinetic & thermal energy

Maxwell distribution

Covers chapter 12 in Rex & Wolfson


Conversions:

Tc = Tk - 273.15

Tf = (9/5)Tc + 32

Helium boils at Tk=4


R

Potential Energy

0

R

2 atom/molecules

-Emin

The curve depends onthe material, e.g. Emin isdifferent for water andiron

Kinetic energy ~ T (temperature)

Binding Forces


Solid (low T)

R

Potential Energy

0

Kinetic energy ~ T

-Emin

The temperature (and thus kinetic energy)is so small that the atoms/molecules can onlyoscillate around a fixed position Rmin

Rmin


Liquid (medium T)

R

Potential Energy

0

Kinetic energy ~ T

-Emin

Rmin

On average, the atoms/molecules like tostick together but sometimes escape andcan travel far.


Gas (high T)

R

Potential Energy

0

Kinetic energy ~ T

-Emin

Rmin

The kinetic energy is much larger thanEmin and the atoms/molecules move aroundrandomly.


What happens if the temperature of a substance is

increased?

R

0

Kinetic energy ~ T

-Emin

Rmin=Rave(T=0)

T=0: Average distance between atoms/molecules: Rmin


What happens if the temperature of a substance is

increased?

R

0

Kinetic energy ~ T

-Emin

Rmin=Rave(T=0)

Rave(T>0) > Rmin

T>To: The average distance between atoms/molecules is larger than Rmin:

the substance expands


Thermal expansionL= Lo T

L0

L

T=T0T=T0+T

A = Ao T = 2

V = Vo T = 3

length

surface

volume

Some examples: = 24x10-6 1/K Aluminum = 1.2x10-4 1/K Alcohol

: coefficient of linear expansion different for each material

MSU Physics 231 Fall 2015 13PHY 231 13

A Heated Ring

A metal ring is heated. What is true:a) The inside and outside radii become largerb) The inside radius becomes larger, the outside

radius becomes smallerc) The inside radius becomes smaller, the outside

radius becomes largerd) The inside and outside radii become smaller


Demo: Bimetallic Strips

Application: contact in a refrigerator

top

bottom

top<bottom if the temperature increases,

The strip curls upward, makes contact and switcheson the cooling.


Water: a special case

Coef. of expansion isnegative: If T dropsthe volume becomeslarger

Coef. Of expansion ispositive: if T drops the volume becomes smaller

Below this ice is formed (it floats on water)


Ice

liquid

ice

(g/cm3)

1

0.917

Phase transformation

Ice takes a larger volume than water!

A frozen bottle of water might explode


Thermal equilibrium

Low temperatureLow kinetic energyParticles move slowly

High temperatureHigh kinetic energyParticles move fast

Thermal contact

Transfer of kinetic energy

Thermal equilibrium: temperature is the same everywhere


Zeroth law of thermodynamics

If objects A and B are both in thermal equilibriumwith an object C, than A and B are also in thermalequilibrium.

There is no transfer of energy between A, B and C


Ideal Gas: properties

Collection of atoms/molecules that

• Exert no force upon each otherThe energy of a system of two atoms/molecules cannot be reduced by bringing them close to each other

• Take no volumeThe volume taken by the atoms/molecules is negligible compared to the volume they are sitting in


R

Potential Energy

0-Emin

Rmin

Ideal gas: we are neglecting the potential energy betweenThe atoms/molecules

R

Potential Energy

0

Kinetic energy


Properties of gases

V = volumeP = pressureT = temperature in K (Kelvin)n = number of moles

Example balloon


Molecular mass

kg 102.00 1002.6

012.0(carbon)

kg 0.0120 g 12.0 (carbon) examplefor

mass (molar)molecular

numbers sAvagodro' 1002.6

molecule)(or atom one of mass

26-23

molar

23

m

M

NmM

N

m

molar

A

A


X

Z

A

Number of electrons

molar mass in grams

Name


Weight of 1 mol of atoms1 mol of atoms weighs A grams (A is the molar mass)

