kinetic theory of gas气体动力学理论

3. Temperature of an Ideal gas
3 ε t = kT 2
2 P = nε t 3
P = n kT
Temperature in Kelvin
The average translational kinetic energy of molecules in an ideal gas is directly proportional to the absolute temperature. The molecular interpretation of temperature: The temperature of a gas is a direct measure of the average translational kinetic energy of its molecules! The higher the temperature, the faster the molecules are moving on the average. It applies reasonably accurately to liquids and solids.
16-1 The Ideal Gas Law and the Molecular Interpretation of Temperature
Our first model of a many-particle system: the Ideal Gas
Models of matter: gas models (random motion of particles)
F root mean square speed vrms
vrms = v =
for a gas, m T ↑
2
3kT = m
v
2
3 RT
µ
1 3 2 ε t = mv = kT 2 2
R k R = = µ m N Am
at a temperature T
µ↑
v2
↓
This is a particular type of average for a statistical process.
− vix
− mvix
v mvi
As a result of a collision the momentum changes by
− m vix − mvix = −2m vix
The impulse exerted on the wall by the collision = 2 m vix
x
v ix
v2 x =
2 n v ∑ i ix i
n
Suppose dA is perpendicular to x axis ，the collision time is dt.
For each (elastic) collision, the impulse on dA
2m vix
The number of molecules that collide with dA in the time interval dt equals the number in the box.
The impulse exerted by all molecules on dA in the time interval dt
dI =
( v ix > 0 )
∑
2 2 mn i v ix dAdt
The impulse exerted by all molecules on dA in the time interval dt
= 6.42 × 10
−21
(J)
Example (a) What is the rms speed of hydrogen molecules when temperature is 0 ℃ , and (b) the rms speed of oxygen molecules at the same temperature. For H2 molecules
Example What is the average translational kinetic energy of molecules in an ideal gas at 37 °C? Solution:
3 ε t = kT 2
3 = × 1.38 × 10 − 23 × ( 37 + 273) 2
2 x 2 1x 2 2x 2 ix
vx x
vy
y
Since the velocity of the molecules in the gas are assumed to be random, there is no preference to one direction or another.
v = vy = v
dN N n= = dV V
)
Velocities of molecules are different. Each molecule has its velocity, which may be changed due to collisions. In an equilibrium state, velocity of each molecule has the same probability to point to any directions. That is, the distribution of velocity of molecules is uniform in direction, which leads to the mean-square speeds of all components of velocity are same.
1 2 d I = ∑ 2 mn i v d A d t = ∑ 2 mn i v ix d A d t 2 ( v ix > 0 ) dI dI 2 dF = P= = ∑ mn i v ix d A d t dt dAdt
2 ix
P=
2 2 mn v = m n 2 x i
2) Statistical assumptions about an ideal gas
) In an equilibrium state, the distribution of molecules on the
position is uniform, which means that the density of number of molecules is the same everywhere,
mvix
dA
Consider a container of volume V containing N molecules, each of mass m moving with speed v. The number of molecules per unit volume is n .
Let ni be the number of molecules that have velocity between r r r vi − vi + dvi per unit volume.
dA x
v v i dt
v ix dt
ni vix dt dA
Volume of the box
The impulse exerted by molecules r r r on dA that have velocity between vi − vi + dvi equals
ni vix dt dA（ 2m vix）
N
+
N
2 2 v2 = v2 + v + v x y z
v
2 x
dN v v n= dV are statistical average values.
2 y
2 z
1 2 v = vy = v = v 3
2 x 2 2 z
2. Calculating the Pressure Exerted by a Gas
2 be the average value of v Let v 2 x x 2 the average value of v y v
2 y
2 the average value of vz2 vz
z vz
2 v ∑ ix
r v
v + v + L+ v = i v = N N 2 2 v v ∑ iy ∑ iz 2 2 vy = i = i vz N N
ni = Ni V
The number of molecules per unit volume equals
n = ∑ ni
v =
2 x
∑
N
k =1
2 v kx
N
Ni 2 N1 2 N2 2 v1x + v2 x + L vix N v + N v + LN v V V = = V N N /V
2 1 1x 2 2 2x 2 i ix
Chapter 16
Kinetic Theory of Gases
16-1 The Ideal Gas Law and the Molecular Interpretation of Temperature
16-2 Distribution of Molecular Speeds
16-6 Mean Free Path
2 x 2
2 z
For a single molecule
v =v +v +v
2 i 2 ix 2 iy
2 v ∑ i 2 v ∑ ix
2 iz
v = vy = v
2 x 2
2 v ∑ iz