Introductory Physics I, phy231, Spring 1995, page 11
GAS LAWS AND KINETIC THEORY
IDEAL GAS LAW
5 2
The pressure of the atmosphere 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 volume 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 Kelvin!!
Ideal gas law: PV = nRT , where R =8:314J= mol K is the universal gas constant 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 mass 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 , Helium He = 4g =mol etc.
2
1 mol e of any gas contains the same numb er of particles.
23
This number is N =6:02 10 M ol ecul es=mol and is called Avogadro's numb er.
A
KINETIC THEORY OF GASES
The temp erature of a gas can b e related to the internal motion of the molecules.
The ideal gas approximation yields: PV =2=3N KE , where KE is the mean kinetic energy of an
individual molecule and N is the total numb er of molecules in the gas.
The internal energy 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 kinetic energy and the temp erature is given by
23
k = R=N is the Boltzmann constant: 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 density 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 formula: 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 ENERGY CONSERVATION
A thermo dynamic system is any collection of ob jects considered together. Everything else is consid-
ered the environment.
th
0 Law: Thermal Equilibrium 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 heat Q added to
the system minus the work 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 times 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 cold 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 eciency of a carnot cycle Heat engine is given by the ratio of the resulting work by
the heat input: Thermal eciency = 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 eciency =1 T =T .
C H
The co ecient of p erformance c:p: of a refrigerator is given by: c:p: = Q =W = T = T T .
C C H C
The co ecient of p erformance c:p: of a heat pump 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 entropyS 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