Examples: 1 mol of Hydrogen weighs 1.0 g 1 mol of Carbon weighs 12.0 g1 mol of Oxygen weighs 16.0 g 1 mol of Zinc weighs 65.4 g

What about molecules?H2O 1 mol of water molecules:

2 x 1.0 g (due to Hydrogen)1 x 16.0 g (due to Oxygen)

Total: 18.0 g


molar

molecules)(or atoms all of mass total

molecules)(or atoms contains mole one so

or moles ofnumber

molecules)(or atoms ofnumber total

MnM

NmM

N

NnNN

Nn

N

A

AA

Number of atoms and moles


Example

A cube of Silicon (molar mass 28.1 g) is 250 g.

A) How much Silicon atoms are in the cube?

B) What would be the mass for the same number of gold atoms (molar mass 197 g)

Total number of moles n = M / Mmolar = 250/28.1 = 8.90

N = n NA = (8.9) (6.02x1023) = 5.4x1024 atoms

M = n Mmolar = (8.90) (197 g) = 1750 g


Question

1) 1 mol of CO2 has a larger mass than 1 mol of CH2

2) 1 mol of CO2 contains more molecules than 1 mol of CH2

a) 1) true 2) true

b) 1) true 2) false

c) 1) false 2) true

d) 1) false 2) false


Properties of gases

V = volumeP = pressureT = temperature in K (Kelvin)n = number of moles

Example balloon


Boyle’s Law (fixed n and T)

P0 V0

2P0 ½V0

½P0 2V0

At constant temperature: P ~ 1/V

implies that PV = constant


Charles’ law (fixed n and P)

V0 T0

2V0 2T0

If you want to maintain a constant pressure, the temperature must be increased linearly with the volume V ~ T implies that (V/T) = constant


Gay-Lussac’s law (fixed n and V)

P0 T0 2P0 2T0

If, at constant volume, the temperature is increased,the pressure will increase by the same factor

P ~ T

implies that (P/T) = constant


Brown’s law (fixed T and P)

n0 V0

2n0 2V0

If you double the number of particles the volume doubles n ~ V implies that (V/n) = constant


Boyle & Charles & Gay-LussacIDEAL GAS LAW

n = number of molesR = universal gas constant 8.31 J/mol·K

If the number of moles is fixed

2

22

1

11or constant T

VP

T

VP

T

PV

nRTPV Does not depend on what type or atom or molecule


ExampleAn ideal gas occupies a volume of 1.0 cm3 at 200 C at 1 atm. A) How many atoms are in the volume?

B) If the pressure is reduced to 1.0x10-11 Pa, while thetemperature drops to 00C, how many atoms remainedin the volume?

PV = nRT, so n = PV/(TR) with R=8.31 J/mol K T=200C=293K, P=1atm=1.013x105 Pa, V=1.0cm3=1x10-6m3

n=4.2x10-5 mol N = n NA = (4.2x10-5) NA=2.5x1019

T = 00C = 273K , P = 1.0x10-11 Pa, V = 1x10-6 m3

n=4.4x10-21 mol N=2.6x103 particles (almost vacuum)


And another!An air bubble has a volume of 1.50 cm3 at 950 m depth (T=7oC). What is its volume when it reaches the surface (T=20oC). (water=1.0x103 kg/m3)?

P950m=P0 + water g h = 1.013 x 105 + (1.0x103)(9.8)(950) = 94.2 x 105 Pa

111

2

2

12

2

22

1

11

)046.1)(0.93(

VVT

T

P

PV

T

VP

T

VP

Vsurface=146 cm3

Expanded by a factor of 97


A volume with dimensions L x W x H is kept underpressure P at temperature T. If the temperature israised by a factor of 2, and the height is made5 times smaller, by what factor does the pressure change, i.e. what is P2/P1? No gas leaks or is added.

a) 0.4 b) 1 c) 2.5 d) 5 e) 10

Use the fact PV/T is constant if no gas is added/leakedP1V1 / T1 = P2V2 / T2

P1V1 / T1 = P2 (V1/5) / (2T1)P2 = (5)(2)(P1 ) = 10 P1 a factor of 10.

