1. PHYSICS: FUN EXCITING SIMPLE
Physics 1 REGULAR
Module 2 Thermal Physics
PRESSURE
IDEAL GAS
EQUATION OF STATE
KINETIC THEORY MODEL
THERMAL PROCESSES
PHYSICS: FUN EXCITING SIMPLE
ap06/p1/thermal/ptE_gases.ppt

2. Overview of Thermal Physics Module:
Thermodynamic Systems: Work, Heat, Internal Energy 0th, 1st and 2nd Law of Thermodynamics
Thermal Expansion
Heat Capacity, Latent Heat
Methods of Heat Transfer: Conduction, Convection, Radiation
Ideal Gases, Kinetic Theory Model
Second Law of Thermodynamics Entropy and Disorder
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 p §18.4 p §19.6 p §19.7 p659)
* First Law of Thermodynamics: Internal Energy, Work, Heat,
Paths between thermodynamic states
(§19.1 p , §19.2 p , § p , §19.4 p )
* Thermal Processes and pV diagrams: Isothermal, Isobaric
Isochoric (constant volume gas thermometer), Adiabatic Cyclic
(§19.5 p , §19.8 p , §17.3 p
References: University Physics 12th ed Young & Freedman

4. HEAT ENGINES & GASES

5. Liquid - intermolecular
Phases of matter
Gas - very weak
intermolecular forces,
rapid random motion
high temp
low pressure
Liquid - intermolecular
forces bind closest neighbours
low temp
high pressure
Solid - strong
intermolecular forces

7. Quantity of a gas number 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

8. 1.00 kg of water vapour H2O
M(H2O) = M(H2) + M(O) = ( ) g = 18 g = 1810-3 kg
n(H2O) = mtot / M(H2O) = 1 / 1810-3 = mol
N(H2O) = n NA = (55.6)(6.0231023) = 3.351025
m(H2O) = M / NA = (1810-3) / (6.0231023) kg = 2.99 kg
1 amu = 1 u = 1.66 kg
m(H2O) = 18 u = (18)(1.6610-27) kg = 2.99 kg

9. Pressure P Is this pressure?
What pressure is applied to the ground if a person stood on one heel?

10. Patmosphere = 1.013105 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. Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg
… and all the king's horses …
What force is required to separate the hemispheres? Is this force significant? ?

13. p = 1x105 Pa R = 0.30 m A = 4R2 F = p A F = (105)(4)(0.3)2 N
Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg
p = 1x105 Pa
R = 0.30 m
A = 4R2
F = p A
F = (105)(4)(0.3)2 N
F = 105 N

14. Gauge and absolute pressures
Pressure gauges measure the pressure above and below
atmospheric (or barometric) pressure. P
atm =
= 1
= 101.3
kPa
= 1013
hPa
millibars
760
torr
= 760 mmHg
Gauge pressure g
Absolute pressure +
100
200
300
400
= 200
= 100
P = 300

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 = 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. All gases contain the same number of molecules when they occupy the same volume under the same conditions of temperature and pressure (Avogadro )
p V = n R T  n = N / NA= p V / R T
Ideal gas, constant mass (fixed quantity of gas)

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

19. p V = n R T U = Q – W = n CV DT p V = N k T k = R / NA mtot = n M
Thermodynamic system
(ideal gas)
work
p V = n R T
p V = N k T
k = R / NA
mtot = n M
N = n NA
internal energy
U = Q – W
= n CV DT
p V T
U S
heat
Q = n C DT
CV or Cp
mtot N n
Q = 0
p V  = constant
T V-1 = constant

20. 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.
Experimental Law z
Kinetic-Molecular Model
(Theory) y x
For the two equations to agree, we must have:

22.  Total kinetic energy for random translational motion of all molecules, Ktr
is the average translational kinetic energy of a single molecule
 Average translational KE of a molecule
For an ideal gas, temperature is a direct measure of the average kinetic energy of its molecules.

