1PHYS1001Physics 1 REGULARModule 2 Thermal Physics

ap06/p1/thermal/ptE_gases.ppt

PRESSURE

IDEAL GAS

EQUATION OF STATE

KINETIC THEORY MODEL

THERMAL PROCESSES

2

Overview of Thermal Physics Module:1. Thermodynamic Systems:

Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics

2. Thermal Expansion

3. Heat Capacity, Latent Heat

4. Methods of Heat Transfer:Conduction, Convection, Radiation

5. Ideal Gases, Kinetic Theory Model

6. Second Law of ThermodynamicsEntropy and Disorder

7. Heat Engines, Refrigerators

3

Kinetic-Molecular Model of an Ideal Gas Thermal Processes

* Ideal gas, Equation of state (§18.1 p611) * Kinetic-molecular model of an ideal (results only – not the mathematical derivations) (§18.3 p619)* Heating a gas: heat capacities, molar heat capacity (§17.5 p582 §18.4 p626 §19.6 p658 §19.7 p659)* First Law of Thermodynamics: Internal Energy, Work, Heat, Paths between thermodynamic states (§19.1 p 723-725, §19.2 p725-728, § 19.3 p728-729, §19.4 p729-735)* Thermal Processes and pV diagrams: Isothermal, Isobaric Isochoric (constant volume gas thermometer), Adiabatic Cyclic (§19.5 p735-737, §19.8 p741-744, §17.3 p644-645

References: University Physics 12th ed Young & Freedman

4

HEAT ENGINES & GASES

5

Phases of matter

Gas - very weak intermolecular forces, rapid random motion

Liquid - intermolecular forces bind closest neighbours

Solid - strong intermolecular forces

low temphigh pressure

high templow pressure

6

Ideal Gas

* Molecules do not exert a force oneach other zero potential energy

* Large number of molecules

* Molecules are point-like

* Molecules are in constant random motion

* Collisions of molecules with walls ofa container and other molecules obey Newton's laws and are elastic

7

Quantity of a gasnumber of particles N

mass of particle m

molar mass M (kg.mol-1) mass of 1 mole of a substance

number of moles n ( mol) 1 mole contains NA particles

Avogadro's constant NA = 6.023x1023 mol-1

1 mole is the number of atoms in a 12 g sample of carbon-12

1 mole of tennis balls would fill a volume equal to 7 Moons

The mass of a carbon-12 atom is defined to be exactly 12 u

u atomic mass units, 1 u = 1.66x10-27 kg

(1 u)(NA) = (1.66x10-27)(6.023x1023) = 10-3 kg = 1 g

mtot = N m

If N = NA mtot = NA m = M M = NA m

n = N / NA = mtot / M

81.00 kg of water vapour H2O

M(H2O) = M(H2) + M(O) = (1 + 1 + 16) g = 18 g = 1810-3 kg

n(H2O) = mtot / M(H2O) = 1 / 1810-3 = 55.6 mol

N(H2O) = n NA = (55.6)(6.0231023) = 3.351025

m(H2O) = M / NA = (1810-3) / (6.0231023) kg = 2.9910-26 kg

1 amu = 1 u = 1.6610-27 kg

m(H2O) = 18 u = (18)(1.6610-27) kg = 2.9910-26 kg

9

Pressure P

pressure !!!

Is this pressure?

What pressure is applied to the ground if a person stood on one heel?

10Pressure P (Pa) Impact of a molecule on the wall of the

container exerts a force on the wall and the wall exerts a force on the molecule. Many impacts occur each second and the total average force per unit area is called the pressure.

P = F / A force F (N)area A (m2)pressure P (Pa)

Patmosphere = 1.013105 Pa

~1032 molecules strike our skin every day with an avg speed ~ 1700 km.s-1

11

Rough estimate of atmospheric pressure

air ~ 1 kg.m-3 g ~ 10 m.s-2 h ~ 10 km = 104 m

p = F / A = mg / A = V g / A = A h / A = g h

Patm ~ (1)(10)(104) Pa

Patm ~ 105 Pa

12

… and all the king's horses …

What force is required to separate the hemispheres? Is this force significant?

?

Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg

13

Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg

p = 1x105 PaR = 0.30 mA = 4R2

F = p AF = (105)(4)(0.3)2 NF = 105 N

14

Gauge and absolute pressures

Pressure gauges measure the pressure above and below atmospheric (or barometric) pressure.

