Phy101 Ch10 Lecture Notes PDF

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Here are Professor Kapila Castoli's Ch. #10 Lecture Notes for Phy101....

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Chapter 10: Thermal Physics Thermal Physics: the study of temperature, heat, and how they affect matter. Temperature – concept of hot and cold are subjective experiences  need a standard To introduce it, we need to define thermal equilibrium first. e.g. Add an ice cube to a cup of hot tea

The ice melts, and the tea gets cooler through exchange of heat. This goes on until thermal equilibrium is reached. i.e. The two have reached a common temperature. Generalizing:

The mercury in the thermometer will go up (or down) until equilibrium is reached. This is how we measure temperare.

 The zeroth law of thermodynamics: If A and B are independently in thermal equilibrium with C, then A and B are in thermal equilibrium with each other.

Temperature scales

Based on melting and freezing point of water and boiling point:  0o C  32o F  100  180 o o 100 C  212 F 

9 5 5 TC  (TF  32) 9 9 Note: TF  TC 5

Conversion: TF  TC  32

The Kelvin (absolute) Scale:

The absolute zero: using a constant volume thermometer with different gases,

we observe that: P



volume

Extrapolating to P = 0, all the curves indicate t° = -273.15°C the absolute zero (theoretical). Experimentally, we have only reached t° ~ 10-7 K. ~ What does t° = 0 K mean?

Zero kinetic energy of the molecules In practice, this cannot be.

TC = T - 273.15°

 20 °C  293 K

room t°

0 °C = 273 K 100 °C = 373 K Conversely, Also,

0 K = -273 °C. K   0 C

Example: Room t°: TC

70 °F

°C ?

5 5  (TF  32)  (70  32) 21 0 C 9 9

Thermal Expansion Ex: expanion joint in a bridge

Ex: railroads

Ex: loosen a tight lid of a jar by holding it under hot water

 no motion

Quantitatively: L  L0 T

where α = coefficient of linear thermal expansion

e.g. α / °C



Aluminum

24 x 10-6

Lead

29 x 10-6

Brass Glass

19 x 10-6 9 x 10-6

Water

2.07 x 10-5

L 1 L0 t

[α] = 1 / °C

Example: Steel reailroad rails are laid when t° = 0 °C. What gap should be left between rail sections so that they just touch when t° = 42 °C? α (steel) = 11 x 10-6 / °C Example: bimetallic strip

α (brass) = 19 x 10-6 / °C α (steel) = 11 x 10-6 / °C Change in area:

L  L 0 T (11 x 10 -6  C-1 )(12.0 m)(42 C) 5.5 mm ( T TC )

A0  L0 ,  2

A L 2 (L 0  L ) 2  (L 0   L 0T ) 2  L20  2L20 T   2 L20 T 2

α2 is very small!

A  L  2 L  T  A0  2 A0  T 2 0

2 0

A  A  A0  2 A0 T

 A0 T

where γ = 2α = coefficient of area expansion

By the same token, change in volume: V  V0 T

where β = 3α = coefficient of volume expansion

Example: On a hot day, an oil trucker loaded 9785 gallons of diesel fuel. He encountered cold weather on the road - 41°F lower t° than before - where he delivered the load. How many gallons did he deliver? β (oil) = 9.5 x 10-4 / °C  V  V0  T  (9785gal )(9.5 x 10 - 4 / C)(-41 F )(5 C /9 F ) -212gal



Vdelivered V0   V  (9785 212) gal  9573gal

N.B.: Unusual Behavior of Water

Max density is at 4 °C.

i.e. It expands when it freezes.

Properties of Ideal Gases Introduce a gas in a container: it expands to occupy the entire volume and exerts a certain pressure

(Force/unit area) on the walls. As the volume expands, there will be lesser amount of pressure.

Also,

case 1 case 2

higher t° lower t°



p, V, t°, amount of gas are interrelated.

For an ideal gas (low density):

pV=nRT

(ideal gas law)

where: p = pressure (absolute) – measured in Pascals or Newtons/m2 V = volume n = number of moles R = universal gas constant = 8.31 J/mol·K T = absolute t° - measured in Kelvins 1 mole: amount of substance that contains as many atoms/molecules as in a 12 gram sample of 12C - How many atoms/molecules are in 1 mole? Lots: NA = 6.02 x 1023 /mole (Avogadro’s number) - Therefore, for 12C: Atomic mass: 12 u = 12 atomic mass units where 1u =1.66 x 10-24 g ≈ mproton ≈ mneutron 

m N A (12u ) (6.02x1023 )(12)(1.66x10 24 g ) 12.0 g

Therefore, the number of moles:

n = m/molar mass

mass of 1 mole of

12

C

where m = mass of substance

Example: A cylinder contains 12L of oxygen at 20°C and 15 atm. The t° is raised to 35°C. The volume is reduced

to 8.5 L. What is the final pressure? 1 atm = 1.013 x 105 Pa

NB:

p iVi nRTi p f V f nRT f



pV = nRT



p V Tf pi Vi T (15atm )(12 L)[(273  35) K ] 22 atm  i or p f  i i  V f Ti p fVf Tf (8.5 L)[(273  20) K ]