23. Theorem of equipartition of energy (James Clerk Maxwell):
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
PE = 0  
Degrees of freedom (T not too high)
monatomic gas, f = 3
diatomic gas, f = 5,
polyatomic gas, f = 6
Only translation possible at very low temp, T  rotation begins, T  oscillatory motion starts

25. Heating a gas
Molar heat capacity

26. Heating a gas at constant volume
1st Law Thermodynamics
U = n CV DT Q
Constant volume process V = 0  W = 0
All the heat Q goes into changing the internal energy U hence temperature T
Larger f  larger CV  smaller T for a given Q

27. Heating a gas at constant pressure W
1st Law Thermodynamics Q
Constant pressure process W = p V
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

29. Thermal processes T1 p1 V1 U1 S1 T2 p2 V2 U2 S2 W n N mtot Q
Reversible processes

30. Isothermal change T = 0
Boyle’s Law ( )
T1= T2 p1V1 = p2 V2
U = pV = n R T

31. isotherm
W is the area under an isothermal curve

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

33. 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

35. adiabat 1 to 2: Q = 0 T1 > T2 W > 0 U < 0 W
An adiabat steeper on a pV diagram than the nearby isotherms since  > 1

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. Atmospheric adiabatic processes
convergence divergence
divergence convergence
HIGH -
more uniform
conditions
inhibits cloud
formation
LOW
less uniform
encourages cloud
sunshine
Atmospheric adiabatic processes
Q = 0
U = - W
T V -1 = constant
Burma Cyclone
5 May killed ?

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?

42. p V = n R T U = Q – W = n CV DT p V = N k T k = R / NA mtot = n M
Thermodynamic system
(ideal gas)
work
p V = n R T
p V = N k T
k = R / NA
mtot = n M
N = n NA
internal energy
U = Q – W
= n CV DT
p V T
U S
heat
Q = n C DT
CV or Cp
mtot N n
Q = 0
p V  = constant
T V-1 = constant

43. CV = (f / 2) R Cp = CV + R = (f / 2 +1) R CV = 5/2 R Cp = 7/2 R
Solution
Identify / 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 = J.mol-1.K-1
CV = 20.8 J.mol-1.K-1
Cp = J.mol-1.K-1
p V = n R T = N k T

44. Execute
At A

45. W1 > 0 since gas expands W1 = p1 V1 = (4.0104)(0.06) = 2.4103 J
1 A to B is isobaric
T1= TB – TA = (400 – 100) K = 300 K
pA = pB = p1 = 40 kPa = 4.00104 Pa
V1 = (0.080 – 0.020) m3 = m3
W1 > 0 since gas expands
W1 = p1 V1 = (4.0104)(0.06) = 2.4103 J
U1 > 0 since the temperature increases
U1 = n CV T1 = (1)(20.8)(300) J = 6.2103 J
Q1 > 0 since U1 > 0 and U1 = Q1 – W1 > 0  Q1 > W1 > 0
Q1 = n Cp T1 = (1)(29.1)(300) J = 8.7103 J
Check: First law
U1 = Q1 - W1 = (8.73103 – 2.4103) J = 6.3103 J Q1

47. W2 = 0 since no change in volume
2 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 = 8.3103 J
Q2 = U2 since W2 = 0
Q2 = 8.3103 J Q2

49. 3 C to A T3 = TA – TC = (100 – 800) K = -700 K
pA = 40 kPa = 4.0104 Pa pC = 80 kPa = 8.0104 Pa pCA = 4.0104 Pa
VA = 0.02 m3 VC = 0.08 m V3 = 0.06 m3
W3 < 0 since gas is compressed
W3 = area under curve = area of rectangle + area of triangle
W3 = - { (0.06)(4.0104) + (½)(0.06)(8.0 104)} J = - 3.6103 J
U3 < 0 since the temperature decreases
U3 = n CV T3 = (1)(20.8)(-700) J
= 103 J
Q3 = U3 + W3
Q3 = (- 14.5 103) J = 103 J Q3

50. Refrigerator cycle: |QH| = |QC| +|W|
Complete cycle U = 0 J 1
A to B 2
B to C 3
C to A
cycle
(total values)
W (kJ)
+ 2.4
- 3.6
- 1.2
Q (kJ)
+ 8.7
+ 8.3
- 18.2
+1.2
U (kJ)
+6.3
+8.3
- 14.6
Q – W (kJ)
+ 6.3
Refrigerator cycle: |QH| = |QC| +|W|
|W| TH TC
|QH|
|QC|

51. Wcycle < 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)(4.0104) J = 103 J (value agrees with table)
Evaluate

52. Problem 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.

53. First Law of Thermodynamics
Solution
Identify / Setup p
T2 > T1 T2
First Law of Thermodynamics T1 V1 V2 V

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.

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.

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