Temperature and Ideal Gas 1 Everything is made of atoms In gases the molecules dont interact with...

Temperature and Ideal Gas

1

•Everything is made of atoms

•In gases the molecules don’t interact with each other. Simple

•How does the atomic (molecular) nature of a gas explain its properties?

•Air in your tires?

•How hot is hot?

•How cold is cold?

Temperature

2

Heat is the flow of energy due to a temperature difference. Heat always flows from objects at high temperature to objects at low temperature.

The quantity indicating how warm or cold an object is relative to some standard is TEMPERATURE (T).

T does not depend on quantity of a substance. A cup and a thimble of boiling water both have same T.

When two objects have the same temperature, they are in thermal equilibrium.

3

The Zeroth Law of Thermodynamics:

If two objects are each in thermal equilibrium with a third object, then the two objects are in thermal equilibrium with each other.

There is no heat flow between objects in thermal equilibrium

Temperature Scales

4

Fahrenheit scale

Water boils* 212 F

Water freezes* 32 F

Absolute zero 459.67 F

(*) Values given at 1 atmosphere of pressure.

Temperature Scales

5

Fahrenheit scale Celsius scale

Water boils* 212 F 100 C

Water freezes* 32 F 0 C

Absolute zero 459.67 F 273.15C

(*) Values given at 1 atmosphere of pressure.

Temperature Scales

6

Absolute or Kelvin scale

Fahrenheit scale

Celsius scale

Water boils* 373.15 K 212 F 100 C

Water freezes* 273.15 K 32 F 0 C

Absolute zero 0 K 459.67 F 273.15C

(*) Values given at 1 atmosphere of pressure.

7

The temperature scales are related by:

F32CF/ 8.1 CF TT

273.15C TT

Fahrenheit/Celsius

Absolute/Celsius

8

Example (text problem 13.3): (a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees?

C 40

328.1

C

CCF

T

TTT

(b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?

F 574

322738.1

322738.1

321.8

F

F

CF

T

T

T

TT

Question

• Which is smaller, a change of 1oF or 1oC?• A) 1oF• B) 1oC• C) they are the same

Molecular Picture of a Gas

• Atoms and molecules are the basic units of matterWe want to explain thermalProperties in terms of atomsand molecules.How many atoms?

Molecular Picture of a Gas

11

We can specify the amount of a substance by giving its mass or by the number of molecules (or atoms) it has.

If we know the mass of a molecule we can go from one description to the other

A golf ball weighs 1.6 ounces.

I have a 20 lb box of golf balls

I have a box of 200 golf balls

Totally equivalent

Molecular Picture of a Gas

12

If a sample contains a single substance (element or compound) the number of particles in the sample is

N = M/m.

N equals the total mass of the sample (M) divided by the mass (m) of the atom (or molecule)

The number density of particles is N/V where N is the total number of particles contained in a volume V.

13

One mole of a substance contains the same number of particles as there are atoms in 12 grams of 12C. The number of atoms in 12 grams of 12C is Avogadro’s number.

123A mol 10022.6 N

It is convenient to have a standard number to facilitate this going back and forth from the two descriptions. Since we deal with human size numbers (gms) this will involve a very large number of atoms

14

A carbon-12 atom by definition has a mass of exactly 12 atomic mass units (12 u or 12 amu).

12g = 12u NA

u = 1g/(6x1023) =1.66x10-24g =1.66x10-27kg

This is the conversion factor between the atomic mass unit and kg (1 u = 1.661027 kg). NA and the mole are defined so that a 1 gram sample of a substance with an atomic mass of 1 u contains exactly NA particles.

A mole of O2 has mass 32 gm., of water m=18 gm

Why 12 gm of Carbon?

