2. Ideal Gas
The number of molecules is large
The average separation between
molecules is large
Molecules moves randomly
Molecules obeys Newton’s Law
Molecules collide elastically with each
other and with the wall
Consists of identical molecules
3. The Ideal Gas Law
PV = nRT in K
n: the number of moles in the ideal gas
N total number
n=
NA of molecules
Avogadro’s number: the number of
atoms, molecules, etc, in a mole of
a substance: NA=6.02 x 1023/mol.
R: the Gas Constant: R = 8.31 J/mol · K
4. Pressure and Temperature
Pressure: Results from collisions of molecules
on the surface
Force
F
Pressure: P=
A Area
dp
Force: F= Rate of momentum
dt given to the surface
Momentum: momentum given by each collision
times the number of collisions in time dt
5. Only molecules moving toward the surface hit
the surface. Assuming the surface is normal to
the x axis, half the molecules of speed vx move
toward the surface.
Only those close enough to the surface hit it
in time dt, those within the distance vxdt
The number of collisions hitting an area A in
time dt is 1 N
⋅ A ⋅ vx ⋅ dt
2V
Average density
The momentum given by each collision to
the surface 2mvx
6. Momentum in time dt:
1 N
dp = (2mv x )⋅ ⋅ ⋅ A ⋅v x dt
2 V
Force: dp 1 N
F= = (2mvx )⋅ ⋅ ⋅ A⋅ v x
dt 2 V
Pressure: F N 2
P = = mv x
A V
Not all molecules have the same v x ⇒ average v2
x
N 2
P = mv x
V
7. 2
vx
1 2 1 2 2
3 3
( 2
= v = v x + v y + vz )
2 1 2 1 2
vx = v = vrms
3 3
vrms is the root-mean-square speed
2 2 2
vx + vy + vz
vrms = v 2 =
3
1N 2 2 N1 2
Pressure: P = mv = mv
3V 3 V 2
Average Translational Kinetic Energy:
1 2 1 2
K = mv = mvrms
2 2
8. 2 N
Pressure: P = ⋅ ⋅K
3 V
2
From PV = ⋅ N ⋅ K and PV = nRT
3
3 nRT 3
Temperature: K= ⋅ = ⋅ k BT
2 N 2
R −23
Boltzmann constant: k B = = 1.38 × 10 J/K
NA
9. 1 2
From PV = ⋅ N ⋅ mvrms
3
N
and PV = nRT = RT
NA
Avogadro’s number
N = nN A
3RT
vrms =
M Molar mass
M = mN A
10. Internal Energy
For monatomic gas: the internal energy = sum
of the kinetic energy of all molecules:
3 3
Eint = N ⋅ K = nN A ⋅ k BT = nRT
2 2
3
Eint = nRT ∝ T
2
11. Mean Free Path
Molecules collide elastically with other
molecules
Mean Free Path λ: average distance between
two consecutive collisions
1
λ= 2
2πd N / V
the bigger the molecules the more molecules
the more collisions the more collisions
12. Q = cm ⋅ ∆T
Molar Specific Heat ∆Eint = Q − W
3
Eint = nRT
2
Definition:
For constant volume: Q = nCV ∆T
For constant pressure: Q = nC p ∆T
The 1st Law of Thermodynamics:
3
∆Eint = nR∆T = Q − W (Monatomic)
2
14. 3
nR∆T = Q − W
2
Constant Pressure (Monatomic)
Q = nC p ∆T
W = P∆V = nR∆T
3
nR∆T = nC p ∆T − nR∆T
2
5
CV = Cp − R Cp = R
2
Cp 5
γ = γ =
CV 3
15. 1st Law
dEint = dQ − dW
Adiabatic Process
Ideal Gas Law
pV = nRT
(Q=0) Eint = nCV T
dEint = −dW = − pdV C p = CV + R
= nCV dT Cp
γ =
pdV CV
pdV + Vdp = nRdT = nR −
nCV
Divide by pV:
dV dp C p − CV dV dV
+ = − = (1 − γ )
V p CV V V
16. dV dp dV Ideal Gas Law
+ = (1 − γ ) pV = nRT
V p V
dp dV
+γ =0
p V
γ γ
ln p + ln V = ln( pV ) = const.
γ
pV = const.
nRT γ
( )V = const.
V
γ −1
TV = const.
17. Equipartition of Energy
The internal energy of non-monatomic
molecules includes also vibrational and
rotational energies besides the
translational energy.
Each degree of freedom has associated with
1
it an energy of k BT per molecules.
2
18. Eint = nCV T
Monatomic Gases
3 translational degrees of freedom:
3 3
Eint = kBT ⋅nN A = nRT
2 2
1 dEint 3
CV = ⋅ = R
n dT 2
19. Eint = nCV T
Diatomic Gases
3 translational degrees of freedom
2 rotational degrees of freedom
2 vibrational degrees of freedom
HOWEVER, different DOFs require different
temperatures to excite. At room temperature,
only the first two kinds are excited:
5 5
Eint = nRT CV = R
2 2
