Phys 250 Ch12 p1

Chapter 12: Gas Laws and Kinetic Theory

Air Pressure

at bottom of column of mercury:

P = gh, h≈76 cm

pressure= atmospheric pressure, which support column of mercury

Boyle’s Law:

At constant temperature, the volume of a gas is inversely proportional to is pressure.

=> the product of pressure and volume for a sample of gas is fixed

p1V1 = p2V2 (T = constant)

Example: A cylinder with a height .20 m and cross sectional area pf 0.40 m2 has a close fitting piston which compresses the cylinder volume to a height of .12 m. If the air started at atmospheric pressure and the temperature remains constant, what is the new pressure of the compressed air within the cylinder?

vacuum

air pressure

Phys 250 Ch12 p2

The Law of Charles and Gay-Lussac

Thermal expansion at constant pressure

V = b V0 T

All gases have the same value for !

starting at 0°C , = 1/273 /°C

Extrapolated: all gases have zero volume at 273°C (460°F)

All gases extrapolate to zero pressure in a constant volume gas thermometers at 273°C

Absolute Temperature Scales are natural scales, and

Example: To what temperature would the air in a hot air balloon have to be heated so that its mass would be 0.980 times that of an equal volume of air at a temperature of 25C?

)constant = (2

2

1

1 pT

V

T

V

Phys 250 Ch12 p3

Ideal Gas Law

Charles’s Law and Boyle’s Law can be combined to relate pressure, volume and absolute temperature for more general changes.

nRTPV

or

nRT

VP

T

VP

=constant =2

22

1

11

n = number of moles

R = Gas Constant

R=8.314 J/mole·K

The Ideal Gas Law is an example of an Equation of State (an equation which relates the variables describing the state of the system).

Phys 250 Ch12 p4

Microscopic

Atoms, simplest and smallest subdivision

Molecule, a combination of atoms bonded together

Different atoms and/or molecules not bonded together

Macroscopic

Elements, chemically simplest materials

Compound, can be chemically reduced

Mixture, can be chemically or physically separated/simplified

The type of atom determines the element, the type of molecule determines the compound, etc.

Atoms and Molecules

mass often expressed in atomic mass units

1 atomic mass unit = 1 u = 1.66x10-27 kg

Isotopes of an element: extremely slight variation in chemical properties => slightly different types of atoms of same element (different masses for different atoms of same element).

The weight of an atom in u is approximately the same as the molecular weight of the corresponding element in grams.

Phys 250 Ch12 p5

Example: A sample of a gas originally has a volume of 0.5 liters at room temperature (23ºC) and pressure (1.00 atm) is transferred to a container where it is cooled to 55C in a volume of .12 L. What is the new pressure of the gas?

Example: What is the density of carbon dioxide gas at a temperature of 23ºC and atmospheric pressure?

Phys 250 Ch12 p6

Example: An automobile tire is filled to a gauge pressure of 240 kPa early in the morning when the temperature is 15ºC. After the car has been driven for the day, the temperature of the iar in the tires is 70ºC. Estimate the new gauge pressure.

Phys 250 Ch12 p7

Kinetic Theory of the Ideal Gas

Container with volume V contains a large number N of identical molecules of mass m.

Molecules act as point particles (size is small compared to intraparticle distances).

Molecules are in constant motion and obey Newton’s Laws of motion. Molecules collide elastically with walls of container.

Walls of container are perfectly rigid.

Pressure from collisions: Each elastic collision exerts an impulse on the wall of the container.

=> Boyle’s Law: pressure is inversely proportional to volume

vx

vy

vvx

vyv

Phys 250 Ch12 p8

Each elastic collision exerts an impulse on the wall of the container

xxx mvmvmvp 2))((

For a molecule with vx to hit wall within a time dt, it must be within vxdt of the wall.

The number of collisions is

The total imulse on the wall is

Impulse is also related to force

Pressure is average force per area

vx

vy

vvx

vyv

A

vxdt

)(2

1dtAv

V

Nx

AdtmvV

N

mvdtAvV

NI

x

xxtotalx

)(

)2)((2

1

2

dtFI xtotalx average

average2

average

)( x

x

mvV

N

AFp

Phys 250 Ch12 p9

av

av2

2222

2222

)(3

2

)(3

13

1

tr

avavzavyavx

zyx

KEV

Np

mvV

Np

vvvv

vvvv

symetry!

M

RT

m

kTvv

nRTNkTpV

KmoleculeJN

Rk

N

RT

nN

nRT

N

nRTKE

nRTpV

KENpV

rms

A

AA

33)(

1038.1

constant sBoltzmann'

2

3

2

3

2

3)(

)(3

2

av2

23

avtr

avtr

Phys 250 Ch12 p10

Example: Estimate the rms speed of oxygen molecules at STP (standard Temperature and Pressure: 0ºC and 1 atm). Compare this with the speed of hydrogen molecules under the same conditions.

Phys 250 Ch12 p11

Internal Energy of an Ideal Gas

average KE of one molecule: KE =3/2 kT

for N molecules KEtot = N 3/2 kT

but N = n NA, and R = NAk so

U = 3/2 n NAk T = 3/2 nRT

this is the Internal Energy of the Gas

Example: A parade balloon contains 368 m3 of helium at a pressure of 115 kPa. What is the internal energy of the helium in the balloon?