Unit 3.3 Kinetic Theory (Derivation Gamble) Flashcards

Question 1

Q

Summarise the main ideas about particles in a solid
(6 things)

Answer

A

Regular pattern
Fixed shape & volume
Particles vibrate in place, low kinetic energy
No space between particles
Intermolecular forces = strong,
they have low potential energy

Question 2

Q

Summarise the main ideas about particles in a liquid
(7 things)

Answer

A

No regular pattern
No fixed shape & volume
Particles move around each other, med. kinetic energy
There’s space between particles,
but are touching
Intermolecular forces = medium.
∴ medium potential energy

Question 3

Q

Summarise the main ideas about particles in a gas
(5 things)

Answer

A

No regular pattern/fixed shape/volume
Particles move fast & free, has high kinetic energy
Space between particles
Intermolecular forces are weak [zero],
they have high potential energy

Question 4

Q

Define the internal energy (U) of a system?
(3 parts)

Answer

A

The sum of the random distributions
of kinetic and potential energies
of all the molecules in the system

Question 5

Q

What is an ideal gas?
(2 parts)

Answer

A

Gas that strictly obeys the equation of state:
Pv = nRT

Question 6

Q

What is the ideal gas equation?
(In data booklet)

Question 7

Q

Ideal gas equation in word form?

Answer

A

Pressure x Volume = moles x molar gas constant x temperature

Question 8

Q

Define P
(ideal gas equation)

Answer

A

Pressure
(Pa)

Question 9

Q

Define v
(ideal gas equation)

Answer

A

Volume
(m³)

Question 10

Q

Define n
(ideal gas equation)

Answer

A

Moles
(mol)

Question 11

Q

Define r
(ideal gas equation) Data booklet

Answer

A

Molar gas constant = 8.31
(J Mol^-1 K^-1)

Question 12

Q

Define T
(ideal gas equation)

Answer

A

Kelvins
(K)

Question 13

Q

How does an ideal gas “strictly” obeys the equation of state PV=nRT?
(3 things)

Answer

A

Must have:
- Negligible vol. of molecules
- Elastic collisions
- No intermolecular forces

Question 14

Q

Features of an ideal gas?
(2 things)

Answer

A

No molecular force
Simply be kinetic energy

Question 15

Q

Charles law & Boyles law proofs?

Answer

A

Take action if needed
(If u see a question on it, then…)

Question 16

Q

What is meant by temperature?
(2-way)

Answer

A

A measure of avg. kinetic energy
of the particles in a substance

Question 17

Q

Define absolute zero?
(2-way)

Answer

A

The temperature of a system when
it has MINIMUM internal energy

Question 18

Q

How do u find the value for absolute zero in a graph?
(Refer to page 9 otherwise…)
(3-way)

Answer

A

rise/run = gradient
V = mx + c, 0 = mx + c
-c/m = absolute zero

Question 19

Q

How to convert Celsius to Kelvins?
(In data booklet kinda)

Answer

A

Add 273.15

Question 20

Q

How to convert Kelvins to Celsius?
(In data booklet kinda)

Answer

A

Minus 273.15

Question 21

Q

What are the 3 gas laws?

Answer

A

Boyle’s law
Charles’ law
Pressure law

Question 22

Q

How can the gas laws only be applied?

Answer

A

For a fixed mass of gas
(n = constant)

Question 23

Q

Define isothermal

Answer

A

When the temperature is constant

Question 24

Q

Define isobaric

Answer

A

When the pressure is constant

Question 25

Q

Define isochoric

Answer

A

When the volume is constant

Question 26

Q

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

Answer

A

PV = constant
P₁V₁ = P₂V₂

Question 27

Q

What is Charles’ law?
(Gas laws, 2 each)

Answer

A

V/T = constant
V₁/T₁ = V₂/T₂

Question 28

Q

What is the Pressure law?
(Gas laws, 2 each)

Answer

A

P/T = constant
P₁/T₁ = P₂/T₂

Question 29

Q

Which gas law is isothermal?

Answer

A

Boyle’s law
P₁V₁ = P₂V₂

Question 30

Q

Which gas law is isobaric?

Answer

A

Charles’ law
V₁/T₁ = V₂/T₂

Question 31

Q

Which gas law is isochoric?

Answer

A

Pressure law
P₁/T₁ = P₂/T₂

Question 32

Q

What equation do u get for combining all 3 gas laws?

Answer

A

P₁V₁/T₁ = P₂V₂/T₂

Question 33

Q

Equation to gain pressure?

Answer

A

Pa = N/m³

Question 34

Q

Pressure equation in word form?

Answer

A

Pressure = force/volume

Question 35

Q

1Pa = ?

Question 36

Q

Define 1 mole?
(2-way)

Answer

A

The n° of atoms
in 12g of carbon-12

Question 37

Q

What is Avagadro’s constant?
(in data booklet)

Answer

A

6.02 x 10²³
(N_a)

Question 38

Q

What is molar mass?

