The document discusses the kinetic theory of gases and its assumptions. It explains Maxwell-Boltzmann distribution curve which shows the distribution of molecular speeds at a given temperature. Higher temperatures result in a greater range of molecular energies. It also discusses how different gases at the same temperature and pressure will have different average molecular speeds depending on their mass. The kinetic energy of particles is the same on average due to the inverse relationship between mass and speed. Real gases deviate from ideal gas behavior more at high pressures and low temperatures due to intermolecular forces and molecular volumes. The document also covers gas laws and using the ideal gas equation to determine relative molecular mass of gases and liquids.

1. KineticTheory of Gases
Maxwell BoltzmannDistributionCurve
- Molecularspeed/energiesat constant temp
- Moleculeat low, most probable,root mean square speed
- Highertemp –greaterspread of energyto right
(total area undercurvethe same)
Straight line Curve line
Vol gas – negligible IMF - (negligible)
Low temp
High temp
KineticTheory simulation
2
2
1
mvKE 
KineticTheory of Gas
5 assumption
- Continuousrandom motion,in straight lines
- Perfectelastic collision
- Ave kineticenergydirectlyproportionalto abs temp ( E α T )
- Vol gas is negligible
- Intermolecularforcesattractiondoesn’texist

2. Kinetic Theory of Gases
Distributionof molecularspeed, Xe, Ar, Ne, He at same temp At same temp
• Xe, Ar, Ne and He have same Ave KE
• Mass He lowest – speed fastest
• Mass Xe highest – speed slowest
He ArNe Xe
Why kineticenergysame for small and large particles?He – mass low ↓ - speed v high ↑ 2
.
2
1
vmKE 
Xe – mass high ↑ - speed v low ↓ 2
.
2
1
vmKE 
Kinetic energy SAME
Maxwell Boltzman DistributionCurve
• Molecularspeed/energy at constant Temp
• Moleculeat low, most probableand high speed
• Highertemp –greater spread of energyto right
• Areaundercurve proportionalto numberof molecules
• Wide range of moleculeswith diff KE at particulartemp
• Y axis – fractionmoleculeshaving a given KE
• X axis – kineticenergy/speedfor molecule
2
2
1
mvKE 

3. Real Gas vs Ideal Gas
Deviation from ideal increase as
Moleculeclose together
Forcesof attractionexist
Kineticenergylow
Moleculecloser together– condensed
Intermolecularforcesstronger
Deviation of Real gas from Ideal
No forces attractionForces attraction
1
RT
PV PV constant at all
pressure, if temp constant
1
RT
PV
1
RT
PV
or
Pressure increase ↑
Pressure exert
on wall less ↓
1
RT
PV
Temp decrease ↓
Temp ↓
PV = nRT
Real Gas
Vol
Intermolecular forces attraction
Ideal Gas
Vol
No intermolecular forces attraction
P
PV
Ideal gas
real gas

4. At Very High Press
Vol gas is significant
At Low temp
Molecules close together – condense
Presence of intermolecular attraction
Molecule will exert lower press at wall
1
RT
PV
Deviation of Real gas from Ideal
At High Pressure
Molecule close together
Forces of attraction exist
Vol for molecule to move abt < observed vol because
molecule occupy space. As pressure increase, free
space formolecule to move become smaller.
Vol gas
significant at
high press
Vol used for
calculation
Actual vol is less
for gas to move
1
RT
PV
Vol used in calculation
is vol of container (too large)
Click here for notes from chemguide
At low temp- greater deviation
Negativedeviation
Presence of intermolecular attraction
Positive deviation
Vol of molecules
PV = nRT
PV
P
Ideal gas
real gas
-( ve) deviation
(IMF dominates)
+(ve) deviation
vol dominates

5. Boiling point
CO2 = - 57oC
CH4 = - 1640C
N2 = - 195oC
H2 = 252oC
Deviation greatest for
CO2 > CH4 > N2 > H2
Ideal gas eqn
At Low pres s+ High Temp
Real gas/Van Der Waals eqn
Ideal Gas vs Real Gas
Ideal gas eqn
At High press + Low Temp
Real gas/Van Der Waals eqn
Correction for press due to
IMF bet molecules Correction for vol due to
vol occupied by gas molecule
Deviation of real gases
Temperature decrease ↓
↓
Higher boiling point
↓
Easier to condense
↓
Intermolecular force
increase
↓
Negative deviation
PV = nRT PV = nRT
Ideal gas Real gas
CO2 polar– IMF greater
more deviationfrom ideal

