Unit 3.3 Kinetic Theory Flashcards

Question

What is Boyle's law? (Gas laws, 2 each)

Answer 1

- PV = constant - P₁V₁ = P₂V₂

Answer 2

- V/T = constant - V₁/T₁ = V₂/T₂

Answer 3

- P/T = constant - P₁/T₁ = P₂/T₂

Answer 4

Boyle's law P₁V₁ = P₂V₂

Answer 5

Charles' law V₁/T₁ = V₂/T₂

Answer 6

Pressure law P₁/T₁ = P₂/T₂

Answer 7

P₁V₁/T₁ = P₂V₂/T₂

Answer 8

Pa = N/m³

Answer 9

Pressure = force/volume

Answer 10

The n° of atoms in 12g of carbon-12

Answer 11

6.02 x 10²³ (N_a)

Answer 12

Mass of one mole

Answer 13

Amount of a substance that contains 6.02 x 10²³ particles

Answer 14

Mr (Relative molecular mass)

Answer 15

Number of moles = mass of gas/Molar mass

Answer 16

- The sum of all the relative atomic masses - for all the atoms in a given formula

Answer 17

M(kg) = Mr/1000

Answer 18

Molar mass (kg) = Relative molecular mass/100

Answer 19

- Particles have volume - Energy lost in collisions - Intermolecular forces

Answer 20

Pressure x Volume = N° of molecules x Boltzmann's constant x Temperature

Answer 21

Pressure (Pa)

Answer 22

Volume of gas (m³)

Answer 23

Number of molecules

Answer 24

Boltzmann's constant (1.38 x 10^-23JK^-1, in data booklet)

Answer 25

- Jerky motion - Random direction - Various speeds

Answer 26

- Random collisions between smoke PTCL & air PTCL - Each collision = change of momentum - Due to Newton's 2nd law, force is applied - Smoke PTCL's bombarded with enough air PTCL's, - force can change smoke PTCL's speed & direction

Answer 27

- Random collisions between gas PTCL's and container walls - Change of momentum for the particles - Means the wall must've applied a force on particles - Due to Newton's 3rd law, - particles must exert equal & opposite force on wall - Force provides pressure as Pa = N/m³

Answer 28

1. State what stays the same 2. State changes and effect it has on rate of collisions 3. Link to force and so pressure

Answer 29

- Constant temperature - Decreased volume - Each collision provides force - Force increases - Therefore more pressure - As volume decreases, pressure increases

Answer 30

- KE constant - Speed constant

Answer 31

- Less distance between collision with wall - More collisions per second

Answer 32

- Constant volume - Increased temperature - Same distance but increased speed - More collisions per second - Increased rate of collisions - As pressure increases, temperature increases

Answer 33

- Constant distance between collision w/ wall - Constant collisions per second

Answer 34

- Increased KE - Mean speed increases

Answer 35

- Constant pressure - Increased temperature - To maintain pressure (rate of collisions - distance between walls increases - Volume increases - As volume increases, temperature increases at constant pressure

Answer 36

- Constant force - Rate of collisions constant

Answer 37

- Increased speed of particles - Increase rate of particles (rate of change of momentum - Unless distance between collisions increase

Answer 38

- They don't move at identical speeds - Distribution of energy between particles = random

Answer 39

- Most probable speed - Mean speed - Root mean square speed

Answer 40

- Most particles move at this speed - Peak of Maxwell-Boltzmann distribution

Answer 41

Average value of all speeds

Answer 42

C̅ = sum of all particles/n° of particles

Answer 43

- The square root of - the mean square speed of the molecules

Answer 44

√C̅² = √(sum of all speeds/n° of particles)

Answer 45

- No intermolecular forces - Negligible volume of molecules - Elastic collisions between molecules

Answer 46

- Duration of collisions very short - compared to time between collisions

Answer 47

Energy distribution for particles is random

Answer 48

Large n° of particles; large n° of collisions

Answer 49

Velocity of particles is uniform between collisions

Answer 50

- Even distribution of particle motion - in all directions

Answer 51

pV = ⅓NmC̅²

Answer 52

Pressure (Nm^-2)

Answer 53

Volume (m³)

Answer 54

Mass/one particle (kg)