Answer

A

Mass of one mole

Question 39

Q

Define molar mass?
(2-way)

Answer

A

Amount of a substance that
contains 6.02 x 10²³ particles

Question 40

Q

But what actually is molar mass?
(Symbol typa thingy)

Answer

A

Mr
(Relative molecular mass)

Question 41

Q

How to calculate n° of moles in a given mass of any gas?
(In data booklet)

Question 42

Q

Word form of n° of mols equation?

Answer

A

Number of moles = mass of gas/Molar mass

Question 43

Q

Define mass of gas?
(Mols equation)

Answer

A

Grams
(g)

Question 44

Q

Define relative molecular mass (Mr)
(2-way)

Answer

A

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

Question 45

Q

The equation to prove Mr is related to M?
(In data booklet)

Answer

A

M(kg) = Mr/1000

Question 46

Q

Word form of M equation

Answer

A

Molar mass (kg) = Relative molecular mass/1000

Question 47

Q

What are the 3 things that define a “real gas”?

Answer

A

Particles have volume
Energy lost in collisions
Intermolecular forces

Question 48

Q

What is the equation linking with the n° of molecules it contains?
(In data booklet)

Question 49

Q

Word form of equation linking with n° of molecules it contains?

Answer

A

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

Question 50

Q

Define P
(2nd ideal gas equation)

Answer

A

Pressure
(Pa)

Question 51

Q

Define V
(2nd ideal gas equation)

Answer

A

Volume of gas
(m³)

Question 52

Q

Define N
(2nd ideal gas equation)

Answer

A

Number of molecules

Question 53

Q

Define k
(2nd ideal gas equation)

Answer

A

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

Question 54

Q

However, how is Boltzmann’s constant gained?
(PV = NkT)
(4-way… whiteboard)

Answer

A

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

Question 55

Q

Describe how smoke particles move?
(3 points)

Answer

A

Jerky motion
Random direction
Various speeds

Question 56

Q

Explain why smoke particles move like that?
(5 points)

Answer

A

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

Question 57

Q

Why do gas particles exert a pressure on their container?
(6 points)

Answer

A

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

Question 58

Q

How do u explain the gas laws
(3 steps)

Answer

A

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

Question 59

Q

Explain why P ∝ 1/V at constant temperature
(Boyle’s Law)
(2 points + 3-way + 1)

Answer

A

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

Question 60

Q

Explain in detail constant temperature?
(Boyle’s Law)
(2 points)

Answer

A

KE constant
Speed constant

Question 61

Q

Explain in detail decreased volume?
(Boyle’s Law)
(2 points)

Answer

A

Less distance between collision with wall
More collisions per second

Question 62

Q

Explain why P ∝ T at constant volume
(Pressure Law)
(2 points, 3-way + 1)

Answer

A

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

Question 63

Q

Explain in detail constant volume?
(Pressure law)
(2 points)

Answer

A

Constant distance between collision w/ wall
Constant collisions per second

Question 64

Q

Explain in detail increased temperature?
(Pressure law)
(2 points)

Answer

A

Increased KE
Mean speed increases

Answer 61

A

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 62

A

Constant force
Rate of collisions constant

Answer 63

A

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

Answer 64

A

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

Answer 65

A

Most probable speed
Mean speed
Root mean square speed

Answer 66

A

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

Answer 67

A

Average value of all speeds

Answer 68

A

C̅ = sum of all particles/n° of particles

Answer 69

A

The square root of
the mean square speed of the molecules

Answer 70

A

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

Answer 71

A

No intermolecular forces
Negligible volume of molecules
Elastic collisions between molecules

Answer 72

A

Duration of collisions very short
compared to time between collisions

Answer 73

A

Energy distribution for particles is random

Answer 74

A

Large n° of particles; large n° of collisions

Answer 75

A

Velocity of particles is uniform between collisions

Answer 76

A

Even distribution of particle motion
in all directions

Answer 77

A

pV = ⅓NmC̅²

Answer 78

A

Pressure
(Nm^-2)

Answer 79

A

Volume
(m³)

Answer 80

A

Mass/one particle
(kg)

Answer 81

A

N° of molecules

Answer 82

A

Mean square speed
(m²s^-2??)

Answer 83

A

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

Answer 84

A

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

Answer 85

A

About 9 stages
(my way)

Answer 86

A

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 87

A

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 88

A

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 89

A

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 90

A

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 91

A

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

Answer 92

A

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

Answer 93

A

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 94

A

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

Answer 95

A

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 96

A

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

Answer 97

A

U = 3/2 nRT

Answer 98

A

Internal energy
(J?)

Answer 99

A

Moles
(mol)

Answer 100

A

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

Answer 101

A

Temperature
(K)

Answer 102

A

4 damn long stages

Answer 103

A

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 104

A

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

Answer 105

A

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 106

A

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 107

A

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 108

A

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

Answer 109

A

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 110

A

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

Answer 111

A

But then the second half is a total disaster wat the fa

… Easier…?