We can rewrite the ideal gas law in terms of the number of molecules in the gas: N

n= N , A 

N  number of molecules

pV nRT 

N R RT N ( )T NA NA

Introduce Boltzmann’s constant: kB 

R 1.38x10  23 J / K NA



pV k B N T

Kinetic Theory of Gases For a complete understanding of the behavior of ideal gases, we need to look at the microscopic level.

-

Random motion! Elastic collisions between molecules Elastic collisions with walls



lots of exchange in Kinetic energy

To study the problem, we need to simplify things further: consider rarefied gases



large distance between molecules



Problem: n moles of ideal gas are confined in a cubical box of volume V. The walls of the box are held at t° T. Find the connection between the pressure, p, exerted by the gas on the walls and the velocity v of the molecules.

Consider a simplified situation: although the molecules move in all directions, study a molecule with vx only.

When the molecule bounces off a wall, p ( p f  pi ) (  mvx )  (mvx )  2mvx . (The collision is elastic.) 2L

Over a time t  v , the molecule travels back and forth (wall to wall and back), so x mv x2  p  2mvx molecule :   L 2L /vx t momentum transferred to the wall :

 p' mvx2  L t

Using the 2nd Law of Newton: p mvx2  t L 2 N mv F  xi L i 1

F

Force exerted on the wall: Generalize to N molecules:

N

This causes a pressure on the wall: N

Substituting

v

2 xi

p

Force F  2  Area of wall L

2

N v x v 2x1  v x22  v x23   gives: p 

i 1

2

2

2

2

2

1 3

2

mv xi2 ) L m N 2 m N 2   v ( ( ) ) ( (v ))  xi V i1 xi L2 L3 i 1

( i1

mN 2 vx V

2 2 2 Generalizing to 3D: v v x  v y  v z  v x  v assuming v x v y v z

So, p  i.e.

2 mN 2 2 N 1 2N KE v  ( mv )  3V 3V 3V 2

The pressure is proportional to - the number of molecules/unit volume and - the average KE of a molecule.

Now, we want to find an interpretation of temperature. 2 1 2 pV  N ( mv ) 2 3



pV k B N T .

Also,

2 2 1 N ( mv )  k B N T 2 3



T

2 2 1 ( mv ) 3k B 2

i.e. The temperature of a gas is a direct measure of the average KE of the gas. In other words, the average translational KE of the molecules is

1 2 3 mv  k B T 2 2

2 3 3 1 KE tot (translational)  N ( mv )  ( Nk B ) T  nRT 2 2 2

Therefore,

i.e. measuring the t° of the gas gives us the KE (translational) of its molecules. N.B.: For a monoatomic gas, all KE is just translational. For a diatomic or polyatomic gas,  KEtranslational, KErotational, KEvibrational

Now, v 2 vrms 

v rms 

root  mean  square speed

3kB T 3RT  m M

M molar mass

Some examples for gases at room temperature:

.

This reflects the unequal molar mass of these molecules. e.g. Oxygen: at ~ 300K molar mass of O2: M=32 x 10-3 kg/mol So, v rms 

m 3 RT 3(8.31 J / mol K )(300 K )   4.8 x 10 2 s M 32 x 10  3 kg / mol

Hydrogen: at ~ 300K molar mass of O2: M=2.0 x 10-3 kg/mol So, v rms 

m 3 RT 3(8.31 J / mol K )(300 K )  1.9 x 10 3 s M 2 x 10 3 kg / mol

So, vrms (H2) ~ 4 vrms (O2). That’s why our atmosphere is depleted of Hydrogen. For Nitrogen (N2): vrms is 5.11 x 102 m/s

similar to O2

Examples of curves calculated for N2 at different temperatures – (Maxwell distribution)

Chapter 10 Quick Quiz

One container is filled with argon gas and another with helium gas. Both containers are at the same temperature. Which atoms have the higher rms speed? 3 RT NB: v rms  M

a) Helium b) Argon c) They have the same speed. d) Not enough information to say

Answer: (a) Helium...