Quiz


K 273.15 C 0 T

Pa101.013 atm 1

oo

5

P

“Standard temperature and pressure” (STP)


number sAvagadro'1002.6

objects ofnumber total

moles ofnumber

23A

A

N

N

N

Nn

Moles


macroscopic to microscopic

B

AB

B

kNRn

N

Rk

TkNPV

TRnPV

constant) s(Boltzman' (J/K)1038.11002.6

31.8 2323

macroscopic quantities

N = number of atoms or molecules (microscopic)


Quiz

Given P1 = 1 atm P2 = 2 atm V1 = 2 m3 V2 = 10 m3

T1 = 100 K N1 = NA N2 = 10 NA

T2 = ? K

A) 200B) 500C) 2000D) 5000E) 100


ExampleHow many air molecules at in the room with a volume of 1000 m3

(assume only molecular nitrogen is present N2)?

PV = N kB T

T = 293P = 1.013x105 Pa V = 1000 m3

N = 2.5x1028


microscopic description: kinetic theory of gases

1) The number of objects is large (statistical model)2) Their average separation is large 3) The objects follow Newton’s laws4) Any particular object can move in any direction with a distribution of velocities5) The objects undergo elastic collision with each other6) The objects make elastic collisions with the walls7) All objects are of the same type


Movie of gas in two dimensions


mean free pathd = average distance between collisions

air at P = 1 atm d = 68 nm = 68 x 10-9 m

high vacuum P = 10-5 Pa d = 1m

in space P = 10-12 Pa d = 108 m


The Maxwell DistributionHowever we can model the distribution of the velocities (& thus the kinetic energies) of the individual gas molecules. The result is the Maxwell Distribution.

The root-mean-square (rms) velocity is 2vvrms


Energy of one object

2

2

1vmK

Objects inside the container have a distribution of velocitiesaround an average – so each object has an average kinetic energygiven by

average squared velocityaverage translation kinetic energy

mass of the object (atom or molecule)


Relationship to ideal gas law

3

2 KNPV

The objects bounce off of each other and the walls of the container (elastic). One can derive the following result

TkK

TkNPV

B

B

2

3get to

with combine

How the average kinetic energy of oneatom is related to temperature


2

2

1vmK with

2

3 combine TkK B

root-mean-square (rms) velocity for one atom or molecule

molar

Brms M

RT

m

Tkvv

332


ExampleWhat is the rms speed of air at 1 atm and room temperature (293 K)? Assume it consist of molecular Nitrogen only (N2)?

molar

Brms M

RT

m

Tkvv

332

R = 8.31 J/mol K

T = 293 K Mmolar = (2 x 14)x10-3 kg/mol

vrms = 511 m/s = 1140 mph !


d is the number of “degrees of freedom” for the motion

d = 3 for an atom (motion in x, y, z directions) like helium gas

d = 5 for a diatomic molecule (motion in x, y, z and two ways to rotate) like nitrogen molecule N2 or hydrogen molecule H2

) (since 2223

nRNkPVd

nRTd

Tkd

NKNd

E BBth

(Homework question for “one degree of freedom” use d = 1)

Total thermal energy


ExampleWhat is the total thermal kinetic energy of the air molecules in thelecture room (assume only molecular nitrogen is present N2)?

Eth = (d/2) PV = 2.5x108 J

d = 5P = 1.013x105 Pa V = 1000 m3

Using KE = (1/2) mv2 this is equivalent to 1000 carswith m=1000 kg each moving with v = 22.3 m/s (50 mph)

Can we use that energy to do work?


Diffusion

msu physics 231 fall 2015 1 physics 231 topic 12: temperature, thermal expansion, and ideal gases...

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