6. PressureLaw
Ideal Gas Equation
PV = nRT
PV = constant
V = constant/P
V ∝ 1/p
Charles’sLaw
PV = nRT
4 diff variables→ P, V, n, T
Avogadro’sLaw
PV = nRT
V = constant x T
V = constant
T
V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
V1 = V2
n1 n2
R = gas constant
Unit - 8.314 Jmol-1
K-1
V = Vol gas
Unit – m3
PV = nRT
Fix 2 variables
↓
changeto diff gas Laws
Boyle’sLaw
n, T fix n, P fix n, V fix
PV = nRT
V = constantx n
V ∝ n
P, T fix
P = Pressure
Unit – Nm-2
/Pa/kPa
n = numberof moles
T = Abs Temp in K
VolPressure
TempVol Temp
Pressure Vol
n
PV = nRT
P = constant x T
P ∝ T
P1 = P2
T1 T2

7. Avogadro’s Law
Gas Helium Nitrogen Oxygen
Mole/mol 1 1 1
Mass/g 4.0 28.0 32.0
Press /atm 1 1 1
Temp/K 273 273 273
Vol/L 22.7L 22.7L 22.7L
Particles 6.02 x 1023 6.02 x 1023 6.02 x 1023
22.7L
“ equal vol of gases at same
temp/press contain
equal numbers of molecules”
T – 0C (273.15 K)
Unit conversion
1 m3
= 103
dm3
= 106
cm3
1 dm3
= 1 litre
StandardMolar Volume
“molar vol of all gases
same at given T and P”
↓
22.7L
22.7L
Video on Avogadro’s Law
1 mole
gas
• 1 mole of any gas at STP (Std Temp/Press)
• occupy a vol of 22.7 dm3
/22 700 cm3
P - 1 atm = 760 mmHg = 100 000 Pa (Nm-2
) = 100 kPa
22.7L

8. Unit conversion
1 atm ↔ 760 mmHg ↔ 100 000 Pa ↔ 100 kPa
1m3
↔ 103
dm3
↔ 106
cm3
1 dm3
↔ 1000 cm3 ↔ 1000 ml ↔ 1 litre
x 103 x 103
cm3 dm3 m3
x 10-3 x 10-3
PressureLaw
Ideal Gas Equation
PV = nRT
PV = constant
V = constant/P
V ∝ 1/p
Charles’sLaw Avogadro’sLaw
PV = nRT
V = constant x T
V = constant
T
V ∝ T
P1V1 = P2V2
V1 = V2
T1 T2
V1 = V2
n1 n2
PV = nRT
Boyle’sLaw
n, T fix n, P fix n, V fix
PV = nRT
V = constantx n
V ∝ n
P, T fix
PV = nRT
P = constant x T
P ∝ T
P1 = P2
T1 T2
Combined
Boyle + Charles + Avogadro
2
22
1
11
T
VP
T
VP

Combined
Boyle + Charles
nRTPV 
P
nT
V
Find R at molar vol
n = 1 mol
T = 273K
P = 100 000 Pa
V = 22.7 x 10-3
m3
R = ?
R = 8.31 JK-1
mol-1
nT
PV
R 
T = 273K
V = 22.7 x 10-3
m3
P = 100 000 Pa

9. Volatile Liquid
(Propanone)
Volatile Gas
(Butane)
Syringe MethodDirect Weighing
Direct Weighing
Heated – convert to gas
RMM calculated- m, T, P, V, ρ are known
n = mass
M
P
RT
M
P
RT
V
m
M
RT
M
m
PV
nRTPV




Density ρ = m (mass)
V (vol)PV
mRT
M
RT
M
m
PV
nRTPV



Molar
mass
RMM using Ideal Gas Eqn
PV = nRT

10. DirectWeighing
PV = nRT
PV = mass x R x T
M
M = m x R x T
PV
= 0.52 x 8.314 x 373
101325 x 2.84 x 10-4
= 56.33
1. Cover top with aluminium foil.
2. Make a hole on aluminium foil
3. Record mass flask + foil
4. Pour 2 ml volatile liq to flask
5. Place flask in water, heat to
boiling Temp and record press
6. Vapour fill flask when heat
7. Cool flask in ice bath –allow
vapour to condense to liquid
8. Take mass flask + foil + liquid
Mass flask + foil 115.15 g
Mass flask + foil +
condensed vapour
115.67 g
Mass condensed vapour 0.52 g
Pressure 101325 Pa
Temp of boiling water 100 0C
373K
Vol of flask 284 cm3
2.84 x 10-4 m3
Data Processing
Vol gas =Vol water in flask = Mass water
Assume density water = 1 g/ml
Click here for lab procedure
Video on RMM determination
RMM (LIQUID) using Ideal Gas Eqn
Procedure Data Collection