Patm = P0 = 1 atm = 101.3 kPa = 1013 hPa = 1013 millibars = 760 torr = 760 mmHg

Gauge pressure Pg

Absolute pressure P

P = Pg + Patm0

100

200

300

400

Pg = 200 kPa Patm= 100 kPaP = 300 kPa

15

Ideal Gases – equation of state (experimental law)

p V = n R T = N k T

R, Universal gas constant (same value for all gases)

R = 8.314 J.mol-1.K-1

Boltzmann constant k = 1.38x10-23 J.K-1

k = R / NA R = k NA

must be in kelvin (K)

16

Ideal gas, constant mass (fixed quantity of gas)

All gases contain the same number of molecules when they occupy the same volume under the same conditions of temperature and pressure (Avogadro 1776 - 1856)

p V = n R T n = N / NA= p V / R T

1 1 2 2

1 2

p V p V

T T

17

Boyle's Law (constant temperature)

p = constant / V

Charles Law (constant pressure)

V = constant T

Gay-Lussac’s Law (constant volume)

p = constant T

18Isothermals pV = constant

0

20

40

60

80

100

120

140

160

180

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

volume V (m3)

pre

ssu

re p

(kP

a)

100 K

200 K

300 K

400 K

n RTp

V

19Thermodynamic system (ideal gas) work

p V T U S

mtot N n

p V = n R T

p V = N k T

k = R / NA

mtot = n M

N = n NA

heat

Q = n C T

CV or Cp

W p dV

dQS

T

internal energy

U = Q – W = n CV T

Q = 0p V = constantT V-1 = constant

20Ideal gas - equipartition of energy classical picture - not valid at low or high temperatures

Degrees of freedom - there is kinetic energy associated with each type of random motion

Translation f = 3

x

y

z

Rotationdiatomic molecule f = 2

Vibrationonly at high T

Provided the temperature is not too high (< 3000 K), a diatomic molecule has 5 degrees of freedom

21

Kinetic–Molecular model for an ideal gas (p619)

Large number of molecules N with mass m randomly bouncing around in a closed container with Volume V.

trKEpV3

2

TkNTRnpV

Kinetic-Molecular Model(Theory)

Experimental Law

For the two equations to agree, we must have:

TkNTRnKEtr 2

3

2

3

x

y

z

22 Total kinetic energy for random translational motion of all molecules, Ktr

Average translational KE of a molecule

2

2

1

2

3

2

3avgtr mvNTkNTRnKE

For an ideal gas, temperature is a direct measure of the average kinetic energy of its molecules.

Tkvvvmmv avgZavgYavgXavg 2

3

2

1

2

1 2,

2,

2,

2

221

avgmv is the average translational kinetic energy of a single molecule

23

At a given temperature T, all ideal gas molecules have the same average translational kinetic energy, no matter what the mass of the molecule

energy stored in each degree of freedom = ½ k T

Theorem of equipartition of energy (James Clerk Maxwell):

The thermal energy kT is an important factor in the natural sciences. By knowing the temperature we have a direct measure of the energy available for initiating chemical reactions, physical and biological processes.

24

Internal energy U of an ideal gas

1

2random random

U KE PE KE N f k T

2

fU N k T

Degrees of freedom (T not too high)

monatomic gas, f = 3 diatomic gas, f = 5, polyatomic gas, f = 6

2

fU N k T

Only translation possible at very low temp, T rotation begins, T oscillatory motion starts

PE = 0

25Heating a gas

tot

totA

Q m c T

N Mm N m n M

N

Q n M c T

Q

C M c

n C T

Molar heat capacity

26

Heating a gas at constant volume

Q

VQ n C T 2

fU N k T

1st Law Thermodynamics

U Q W

Constant volume process V = 0 W = 02 V

fN k T nC T

1 A An N N N k R

2V

fC R

Larger f larger CV smaller T for a given Q

All the heat Q goes into changing the internal energy U hence temperature T

U = n CV T

27

Heating a gas at constant pressure

Q

pQ n C T 2 Vf

U N k T nC T

1st Law Thermodynamics

U Q W

Constant pressure process W = p V

V pn C T n C T n R T

VCp C R

W

pV n RT p V n R T

It requires a greater heat input to raise the temp of a gas a given amount at constant pressure c.f. constant volume Q = U + W W > 0

