Of the molecular-kinetic theory of gases Presentation

Fundamental equation of the molecular-kinetic theory of gases

Today’s Theme:

Learning Objectives

To apply the fundamental equation of the MKT in problem solving

Notation Guidelines

p – pressure (Pa)
T – absolute temperature (K)
V – Volume (m3)
v – velocity (m/s) (can be written as c)
- average velocity (m/s)
Na – Avogadro constant (6.02 x 1023)
Mr – relative molecular mass (mass in grams of a mole of the molecule)
n – amount of substance in moles
N – number of molecules
m0 – mass of one molecule
R – universal gas constant (8.31 J K-1 mol-1)
k – the Boltzmann constant (1.38 x 10-23 J K-1)

Combined Gas Law

These gas laws can be combined together in the combined gas law, which applies for any fixed mass of gas.
𝑝𝑝𝑉𝑉∝𝑇𝑇
Or
𝑝 1 𝑉 1 𝑇 1 𝑝 1 𝑝𝑝 𝑝 1 1 𝑝 1 𝑉 1 𝑉𝑉 𝑉 1 1 𝑉 1 𝑝 1 𝑉 1 𝑇 1 𝑇 1 𝑇𝑇 𝑇 1 1 𝑇 1 𝑝 1 𝑉 1 𝑇 1 = 𝑝 2 𝑉 2 𝑇 2 𝑝 2 𝑝𝑝 𝑝 2 2 𝑝 2 𝑉 2 𝑉𝑉 𝑉 2 2 𝑉 2 𝑝 2 𝑉 2 𝑇 2 𝑇 2 𝑇𝑇 𝑇 2 2 𝑇 2 𝑝 2 𝑉 2 𝑇 2

Ideal Gas Equation

We can take the combined gas law for a fixed mass of gas and determine the proportionality constant
𝑝𝑝𝑉𝑉=(𝑝𝑝𝑟𝑟𝑜𝑜𝑝𝑝𝑜𝑜𝑟𝑟𝑡𝑡𝑖𝑖𝑜𝑜𝑛𝑛𝑎𝑎𝑙𝑙𝑖𝑖𝑡𝑡𝑦𝑦 𝑐𝑐𝑜𝑜𝑛𝑛𝑠𝑠𝑡𝑡𝑎𝑎𝑛𝑛𝑡𝑡)𝑇𝑇
Since changing the amount of substance will change the pressure, volume, and temperature, the amount of substance is part of this constant. The other part is experimentally determined. This constant, n, is known as the molar gas constant or universal gas constant (because it is the same for all gases).
𝑝𝑝𝑉𝑉=𝑛𝑛𝑅𝑅𝑇𝑇
R = 8.31 J K-1 mol-1

Ideal Gas Equation (continued)

A second version of this equation exists:
𝑝𝑝𝑉𝑉=𝑁𝑁𝑘𝑘𝑇𝑇
Where N is the number of molecules, and k is the Boltzmann constant. k=1.38 x 10-23 J K-1

The equation you will use depends on what information you have!

Average Kinetic Energy of Gas Molecules

We can use the equation of MKT and the equation of state for an ideal gas to find the average kinetic energy of a gas molecule:
𝑝𝑝= 𝑁𝑘𝑇 𝑉 𝑁𝑁𝑘𝑘𝑇𝑇 𝑁𝑘𝑇 𝑉 𝑉𝑉 𝑁𝑘𝑇 𝑉 and 𝑝𝑝= 1 3 1 1 3 3 1 3 ∙ 𝑁 𝑉 𝑁𝑁 𝑁 𝑉 𝑉𝑉 𝑁 𝑉 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 < 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 >
𝑁𝑘𝑇 𝑉 𝑁𝑁𝑘𝑘𝑇𝑇 𝑁𝑘𝑇 𝑉 𝑉𝑉 𝑁𝑘𝑇 𝑉 = 1 3 1 1 3 3 1 3 ∙ 𝑁 𝑉 𝑁𝑁 𝑁 𝑉 𝑉𝑉 𝑁 𝑉 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 < 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 >
3𝑘𝑘𝑇𝑇= 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 < 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 >
Knowing that <𝐸 𝑘 <𝐸𝐸 <𝐸 𝑘 𝑘𝑘 <𝐸 𝑘 >= 𝑚 0 <𝑣> 2 2 𝑚 0 <𝑣> 2 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 <𝑣𝑣> 𝑚 0 <𝑣> 2 2 𝑚 0 <𝑣> 2 𝑚 0 <𝑣> 2 2 2 𝑚 0 <𝑣> 2 2 , we can write
𝑚 0 < 𝑣 2 > 2 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 < 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 > 𝑚 0 < 𝑣 2 > 2 2 𝑚 0 < 𝑣 2 > 2 = 3 2 3 3 2 2 3 2 𝑘𝑘𝑇𝑇 𝑜𝑜𝑟𝑟 <𝐸 𝑘 <𝐸𝐸 <𝐸 𝑘 𝑘𝑘 <𝐸 𝑘 >= 3 2 3 3 2 2 3 2 𝑘𝑘𝑇𝑇

r.m.s. speed

From our equation for average kinetic energy, we can derive a second equation for the r.m.s. speed:
𝑚 0 < 𝑣 2 > 2 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 < 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 > 𝑚 0 < 𝑣 2 > 2 2 𝑚 0 < 𝑣 2 > 2 = 3 2 3 3 2 2 3 2 𝑘𝑘𝑇𝑇

𝑣 𝑟𝑚𝑠 𝑣𝑣 𝑣 𝑟𝑚𝑠 𝑟𝑟𝑚𝑚𝑠𝑠 𝑣 𝑟𝑚𝑠 = <𝑣 2 > <𝑣 2 > <𝑣 2 <𝑣𝑣 <𝑣 2 2 <𝑣 2 > <𝑣 2 > = 3𝑘𝑇 𝑚 0 3𝑘𝑇 𝑚 0 3𝑘𝑇 𝑚 0 3𝑘𝑘𝑇𝑇 3𝑘𝑇 𝑚 0 𝑚 0 𝑚𝑚 𝑚 0 0 𝑚 0 3𝑘𝑇 𝑚 0 3𝑘𝑇 𝑚 0