11. DirectWeighing
1. Fill flask with water and invert it .
2. Record press + temp of water
3. Mass of butane + lighter (ini)
4. Release gas into flask
6. Measure vol gas
7. Mass of butane + lighter (final)
Total Press (atm) = partial P(butane) + partial P(H2O)
P butane = P(atm) – P(H2O)
= (760 – 19.32) mmHg
P butane = 743.911 mmHg → 99.17Pa
Dalton’s Law of Partial Press:
Total press of mix of gas = sum of partial press of all individual gas
5. Adjust water level in flask
until the same as atm pressure
RMM butane RMM butane Collection gas
RMM (GAS) using Ideal Gas Eqn
Procedure Data Collection
Mass butane + lighter 87.63 g
Mass butane + lighter
(final)
86.98 g
Mass butane 0.65 g
Pressure 99.17 Pa
Temp of boiling water 21.7 0C
294 K
Vol of flask 276 cm3
2.76 x 10-4 m3
Data Processing
PV = nRT
PV = mass x R x T
M
M = m x R x T
PV
= 0.65 x 8.314 x 294
99.17 x 2.76 x 10-4
= 58.17

12. SyringeMethod
1. Set temp furnace to 98C.
2.Put 0.2ml liq into a syringe
3. Record mass syringe + liq
5. Inject liq into syringe
6. Liq will vaporise ,
Record vol of heat vapour + air
4. Record vol of heated air.
Mass syringe + liq
bef injection
15.39 g
Mass syringe + liq
after injection
15.27 g
Mass of vapour 0.12 g
Pressure 100792Pa
Temp of vapour 371 K
Vol heated air 7 cm3
Vol heated air + vapour 79 cm3
Vol of vapour 72 – 7 = 72 cm3
7.2 x 10-5 m3
Video on RMM determination
RMM (LIQUID)using Ideal Gas Eqn
Data CollectionProcedure
Data Processing
PV = nRT
PV = mass x R x T
M
M = m x R x T
PV
= 0.12 x 8.314 x 371
100792 x 7.2 x 10-5
= 51.1

13. P = 101 kNm-2
= 101 x 103 Nm-2
Calculate RMM of gas
Mass empty flask = 25.385 g
Mass flask fill gas = 26.017 g
Mass flask fill water = 231.985 g
Temp = 32C, P = 101 kPa
Find molar mass gas by direct weighing, T-23C , P- 97.7 kPa
Mass empty flask = 183.257 g
Mass flask + gas = 187.942 g
Mass flask + water = 987.560 g
Mass gas = (187.942 – 183.257) = 4.685 g
Vol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3
RMM determination
PV = nRT
PV = mass x R x T
M
M = mass x R x T
PV
= 4.685 x 8.314 x 296
97700 x 804.303 x 10-6
= 146.7
Vol gas = 804.303 cm3
= 804.303 x 10-6 m3
P = 97.7 kPa
= 97700 Pa
Density water = 1g/cm3
M = m x RT
PV
= 0.632 x 8.314 x 305
101 x 103 x 206 x 10-6
= 76.8
m gas = (26.017 – 25.385)
= 0.632 g
vol gas = (231.985 – 25.385)
= 206 x 10-6 m3
X containC, H and O. 0.06234 g of X combusted,
0.1755 g of CO2 and0.07187 g of H2O produced.
Find EF of X
Element C H O
Step 1 Mass/g 0.0479 0.00805 0.006384
RAM/RMM 12 1 16
Step 2 Number
moles/mol
0.0479/1 2
= 0.00393
0.00805/1
= 0.00797
0.006384/16
= 0.000393
Step 3 Simplest ratio 0.00393
0.000393
= 10
0.00797
0.000393
= 20
0.000393
0.000393
= 1
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
CHO + O2 CO2 + H2O
Mol C atom in CO2
= 0.1755 = 0.00393 mol
44
Mass C = mol x RAM C
= 0.00393 x 12
= 0.0479 g
Mol H atom in H2O
= 0.07187 = 0.0039 x 2 = 0.00797 mol
18
Mass H = mol x RAM H
= 0.00797 x 1.01
= 0.00805 g
Mass of O = (Mass CHO – Mass C – Mass H)
= 0.06234 – 0.0479 - 0.00805 = 0.006384 g
0.06234 g 0.1755 g 0.07187 g
Empirical formula – C10H20O1

14. Find EF for X with composition by mass. S 23.7 %, O 23.7 %, CI 52.6 %
Given, T- 70 C, P- 98 kNm-2 density - 4.67g/dm3
What molecular formula?
Empirical formula - SO2CI2
Density ρ = m (mass)
V (vol)
Element S O CI
Composition 23.7 23.7 52.6
Moles 23.7
32.1
= 0.738
23.7
16.0
= 1.48
52.6
35.5
= 1.48
Mole ratio 0.738
0.738
1
1.48
0.738
2
1.48
0.738
2
P
RT
M
P
RT
V
m
M
RT
M
m
PV
nRTPV




