Prepared
by Dr. A. Mekki
Summary
of chapter 20
where N is the number of
molecules, m is the mass of the substance, M is the molar mass of the substance
and NA is Avogadros number.
NA = 6.02 x 10-23
molecules/mole (is the number of molecules in ONE mole of the substance).
The ideal gas law
R = 8.31 J/mole K, is the
gas constant and n is the number of moles of the gas.
We have three situations
(i) Temperature constant (isothermal process)
(ii) Volume constant (isochoric process)
(iii) Pressure constant (isobaric process)
In the above equations T in Kelvin.
T
in Kelvin
For an isobaric
process, the work
done on or by the gas is
W =
P DV
If Vf > Vi
(expansion), then W > 0, the gas do work
If Vf < Vi (compression),
then W < 0, external work is done on the gas.
where vrms
is the root-mean-square speed of the gas molecules =
This
speed is related to the molar mass and the temperature of the gas as
follows:
T in Kelvin
T in Kelvin
k =
1.38 x 10-23 J/K is Boltzman constant.
T in Kelvin
Therefore
the change in internal energy is
So: for an isothermal process the change in
internal energy of the gas is ZERO because DT = 0.
(i)
for a constant volume
process (isochoric) the heat is given by
(ii)
for a constant pressure
process (isobaric) the heat is given by
T
in Kelvin
where is the specific heat ratio (constant).
Process |
P-V diagram |
W |
Q |
DEint |
Isothermal |
|
|
|
0 |
Isobaric |
|
P DV |
n Cp DT |
n Cv DT |
Isochoric |
|
0 |
n Cv DT |
n Cv DT |
Adiabatic |
|
- n Cv DT |
0 |
n Cv DT |
Cyclic |
|
Area enclosed |
Area enclosed |
0 |
Note:
This is ALWAYS
true, for all processes!
Gas |
Cv
|
Cp |
g
= Cp/Cv |
Monoatomic |
3/2 R |
5/2 R |
1.67 |
Diatomic |
5/2 R |
7/2 R |
1.4 |
Cp = Cv
+ R