15

Example (text problem 13.37): Air at room temperature and atmospheric pressure has a mass density of 1.2 kg/m3. The average molecular mass of air is 29.0 u. How many air molecules are there in 1.0 cm3 of air?

moleculeair per mass average

cm 1.0in air of mass total particles ofnumber

3

The total mass of air in the given volume is:

kg 102.1cm 100

m 1

1

cm 0.1

m

kg 2.1 633

3

Vm

16

particles 105.2

kg/u 1066.1u/particle 0.29

kg 102.1

moleculeair per mass average

cm 1.0in air of mass total particles ofnumber

19

27

6

3

Example continued:

Question

• Which contains more atoms, 5 mol. of helium• (mass He =4amu) or 1 mol of neon (m Ne

=20amu)• A) Helium• B) Neon• C) both have same number of atoms

17

Question

• Which contains more atoms, 1 mol of helium• or 1 mol of Steam (water)• A) Helium• B) water• C) both have same number of atoms

18

Decrease the volume Increase the pressure

19

Constant T: P ~1/V

Increase the number of moleculesIncrease the pressure

20

Constant V,T: P ~ N

We also know that, as you drive, tire pressure increases with T

21

Constant V,N: P ~ T

Absolute Temperature and the Ideal Gas Law

22Constant P: V ~T

Absolute Temperature and the Ideal Gas Law

23

TV

Experiments done on dilute gases (a gas where interactions between molecules can be ignored) show that:

For constant pressure

Charles’ Law

For constant volume

TP Gay-Lussac’s Law

24

Boyle’s LawFor constant temperature V

P1

For constant pressure and temperature

NV Avogadro’s Law

What Temperature do we use?

25

There is a lowest possible T V→0

Absolute Temperature

• There is a coldest possible temperature Absolute zero.

• All objects will transfer heat to an object at absolute 0.

• Experiment show (e.g. V→0 ) that the coldest possible T is -273.15oC.

• Kelvin scale measures T from Absolute 0 in units of 1oC

TK=Tc+273o26

27

Putting all of these statements together gives the ideal gas law (microscopic form):

NkTPV k = 1.381023 J/K is Boltzmann’s constant

The ideal gas law can also be written as (macroscopic form):

nRTPV R = NAk = 8.31 J/K/mole is the universal gas constant and n is the number of moles.

28

Example (text problem 13.41): A cylinder in a car engine takes Vi = 4.50102 m3 of air into the chamber at 30 C and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume and to 20.0 times the original pressure. What is the new temperature of the air?

Here, Vf = Vi/9, Pf = 20.0Pi, and Ti = 30 C = 303 K.

iii NkTVP

fff NkTVP

The ideal gas law holds for each set of parameters (before compression and after compression).

29

Example continued:

Take the ratio:i

f

i

f

ii

ff

T

T

NkT

NkT

VP

VP

The final temperature is

K 673K 30390.20

i

i

i

i

ii

f

i

ff

V

V

P

P

TV

V

P

PT

The final temperature is 673 K = 400 C.

30

Putting all of these statements together gives the ideal gas law (microscopic form):

NkTPV k = 1.381023 J/K is Boltzmann’s constant

The ideal gas law can also be written as (macroscopic form):

nRTPV R = NAk = 8.31 J/K/mole is the universal gas constant and n is the number of moles.

Question

When the temperature of a quantity of gas is increasedA) the pressure must increase.B) the volume must increase.C) the pressure and/or the volume must increase.D) none of the above

Question

A pot of water on the stove is heated from 25oC to 100oC. By what factor does the

• temperature in Kelvin change?• A) T2 = 4T1

• B) T2 = 1.25T1

• C) T2 = 0.80T1

• D) T2 = 0.20T1

Question

• Inside your air-conditioned apartment, you blow up a balloon as large as possible and then take it outside on a hot summer day. The balloon is most likely to thenA) shrink.B) remain the same size.C) expand and pop.

Question

The Kelvin temperature of an ideal gas is doubled and the volume is halved. How is the pressure affected? A) increases by a factor of 2

B) increases by a factor of 4 C) stays the same D) decreases by a factor of 2

E) decreases by a factor of 4

Kinetic Theory of the Ideal Gas

35

An ideal gas is a dilute gas where the particles act as point particles with no interactions except for elastic collisions.

Point particles can have only KE, no internal PE

Add heat (energy) to gas, energy increases

KE increases.

But if we add heat, temperature also increases.

T depends on KE

Kinetic Theory of Ideal Gas

36

T~ KEav/molecule

if T =absolute 0, molecules don’t move.

