22
Kinetic Theory of Gases Dr. Richard C. Sobers Jr.
Kinetic Theory of GasesDr. Richard C. Sobers Jr.
Kinetic Energy of a Moving Particle
Mass m (kg)
Velocity v (m/s)
Kinetic Energy, KE, = 1/2 mv2
Unit of Energy = Joule (J)1J = 1 kg • m2/v2
Joules are also unit of work and heat
Also 1 J = 4.184 cal KJ and Kcal are often used
Kinetic Energy of a Moving Particle
Mass m (kg)
Velocity v (m/s)
Kinetic Energy, KE, = 1/2 mv2
Unit of Energy = Joule (J)1J = 1 kg • m2/v2
Joules are also unit of work and heat
Also 1 J = 4.184 cal KJ and Kcal are often used
Maxwell Velocity Distribution
A weighting factor used to
f (u)du = 4πu2 m2πRT
⎛⎝⎜
⎞⎠⎟3/2
e−mu2 /2RTdu
Maxwell Velocity Distribution
A weighting factor used to
f (u)du = 4πu2 m2πRT
⎛⎝⎜
⎞⎠⎟3/2
e−mu2 /2RTdu
Maxwell Velocity Distribution
A weighting factor used to
f (u)du = 4πu2 m2πRT
⎛⎝⎜
⎞⎠⎟3/2
e−mu2 /2RTdu
Maxwell Velocity Distribution
A weighting factor used to
f (u)du = 4πu2 m2πRT
⎛⎝⎜
⎞⎠⎟3/2
e−mu2 /2RTdu
most probable speed: ump = 395m/s
root mean square speed: urms = 484m/s
average speed: urms = 446m/s
Maxwell Energy Distribution
A weighting factor used to
f (E)dE = 2 Eπ
⎛⎝⎜
⎞⎠⎟1/2 1
RT⎛⎝⎜
⎞⎠⎟3/2
e−E /RTdE
Kinetic Energy
Minimal Energy Requirement
A weighting factor used to
Suppose a reaction has a 15kJ activation energy.
The Boltzmann Factor
e− EkBT
A weighting factor used to predict the number of particles with an energy E at a temperature T.
Factor decreases with increasing E and increases with increasing T.
A weighting factor used to
f (E)dE = 2 Eπ
⎛⎝⎜
⎞⎠⎟1/2 1
RT⎛⎝⎜
⎞⎠⎟3/2
e−E /RTdE
The Boltzmann Factor
e− EikBT
Boltzmann Constant: kB = 1.381x10-23 J/K
A weighting factor used to
For one mole of particles: 1.381x10-23J/K • 6.022x1023mol-1 =
8.314 J/(K•mol) The gas constant R
Translational Kinetic Energy of Particles
A weighting factor used to
Average Kinetic Energy of a mole of particles in one dimension: 1/2 RT (1/2kBT per molecule)
In three dimensions: KEave = 1/2RT + 1/2RT + 1/2RT = 3/2RT
Another factor derived from kinetic theory of gases is that each degree of freedom provides 1/2RT energy.
Translational Kinetic Energy of Particles
A weighting factor used to
KEave = 3/2kBTPer molecule molecule:
KEave =12mu2
Or for one mole:
KEave =12Mu2
m = mass of one molecule, M = molar mass
KEave = 3/2RT
Translational Energy of Particles
A weighting factor used to
For one mole of particles:
32RT = 1
2mu2
u2 = urms =3RTM
Sample Problem
Calculate the root mean square speed of molecular chlorine at 20oC.
urms =3RTM
T = 293KM = 7.090x10-2 kg/mol
urms =3(8.314J •K −1 •mol−1)293K
7.090x10−2 kg /mol= 321m / s
Effusion and DiffusionThe spreading out of a gas from high concentration to low concentration is diffusion.
Effusion is the process by which molecules escape through a small hole into a vacuum.
EffusionRelative rates of effusion and diffusion are due to relative molecular speeds.
rate1 ∝3RTM1
rate2 ∝3RTM 2
Compare two gases:urms =
3RTM
Graham’s Law of DiffusionRelative rates of effusion and diffusion are due to relative molecular speeds.
rate1rate2
=
3RTM1
3RTM 2
Compare two gases:
rate1 ∝3RTM1
rate2 ∝3RTM 2
rate1rate2
= M 2
M1
Graham’s Law of DiffusionRelative rates of effusion and diffusion are due to relative molecular speeds.
rate1rate2
= M 2
M1
Graham’s law of diffusion (1832)
The relative rates of diffusion of two gases at the same temperature and pressure are inversely proportional to the molar masses of the gases.
Graham’s Law of Diffusion
Example: It takes 192s for an unknown gas to effuse through a porous barrier and 84s for N2 to effuse through the same barrier. Calculate the molar mass of the unknown gas.
RateX = 1/192s = 0.00521s-1
RateN2 = 1/84s = 0.012s-1
MX = ?
MN2 = 28.02g/mol
Rates are inversely proportional to time taken!
Graham’s Law of Diffusion
Example: It takes 192s for an unknown gas to effuse through a porous barrier and 84s for N2 to effuse through the same barrier. Calculate the molar mass of the unknown gas.
rateXrateN 2
= MN 2
MX
0.00521s−1
0.021s−1= 28.02g /mol
MX
Graham’s Law of Diffusion
Example: It takes 192s for an unknown gas to effuse through a porous barrier and 84s for N2 to effuse through the same barrier. Calculate the molar mass of the unknown gas.
0.442 = 28.02g /molM1
M1 = 146g/mol
