6, Module Lecture 4 Molecular Basis of Thermal Physics
Module 6, Lecture 4: Average Values in a Gas
We will work on a __________ basis, or use ______ values. Average will be denoted by: Internal energy consists mainly of: For N molecules:
1 U = N KE = N m v 2 2
Use the average of the square of the velocity, not the square of the average velocity, or even the square of the average speed.
v=
v1 = -2 v2 = 4
EF 152 Spring, 2009 Lecture 6-4 1 EF 152 Spring, 2009 Lecture 6-4
v= v2 = v
2
=
A. B. C. D.
1 3 9 10
2
Average Values in a Gas
2 2 2 2 v 2 = v x + v y + v z2 = v x + v y + v z2
Pressure (from a microscopic viewpoint)
Pressure: arises from collisions of molecules of gas with walls of container
Orient wall parallel to y-z plane. Collision changes x velocity.
Since there is no preferred direction of motion:
vi = v x i + v y + v z k j vf = i+ + k j
P =
Combining with:
1 U = N m v2 2
or U = N
Consider a thin slice by the wall, with volume Adx Write dx in terms of vx:
Number of molecules: Number of collisions:
3 2 m vx 2
3
N coll =
EF 152 Spring, 2009 Lecture 6-4 EF 152 Spring, 2009 Lecture 6-4
1N v x dtA 2V
4
Pressure (from a microscopic viewpoint)
Combine everything: 1 N dPx v dtA (2mv x ) Fx N coll (2mvx ) 1 2 V x 1N 2 p= = mv x = dt = = A A dt A dt AV
Recall we are working on a statistical basis, so use averages. Combine with:
Example: Kinetic Energy of 1 mole of Gas
Given: Recall that 1 mole of gas at STP (0C, 1.00 atm) occupies 22.4 L (0.79 ft3). Required: Kinetic energy of a mole gas
2 of pV = Nm v x = nRT = NkT
3 2 U = N m vx 2
U=
EF 152 Spring, 2009 Lecture 6-4
3 nRT 2
5 EF 152 Spring, 2009 Lecture 6-4 6
Molecular Speeds
U=
Solve for v:
m = mass of single molecule M = molecular mass (grams/mole)
Example: Average Speed of an Oxygen Molecule
Given: O2 molecule at STP conditions. Required: Root-mean-square speed
3 1 1 nRT = Nm v 2 = nM v 2 2 2 2
vrms =
3RT 3kT = M m
What does the subscript rms mean? Calculating rms values: 1. _______ each speed 2. Add em up 3. Divide by number of molecules 4. Take __________
EF 152 Spring, 2009 Lecture 6-4 7 EF 152 Spring, 2009 Lecture 6-4 8
Heat Capacity of Gases
Isochoric: Constant Volume (isovolumetric), W=0
Additional Forms of Energy
Q = U = ncv T
Combine with:
U=
3 nRT 2
Type of Gas Monatomic Diatomic
Gas He Ar H2 N2 O2 CO CO2 SO2 H2S
cv
12.47 12.47 20.42 20.76 21.10 20.85 28.46 31.39 25.95
9
3 cv = R = 12.47 J /(mol K ) 2
Diatomic will have two additional rotational energies. Vibrational motion only occurs above about 600K.
Polyatomic
EF 152 Spring, 2009 Lecture 6-4
EF 152 Spring, 2009 Lecture 6-4
10
Equipartition of Energy
Each velocity component (either translational or angular) has, on average, an associated energy per molecule of kT. Degrees of Freedom: number of velocity components to describe the motion. Diatomic gas: ___ degrees of freedom
cv =
Gas Monotomic Diatomic
EF 152 Spring, 2009 Lecture 6-4
5 R = 20.79 J /( mol K ) 2
cp = R + cv (5 2)R (7 2)R
cv (3 2)R (5 2)R
(5 3) = 1.67 (7 5) = 1.4
11
= cp cv