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Introductory I, phy231, 1995, page 11

GAS LAWS AND KINETIC THEORY

IDEAL LAW

5 2

 The of the at sea level is 1.01  10 N=m .

 Boyle's Law: For a gas at a constant temp erature the pro duct of pressure and is constant:

PV = constant,or P V = P V .

1 1 2 2

 Law of Charles and Gay-Lussac: For a gas at a constant pressure the volume is prop ortional to the

temp erature: V=T = constant,or V =V = T =T .

1 2 1 2

Rememb er: The temp erature MUST b e in !!

 law: PV = nRT , where R =8:314J=mol  K  is the universal and n

corresp onds to the quantity of gas in mol es.

 The number of mol es is de ned as the ratio of the total M of the gas and the Molecular Mass

in g =mol  of the gas: n = M=M ol ecul ar M ass. The Molecular mass of a gas is di erent for each

gas. For example: Molecular oxygen O =32g =mol , He= 4g =mol etc.

2

 1 mol e of any gas contains the same numb er of .

23

This number is N =6:02  10 M ol ecul es=mol and is called Avogadro's numb er.

A

KINETIC THEORY OF

 The temp erature of a gas can b e related to the internal of the molecules.

 The ideal gas approximation yields: PV =2=3N KE , where KE is the mean kinetic of an

individual molecule and N is the total numb er of molecules in the gas.

 The U of a gas is de ned as U = N KE and thus PV =2=3U .

KE =3=2kT , where  The relation b etween the mean and the temp erature is given by

23

k = R=N is the : k =1:3807  10 J=K .

A

q

p

2

 The root mean square velo city is the square ro ot of the mean velo city squared: v = v = 3P=,

rms

where  is the of the gas.

q

3kT =m, where m is the mass of one individual molecule.  v can also b e expressed as v =

rms rms

 The mass of one molecule can b e calculated from Avogadro's numb er and the Molecular mass of the

gas: m = M ol ecul ar M ass=N .

A

 Other helpful relations: The total mass M of a gas is the pro duct of the Molecular Mass and the

numb er of moles n: M = M ol ecul ar M ass  n

The total numb er of gas molecules in a gas N is the pro duct of the numb er of moles n and Avogadro's

number N : N = nN .

A A

mg h=k T

 Barometric : The pressure at height h ab ove height zero is given by P = P e .

0 11

Introductory Physics I, phy231, Spring 1995, page 12

THERMODYNAMICS

THERMODYNAMIC SYSTEMS AND

 A thermo dynamic system is any collection of ob jects considered together. Everything else is consid-

ered the environment.

th

 0 Law:  If two ob jects are in thermal equilibrium with a third ob ject they

are also in thermal equilibrium with each other.

 The internal energy is de ned as the total energy within the system. If no phase changes are involved

the internal energy is prop ortional to the temp erature of the system.

st

 1 Law: Energy Conservation The change of internal energy U is equal to the Q added to

the system minus the W done by the system: U = Q W .

 Adiabatic: In an adiabatic pro cess no heat is exchanged with the environment fast pro cess:

Q =0! U = W .

 Isothermal: In an isothermal pro cess the temp erature internal energy do es not change:

U =0! Q = W .

 Isobaric: In an isobaric pro cess the pressure do es not change. The work done by the system is then

equal to the pro duct of the pressure the change of the volume: W = P V .

 In general, the work is equal to the area under the path in the P-V diagram. If a system changes

from P and V straighttoP and V the work done is W =1=2P + P V V .

1 1 2 2 2 1 2 1

 Iso choric: In an iso choric pro cess the volume of a system do es not change. If the volume is constant

no work is done: W =0! U = Q.

CARNOT ENGINE

 In a reversible pro cess there is no wasted energy and the system is very nearly in equilibrium at all

times.

 A carnot cycle is an idealized reversible pro cess of an ideal gas b etween a and hot reservoir and

consists of four reversible pro cesses two adiabatic and two isothermal. Heat Q is taken out of a

H

hot reservoir and some of this energy is converted to work and the rest is added as heat Q to the

C

cold reservoir.

 The thermal eciency of a carnot cycle Heat engine is given by the ratio of the resulting work by

the heat input: Thermal eciency = W=Q . The work is equal to the heat ow from the hot to the

H

cold reservoir: W = Q Q .

H C

 The ratio Q =Q is prop ortional to the ratio of the temp eratures T =T in Kelvin!! and thus:

C H C H

Thermal eciency =1 T =T .

C H

 The co ecient of p erformance c:p: of a refrigerator is given by: c:p: = Q =W = T =T T .

C C H C

The co ecient of p erformance c:p: of a heat is given by: c:p: = Q =W = T =T T .

H H H C

ENTROPY

nd

 2 Law: Heat cannot, by itself, pass from a colder to a warmer b o dy.

 The S is de ned as the ratio of absorb ed Heat Q divided by the Temp erature T in a

reversible pro cess: S = Q=T . 12