Density = 4.67 gdm-3
= 4.67 x 10-3 gm-3
M = (4.67 x 10-3) x 8.31 x (273 +70)
9.8 x 104
M = 135.8
135.8 = n [ 32 + (2 x 16)+(2 x 35.5) ]
135.8 = n [ 135.8]
n = 1
MF = SO2CI2
P = 98 kN-2
= 9.8 x 104 Nm-2
3.376 g gas occupies 2.368 dm3 at T- 17.6C, P - 96.73 kPa.
Find molar mass
PV = nRT
PV = mass x RT
M
M = mass x R x T
PV
= 3.376 x 8.314 x 290.6
96730 x 2.368 x 10-3
= 35.61
Vol = 2.368 dm3
= 2.368 x 10-3 m3
P – 96.73 kPa → 96730Pa
T – 290.6K
6.32 g gas occupy 2200 cm3, T- 100C , P -101 kPa.
Calculate RMM of gas
PV = nRT
n = PV
RT
n = (101 x 103) (2200 x 10-6)
8.31 x ( 373 )
n = 7.17 x 10-2 mol
Vol = 2200 cm3
= 2200 x 10-6 m3
RMM = mass
n
RMM = 6.32
7.17 x 10-2
= 88.15

15. Sodiumazide, undergoesdecompositionrxn to produceN2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below
i. State number of sig figures for Temp, Mass, and Pressure
i. Temp – 4 sig fig Mass – 3 sig fig Pressure – 3 sig fig
Temp/C Mass NaN3/kg Pressure/atm
25.00 0.0650 1.08
ii. Find amt, mol of NaN3 present
ii.
iii. Find vol of N2, dm3 produced in these condition
RMM NaN3 – 65.02
molMol
RMM
mass
Mol
00.1
02.60
0.65


P
nRT
V
nRTPV

 n = 1.50 mol
P – 1.08 x 101000 Pa
= 109080 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 25.00 + 273.15
= 298.15K
2 mol – 3 mol N2
1 mol – 1.5 mol N2
33
1.340341.0
109080
15.29831.850.1
dmmV
V
P
nRT
V




Densitygas is 2.6 gdm-3 , T- 25C , P – 101 kPa
Find RMM of gas
P
RT
M
P
RT
V
m
M
RT
M
m
PV
nRTPV




Density ρ = m (mass)
V (vol)
M = (2.6 x 103) x 8.31 x (298)
101 x 103
M = 63.7

16. Sodiumazide, undergoesdecompositionrxn to produceN2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below
Temp/C Volume N2/L Pressure/atm
26.0 36 1.15
Find mass of NaN3 needed to produce 36L of N2
RMM NaN3 – 65.02
RT
PV
n
nRTPV


1.1 x 65.02 = 72 g NaN3
P – 1.15 x 101000 Pa
= 116150 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 26.0 + 273.15
= 299.15K
3 mol N2 – 2 mol NaN3
1.7 mol N2 – 1.1 mol NaN3
moln
n
7.1
15.29931.8
1036116150 3





Vol = 36 dm3
= 36 x 10-3 m3
Convert mole NaN3 → Mass /g
Densitygas is 1.25g dm-3 at T- 25C ,P- 101 kPa. Find
RMM of gas
P
RT
M
P
RT
V
m
M
RT
M
m
PV
nRTPV




Density ρ = m (mass)
V (vol)
M = (1.25 x 103) x 8.31 x (298)
101 x 103
M = 30.6

17. PV
mRT
M
RT
M
m
PV
nRTPV



Copper carbonate, CuCO3, undergo decomposition to produce a gas.
Determine molar mass for gas X
CuCO3(s) → CuO(s) + X (g)
Temp, mass, vol and pressure was collected in table below
Temp/K Vol gas/ cm3 Pressure/kPa Mass gas/g
293 38.1 101.3 0.088
Find Molar mass for gas X
P – 101300 Pa
T – 293 K
Vol = 38.1 cm3
= 38.1 x 10-6 m3
5.55
101.38101300
29331.8088.0
6



 
M
M
Potassium chlorate, KCIO3, undergo decomposition to produce a O2.
Find amt O2 collected and mass of KCIO3 decomposed
KCIO3
Temp/K Vol gas/ dm3 Pressure/kPa
299 0.250 101.3
2KCIO3(s) → 2KCI(s) + 3O2 (g)
RT
PV
n
nRTPV


2
3
.010.0
29931.8
10250.0101300
Omoln
n





Vol = 0.250 dm3
= 0.250 x 10-3 m3
P – 101300 Pa
Convert mole KCIO3 → Mass
2KCIO3 → 2KCI+ 3O2
2 mol – 3 mol O2
0.0066 mol – 0.01 mol O2
0.0066 x 122.6 = 0.81 g KCIO3
RMM KCIO3 – 122.6