28

22

22

2

2

1

monatomic 3 1.67

diatomic 5 1.4

21

21

32

501

V

p V

p V

V V

V V

V

monatomic

diatomic

fC R

fC C R R

fRC C R

fC CR

C C R

f

f

f

RC

29

n N mtot

Thermal processes

Q

W

T1

p1

V1

U1

S1

T2

p2

V2

U2

S2

VU WQ nC T p VC C R

21

dQS

T

pV n RT N k T

Reversible processesp VQ nC T Q nC T

30Isothermal change T = 0

U = 0 pV = n R T

22 21 1 1

2 1 11 1 2 2

1 2 2

2 2 2

1 1 1

ln

ln

ln

V VV V

W WnRT nRT

nRT VQ W pdV dV nRT

V V

V p pp V p V Q W n RT

V p p

V W V pe e

V nRT V p

Boyle’s Law (1627 -1691)T1= T2 p1V1 = p2 V2

31

Isothermals pV = constant

0

20

40

60

80

100

120

140

160

180

200

0.00 0.05 0.10 0.15 0.20 0.25

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 K

1

2W

Isothermal process

W is the area under an isothermal curve isotherm

32Isochoric (V = 0) W = 0 U = Q = n CV T

Isothermals pV = constant

0

20

40

60

80

100

120

140

160

180

200

0.00 0.05 0.10 0.15 0.20 0.25

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 K

1

2

W = 0

Isochoric process

1 to 2: Q > 0 T 1 < T 2 T > 0 U > 0

isochor

33

Isothermals pV = constant

0

20

40

60

80

100

120

140

160

180

200

0.00 0.05 0.10 0.15 0.20 0.25

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 K

1

2

W

Isobaric process

Isobaric (p = 0) W = p V Q = n Cp T

U = Q – W = n CV T

T2 > T1 U > 0 W > 0 Q > 0 W < Q

V1/T1=V2/T2

isobar

34

Adiabatic (Q = 0)

U = - W CV = (f / 2) R Cp = CV + R = (f / 2 +1)R

= Cp / CV = (f + 2) / f

p V = constant diatomic gas f = 5

T V -1 = constant = 7 / 5 = 1.4

V1 1 2 2 1 1 2 2

1

1

CW p V p V p V p V

R

35

Isothermals pV = constant

0

20

40

60

80

100

120

140

160

180

200

0.00 0.05 0.10 0.15 0.20 0.25

volume V (m3)

pre

ssu

re

p

(kP

a)

100 K

400 K

800 K

1

2

W

Adiabatic process

W

An adiabat steeper on a pV diagram than the nearby isotherms since > 1

1 to 2: Q = 0 T1 > T2 W > 0 U < 0

adiabat

36

Adiabatic processes can occur when the system is well

insulated or a very rapid process occurs so that there is not

enough time for a significant heat to be transferred eg rapid

expansion of a gas; a series of compressions and

expansions as a sound wave propagates through air.

Atmospheric processes which lead to changes in

atmospheric pressure often adiabatic: HIGH pressure cell,

falling air is compressed and warmed. LOW pressure cell,

rising air expands and cooled condensation and rain.

37

Q = 0U = - WT V -1 = constant

convergence divergence

divergence convergenceHIGH - more uniform conditions - inhibits cloud formation

LOW - less uniform conditions - encourages cloud formation

sunshine

Atmospheric adiabatic processes

Burma Cyclone5 May 2008 +50 000 killed ?

38

V

2 2 V2 1 2 1 1 21 1V V

( 1)p 2 1

p VV V 1 2

( 1)1 1 2 2 2 2 2 1

1 2 1 1 1 2

2 1

1

( 1)

2

ln( / ) ln( / ) ln (

con

/

stan

)

1 1

t

RT V CT V

nRTU W nC dT p dV dV

V

dT R dV RT T V V V V

T C V C

CR T VR C C

C C T V

p V p V T p V V

T T T p V V

p V

p V

TV

constantpV

39

Cyclic Processes:

U = 0 reversible cyclic process

40

Problem E.1

Oxygen enclosed in a cylinder with a movable piston

(assume the gas is ideal) is taken from an initial state A

to another state B then to state C and back to state A.

How many moles of oxygen are in the cylinder? Find

the values of Q, W and U for the paths A to B; B to C;

C to A and the complete cycle A to B to C to A and

clearly indicate the sign + or – for each process.

Does this cycle represent a heat engine?