Answer 55

N° of molecules

Answer 56

Mean square speed (ms^-2)

Answer 57

- Newton's 3rd law - Force of molecules - colliding with the container

Answer 58

- Newton's 2nd law - Depends on the change of momentum - of particles due to collisions

Answer 59

About 9 stages (my way)

Answer 60

- Molecules move in all directions - 1 molecule of mass (m) - Travels with velocity (c_{x)
- Collides with walls of container
- Each wall has a length of L}

Answer 61

- Calculate the change in momentum - Before it moves with velocity v_x - and after the collision it moves with -v_x - △mc_x = (mc_x) - (-mc_x) -> **△mc_x = 2mc_x** - Equation 1

Answer 62

- Time given by distance/speed - Speed is c_x - Distance is twice the length of box - (distance to collide, then collide again with same wall) - t = 2L/c_x - Equation 2

Answer 63

- Calculate force by: - Force = change in momentum/time - Sub in equation 1 & 2 - F = 2mc_x/2L/c_x -> **mc²x_x/L - Equation 3

Answer 64

- Equation 3 gives force of one molecule - acting on the side of the container - Can now calculate pressure one molecule causes in x direction - p = F/A, sub in equation 3 - p = mc²_x/2L/c_x -> mc²_x/L³ - Equation 3.5

Answer 65

- Assume box is a cube - Can replace L³ with V (equation 3.5) - Both units are m³ - p = mc²_x/V - Equation 4

Answer 66

- Gives pressure of one molecule - acting on the side of the container in 1 direction

Answer 67

- Must find pressure of all particles in all directions - For total pressure: - use n° of particles x mean pressure per PTCL - Energy distribution between particles = random - All molecules of gas = different speeds in x direction - Find mean pressure per PTCL using rms speed - Then multiplying by N (total n° of molecules)

Answer 68

- p = mc_x²/V - p = c̅_x²/V - p = Nmc̅_x²/V - Equation 5

Answer 69

- Equation 5 gives pressure in x direction - Mean speed in all directions given by: - c̅² = c̅²_x + c̅²_y + c̅²_z - But average velocities in all directions are equal: - c̅² = 3c̅²_x - c̅²₂ = 1/3 c̅²

Answer 70

- Sub in c̅²₂ = 1/3 c̅² into equation 5: - p = Nmc̅_x²/V - pV = Nmc̅_x² - pV = Nm c̅²/3 - **pV = 1/3Nmc̅²**

Answer 71

U = 3/2 nRT

Answer 72

Internal energy (J?)

Answer 73

Moles (mol)

Answer 74

Molar gas constant (m² kg s^-2 K^-1 mol^-1) (Psh)

Answer 75

Temperature (K)

Answer 76

4 damn long stages

Answer 77

- For an ideal gas, - internal energy = all KE energy - It's the sum of KE of all the particles - Internal energy U = N x mean KE of a particle

Answer 78

- Start with these equations: - pV = nRT - pV = 1/3 Nmc̅² - KE = 1/2 mv² - Equate them all - 1/3 Nmc̅² =nRT

Answer 79

- Multiply 2/3 to get the 1/2 needed for KE: - 1/2 Nmc̅² = 3/2nRT - 1/2 mc̅² = 3/2 n/N RT - Equation 1

Answer 80

- Equation 1 gives mean KE of a molecule - Simplify: - Sub n = N/N_A into equation 1: - 1/2 mc̅² = 3/2 N/N_A/N RT - 1/2 mc̅² = 3/2 R/N_A T - Sub k = R/N_A - 1/2 mc̅² = 3/2 kT

Answer 81

- Internal U = N x mean KE of particle: - U = 3/2 NkT - Expressed in terms of n° of moles - Sub k = R/N_A - Use n = N/N_A: - U = 3/2 N R/N_A T - **U = 3/2 nRT**

Answer 82

- A mole has n = 1 - U = 3/2 RT

Answer 83

- Divide energy for a mole by N_A - KE_{one molecule} = 3/2 R/N_A T - Simplify using R = kN_A: - KE_{one molecule} = (3/2 kN_A/N_A) T - **KE_{one molecule} = 3/2 kT**

Answer 84

A way easier method, just keep writing it