Unit 3.3 Kinetic Theory (Derivation Gamble) Flashcards

Question

Define isochoric

Answer 1

When the volume is constant

Answer 2

- PV = constant - P₁V₁ = P₂V₂

Answer 3

- V/T = constant - V₁/T₁ = V₂/T₂

Answer 4

- P/T = constant - P₁/T₁ = P₂/T₂

Answer 5

Boyle's law P₁V₁ = P₂V₂

Answer 6

Charles' law V₁/T₁ = V₂/T₂

Answer 7

Pressure law P₁/T₁ = P₂/T₂

Answer 8

P₁V₁/T₁ = P₂V₂/T₂

Answer 9

Pa = N/m³

Answer 10

Pressure = force/volume

Answer 11

- The n° of atoms - in 12g of carbon-12

Answer 12

6.02 x 10²³ (N_a)

Answer 13

Mass of one mole

Answer 14

- Amount of a substance that - contains 6.02 x 10²³ particles

Answer 15

Mr (Relative molecular mass)

Answer 16

Number of moles = mass of gas/Molar mass

Answer 17

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

Answer 18

M(kg) = Mr/1000

Answer 19

Molar mass (kg) = Relative molecular mass/1000

Answer 20

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

Answer 21

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

Answer 22

Pressure (Pa)

Answer 23

Volume of gas (m³)

Answer 24

Number of molecules

Answer 25

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

Answer 26

- PV = nRT, n = N/N_A - PV = (N/N_A)RT - k = R/N_A - HENCE, PV = NkT

Answer 27

- Jerky motion - Random direction - Various speeds

Answer 28

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

Answer 29

- 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 30

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 31

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

Answer 32

- KE constant - Speed constant

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

- Increased KE - Mean speed increases

Answer 37

- **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 38

- Constant force - Rate of collisions constant

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

Average value of all speeds

Answer 44

C̅ = sum of all particles/n° of particles

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

Energy distribution for particles is random

Answer 50

Large n° of particles; large n° of collisions

Answer 51

Velocity of particles is uniform between collisions

Answer 52

- Even distribution of particle motion - in all directions

Answer 53

pV = ⅓NmC̅²

Answer 54

Pressure (Nm^-2)

Answer 55

Volume (m³)

Answer 56

Mass/one particle (kg)

Answer 57

N° of molecules

Answer 58

Mean square speed (m²s^-2??)

Answer 59

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

Answer 60

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

Answer 61

About 9 stages (my way)

Answer 62

- 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 63

- 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 64

- 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 65

- Calculate force by: - Force = change in momentum/time - Sub in equation 1 & 2 - F = 2mc_x/2L/c_x -> **mc²_x/L** - Equation 3

Answer 66

- 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 67

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

Answer 68

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

Answer 69

- 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 70

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

Answer 71

- 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 - **c̅²_x = 1/3 c̅²**

Answer 72

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

Answer 73

U = 3/2 nRT

Answer 74

Internal energy (J?)

Answer 75

Moles (mol)

Answer 76

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

Answer 77

Temperature (K)

Answer 78

4 damn long stages

Answer 79

- 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 80

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

Answer 81

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

Answer 82

- 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 **- n/N = 1/N_a & R/N_a = k - K.E for 1 particle = 3/2 x 1/N_a RT - Hence: = 3/2 kT**

Answer 83

- **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** - Sub k = R/N_A - Use n = N/N_A: - U = 3/2 N R/N_A T - **U = 3/2 nRT**

Answer 84

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

Answer 85

- 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 86

A way easier method, just keep writing it.... .... we had to optimise it and now we're gambling!

Answer 87

But then the second half is a total disaster wat the fa ... Easier...?