Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 13: Temperature and Ideal Gas

1

Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

Chapter 13: Temperature and Ideal Gas

•What is Temperature?

•Temperature Scales

•Thermal Expansion

•Molecular Picture of a Gas

•The Ideal Gas Law

•Kinetic Theory of Ideal Gases

•Chemical Reaction Rates

•Collisions Between Molecules

2



§13.1 Temperature

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

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.

4



§13.2 Temperature Scales

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.

5



The temperature scales are related by:

F32CF/ 8.1 CF TT

273.15C TT

Fahrenheit/Celsius

Absolute/Celsius

6



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 Kelvin equal the numerical value of Fahrenheit degrees?

F 574

322738.1

322738.1

321.8

F

F

CF

T

T

T

TT

7



§13.3 Thermal Expansion of Solids and Liquids

Most objects expand when their temperature increases.

8



An object’s length after its temperature has changed is

01 LTL is the coefficient of thermal expansion

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

9



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

1-6211

N/m 108.4

K 20K 1012N/m100.2

A

F

TYL

LY

Using Hooke’s Law:

10



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

The blue square has an area of L02.

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

L0

L0+L

11



TA

A

TA

TLL

LTTLL

TLLTLL

2

21

2

2

Aarea new

0

0

20

20

20

2220

20

0000

The fractional change in area is:

12



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

TV

V

0

For solids =3

13



§13.4 Molecular Picture of a Gas

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

If a sample contains a single element, the number of particles in the sample is N = M/m. N is the total mass of the sample (M) divided by the mass per particle (m).

14



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

15



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

kg 1066.1

g 1000

kg 1

106.022

mole 1

mole 12

g 12

27

23

-

This is the conversion factor between the atomic mass unit and kg (1 u = 1.6610-27 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.

16



Example (text problem 13.39): 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 102.1

cm 100

m 1

1

cm 0.1

m

kg 2.1

6

33

3

Vm

17



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:

18



§13.5 Absolute Temperature and the Ideal Gas Law

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

For constant pressure TV Charles’ Law

For constant volume TP Gay-Lussac’s Law

19



Boyle’s LawFor constant temperature V

P1

For constant pressure and temperature

NV Avogadro’s Law

20



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

NkTPV k = 1.3810-23 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.

21



Example (text problem 13.41): A cylinder in a car engine takes Vi = 4.5010-2 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).

22



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.

23



§13.6 Kinetic Theory of the Ideal Gas

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

24



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.

The pressure will depend on:

•The number of gas particles

•Frequency of collisions with the walls

•Amount of momentum transferred during each collision

25



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

26



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

27



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

28



Example (text problem 13.60): What is the temperature of an ideal gas whose molecules have an average translational kinetic energy of 3.2010-20 J?

K 15503

22

3

tr

tr

k

KT

kTK

29



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 6.6410-27 1360

H2 3.32 10-27 1930

N2 4.64 10-26 515

O2 5.32 10-26 482

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

30



§13.7 Temperature and Reaction Rates

For a chemical reaction to proceed, the reactants must have a minimum amount of kinetic energy called activation energy (Ea).

31



If kTEa 2

3

then only molecules in the high speed tail of Maxwell-Boltzmann distribution can react. When this is the situation, the reaction rates are an exponential function of T.

kTEa

eratesreaction

32



Example (text problem 13.76): The reaction rate for the hydrolysis of benzoyl-l-arginine amide by trypsin at 10.0 C is 1.878 times faster than at 5.0 C. Assuming that the reaction rate is exponential, what is the activation energy?

1

1rkT

Ea

e

2

2rkT

Ea

e

where T1 = 10.0 C = 283 K and T2 = 5 C = 278 K; and r1 = 2.878 r2.

2

a

1

a

2

1

kTkTexp

r

r EEThe ratio of the reaction rates is

33



Solving for the activation energy gives:

J 1037.1

K 2831

K 2781

878.1lnJ/K 1038.1

11

ln

1923

12

2

1

a

TT

rr

k

E

Example continued:

34



§13.8 Collisions Between Gas Molecules

On average, a gas particle will be able to travel a distance

VNd /2

12

before colliding with another particle. This is the mean free path. The quantity d2 is the cross-sectional area of the particle.

35



After a collision, the molecules involved will have their direction of travel changed. Successive collisions produce a random walk trajectory.

36



Substances will move from areas of high concentration to areas of lower concentration. This process is called diffusion.

In a time t, the rms displacement in one direction is:

Dtx 2rms D is the diffusion constant (see table 13.3).

37



Example (text problem 13.81): Estimate the time it takes a sucrose molecule to move 5.00 mm in one direction by diffusion in water. Assume there is no current in the water.

s 25000

/sm 100.52

m 1000.5

2D 210

232rms

xt

Dtx 2rms

Solve for t

38



Summary

•Definition of Temperature

•Temperature Scales (Celsius, Fahrenheit, Absolute)

•Thermal Expansion

•Origin of Pressure in a Gas

•Ideal Gas Law

•Exponential Reaction Rates

•Mean Free Path

Documents

Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 13: Temperature and Ideal Gas