41

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

42Thermodynamic system (ideal gas) work

p V T U S

mtot N n

p V = n R T

p V = N k T

k = R / NA

mtot = n M

N = n NA

heat

Q = n C T

CV or Cp

W p dV

internal energy

U = Q – W = n CV T

Q = 0p V = constantT V-1 = constant

43

SolutionIdentify / Setup

oxygen diatomic f = 5

CV = (f / 2) R Cp = CV + R = (f / 2 +1) R

CV = 5/2 R Cp = 7/2 R

R = 8.315 J.mol-1.K-1

CV = 20.8 J.mol-1.K-1 Cp = 29.1 J.mol-1.K-1

p V = n R T = N k T

area under graph

V

V

p

U nC T

U Q W

W p dV pV

Q nC T

Q nC T

44

Execute

At A

4 2

2

4 10 2 10mol 0.96 mol 1 mol

8.31 10A A

A

p Vn

RT

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

451 A to B is isobaric

T1= TB – TA = (400 – 100) K = 300 K

pA = pB = p1 = 40 kPa = 4.00104 Pa

V1 = (0.080 – 0.020) m3 = 0.060 m3

W1 > 0 since gas expands

W1 = p1 V1 = (4.0104)(0.06) = 2.4103 J

U1 > 0 since the temperature increases

U1 = n CV T1 = (1)(20.8)(300) J = 6.2103 J

Q1 > 0 since U1 > 0 and U1 = Q1 – W1 > 0 Q1 > W1 > 0

Q1 = n Cp T1 = (1)(29.1)(300) J = 8.7103 J

Check: First law

U1 = Q1 - W1 = (8.73103 – 2.4103) J = 6.3103 J

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

BQ1

46

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

472 B to C is isochoric

T2 = TC – TB = (800 – 400) K

= 400 K

V2 = 0 m3

W2 = 0 since no change in volume

U2 > 0 since the temperature increases

U2 = n CV T2 = (1)(20.8)(400) J = 8.3103 J

Q2 = U2 since W2 = 0

Q2 = 8.3103 J

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

Q2

48

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

493 C to AT3 = TA – TC = (100 – 800) K = -700 K

pA = 40 kPa = 4.0104 Pa pC = 80 kPa = 8.0104 Pa pCA = 4.0104 Pa

VA = 0.02 m3 VC = 0.08 m3 V3 = 0.06 m3

W3 < 0 since gas is compressed

W3 = area under curve = area of rectangle + area of triangle

W3 = - { (0.06)(4.0104) + (½)(0.06)(8.0104- 4.0104)} J = - 3.6103 J

U3 < 0 since the temperature decreases

U3 = n CV T3 = (1)(20.8)(-700) J

= - 14.6103 J

Q3 = U3 + W3

Q3 = (- 14.5103 - 3.6103) J = - 18.2103 J

0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

Q3

50

1A to B

2B to C

3C to A

cycle(total values)

W (kJ) + 2.4 0 - 3.6 - 1.2

Q (kJ) + 8.7 + 8.3 - 18.2 +1.2

U (kJ) +6.3 +8.3 - 14.6 0

Q – W (kJ) + 6.3 + 8.3 - 14.6 0

Complete cycle U = 0 J

Refrigerator cycle: |QH| = |QC| +|W|

THTC

|QH| |QC|

|W|

51Wcycle < 0 work is done on the system the system is not a heat engine because a heat engine needs to do a net amount of work on the surroundings each cycle.

The net work corresponds to the area unenclosed i.e. the area of the triangle:

Wcycle = - (1/2)(0.06)(4.0104) J = - 1.2103 J (value agrees with table)

Evaluate0

10

20

30

40

50

60

70

80

90

100

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

volume V (m3)

pre

ss

ure

p (

kP

a)

100 K

400 K

800 KA

C

B

52Problem E.2 Typical 5 mark exam question

An ideal gas is enclosed in a cylinder which has a movable piston. The gas is heated resulting in an increase in temperature of the gas and work is done by the gas on the piston so that the pressure remains constant.

(a) Is the work done by the gas positive, negative or zero. Explain (b) From a microscopic view, how are the gas molecules effected? Explain.

(c) From a microscopic view, how is the internal energy of the gas molecules effected?

(d) Is the heat less than, greater than or equal to the work? Explain.

53SolutionIdentify / Setup

2

12 1 0

V

VW p dV p V V

U Q W First Law of Thermodynamics

V

p

T1

T2

T2 > T1

V1 V2

VU nC T

54

(a)Since work is done by the gas on the piston, the system expands as the volume increases (pressure remains constant)

The work done by the gas on the piston is positive.

(b)Since the temperature increases, the average translational kinetic energy of the gas molecules increases.

2

12 1 0

V

VW p dV p V V

55

(c)

The change in internal energy, U of an ideal gas is given by

where n is the number of moles of the gas and CV is the molar heat capacity of the gas at constant volume. Since the temperature increases, the internal energy must increase. Therefore, the total kinetic energy due to random, chaotic motion of the gas molecules increases.

VU nC T

56

(d)

Heat Q refers to the amount of energy transferred to the gas due to a temperature difference between the system and surroundings.

First law of thermodynamics

0 0 0U Q W U Q W Q W

The heat Q is greater the work done by the gas W.