T T TMore total energy, but same T, same average KE

Temperature is related to average KE

• This is true even for liquids and solids

37

k

KT

kTK

3

22

3

tr

tr

Pressure is caused by collisions

38

Gas particles have random motions. Each time a particle collides with the walls of its container there is a force exerted on the wall. The force per unit area on the wall is equal to the pressure in the gas.

39

The pressure will depend on:

•The number of gas particles ~N

•Frequency of collisions with the walls ~ v

•Amount of momentum transferred during each collision ~ mv

• P~ Nmv2 ~N KEmol

40

The pressure in the gas is

tr3

2K

V

NP

Where <Ktr> is the average translational kinetic energy of the gas particles; it depends on the temperature of the gas.

kTK2

3tr

• Typical air molecule is moving more than • 1,000 Miles/hr. Some move faster some slower

42

The average kinetic energy also depends on the rms speed of the gas

2rms

2tr 2

1

2

1mvvmK

where the rms speed is

m

kTv

mvkTK

3

2

1

2

3

rms

2rmstr

What is rms?root mean square

• v2 rms is the average of the square of the

velocities. It is not the square of the average

43

N

vvvv Nav

....21

N

vvvv Nrms

222

212 ...

Average of squares

N

vvvv Nrms

221

21 ...

rms

rms Examplev1 = 4 v2 = 8 vav = 6

v12 = 16 v2

2 = 64

45

3.6402

6416

rmsv

2rms

2tr 2

1

2

1mvvmK

46

The distribution of speeds in a gas is given by the Maxwell-Boltzmann Distribution.

47

Example (text problem 13.60): What is the temperature of an ideal gas whose molecules have an average translational kinetic energy of 3.201020 J?

K 15503

22

3

tr

tr

k

KT

kTK

48

Example (text problem 13.70): What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at 25 C?

m

kTv

3rms

Element Mass (kg) rms speed (m/s)

He 6.641027 1360

H2 3.321027 1930

N2 4.641026 515

O2 5.321026 482

On the Kelvin scale T = 25 C = 298 K.

Question

At a given temperature, a hydrogen molecule has a speed of 800 m/s. At the same

temperature, an oxygen molecule has a speed of A) 800 m/s. B) 400 m/s. C) 200 m/s. D) 100 m/s.

50

Question

When will a real gas behave most like an ideal gas?A) at high temperatures and high pressuresB) at low temperatures and high pressuresC) at low temperatures and low pressuresD) at high temperatures and low pressures

Question

• The rms speed of a box of molecules which are moving at non uniform speeds is

• greater than the average speed.

• A) always• B) sometimes• C) never

52

Thermal expansion

53

Most objects including liquids and solids expand when their Temperature increases

54

An object’s length after its temperature has changed is

01 LTL is the coefficient of linear expansion

where T = TT0 and L0 is the length of the object at a temperature T0.

55

Example (text problem 13.84): An iron bridge girder(Y = 2.01011 N/m2) is constrained between two rock faces whose spacing doesn’t change. At 20.0 C the girder is relaxed. How large a stress develops in the iron if the sun heats the girder to 40.0 C?

27

16211

N/m 108.4

K 20K 1012N/m100.2

A

F

TYL

LY

Using Hooke’s Law:

= 12106 K1 (from Table 13.2)

56

How does the area of an object change when its temperature changes?

The blue square has an area of L0

2.

With a temperature change T each side of the square will have a length change of L = TL0.

L0

L0+L

57

20

20

20

2220

20

0000

2

2

areanew

TLL

LTTLL

TLLTLLA

The fractional change in area is:

TA

A

TAAA

TAA

2

)2(

21areanew

0

00

0

58

The fractional change in volume due to a temperature change is:

TV

V

0

For solids = 3

Question A metal plate with a hole cut in it is heated. As

the plate expands,A) the hole expands.

B) the hole shrinks. C) the hole stays the same size.

Question • You can loosen the metal lid on a glass jar by

running it under hot water. Given that the lid and the jar have roughly the same diameter, compare the expansion of the diameter of the lid to that of the jar. (steel = 12 x 10-6 K-1, glass = 3.25 x 10-6 K-1)

• A) Llid = 3.7 Ljar

• B) Llid = 0.27Ljar

• C) Llid = 7.4 Ljar

• D) Llid = 1.9 Ljar