Kinetic Theory of Gases

Maxwell Boltzmann Distribution Curve - Molecular speed/energies at constant temp

- Molecule at low, most probable, root mean square speed - Higher temp –greater spread of energy to right

(total area under curve the same)

Straight line Curve line

Vol gas – negligible IMF - (negligible)

Low temp

High temp

Kinetic Theory simulation

2

2

1mvKE

Kinetic Theory of Gas 5 assumption - Continuous random motion, in straight lines - Perfect elastic collision - Ave kinetic energy directly proportional to abs temp ( E α T ) - Vol gas is negligible - Intermolecular forces attraction doesn’t exist

Kinetic Theory of Gases

Distribution of molecular speed, 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 Ar Ne Xe

Why kinetic energy same for small and large particles? He – mass low ↓ - speed v high ↑ 2.2

1vmKE

Xe – mass high ↑ - speed v low ↓ 2.2

1vmKE

Kinetic energy SAME

Maxwell Boltzman Distribution Curve • Molecular speed/energy at constant Temp • Molecule at low, most probable and high speed • Higher temp –greater spread of energy to right • Area under curve proportional to number of molecules • Wide range of molecules with diff KE at particular temp • Y axis – fraction molecules having a given KE • X axis – kinetic energy/speed for molecule

2

2

1mvKE

Real Gas vs Ideal Gas

Deviation from ideal increase as

Molecule close together Forces of attraction exist

Kinetic energy low Molecule closer together – condensed Intermolecular forces stronger

Deviation of Real gas from Ideal

No forces attraction Forces attraction

1RT

PVPV constant at all

pressure, if temp constant

1RT

PV1

RT

PVor

Pressure increase ↑

Pressure exert on wall less ↓

1RT

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

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

1RT

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

1RT

PV

Vol used in calculation is vol of container (too large)

Click here for notes from chemguide

At low temp- greater deviation

Negative deviation 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

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 deviation from ideal

Pressure Law

Ideal Gas Equation

PV = nRT PV = constant

V = constant/P

V ∝ 1/p

Charles’s Law

PV = nRT

4 diff variables → P, V, n, T

Avogadro’s Law

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-1K-1

V = Vol gas Unit – m3

PV = nRT Fix 2 variables

↓ change to diff gas Laws

Boyle’s Law

n, T fix n, P fix n, V fix

PV = nRT V = constant x n

V ∝ n

P, T fix

P = Pressure Unit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

Vol Pressure

Temp Vol Temp

Pressure Vol

n

PV = nRT P = constant x T

P ∝ T

P1 = P2

T1 T2

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

Standard Molar 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

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

Pressure Law

Ideal Gas Equation

PV = nRT PV = constant

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

PV = nRT V = constant x T V = constant

T

V ∝ T

P1V1 = P2V2 V1 = V2

T1 T2

V1 = V2

n1 n2

PV = nRT

Boyle’s Law

n, T fix n, P fix n, V fix

PV = nRT V = constant x 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

nTV

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

PVR

T = 273K

V = 22.7 x 10-3 m3

P = 100 000 Pa

Volatile Liquid (Propanone)

Volatile Gas (Butane)

Syringe Method Direct Weighing

Direct Weighing

Heated – convert to gas

RMM calculated - m, T, P, V, ρ are known

n = mass M

P

RTM

P

RT

V

mM

RTM

mPV

nRTPV

Density ρ = m (mass) V (vol) PV

mRTM

RTM

mPV

nRTPV

Molar mass

RMM using Ideal Gas Eqn

PV = nRT

Direct Weighing

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

Direct Weighing

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

Syringe Method

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 Collection Procedure

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

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 contain C, H and O. 0.06234 g of X combusted, 0.1755 g of CO2 and 0.07187 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

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

RTM

P

RT

V

mM

RTM

mPV

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

Sodium azide, undergoes decomposition rxn to produce N2 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

massMol

00.102.60

0.65

P

nRTV

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

nRTV

Density gas is 2.6 gdm-3 , T- 25C , P – 101 kPa Find RMM of gas

P

RTM

P

RT

V

mM

RTM

mPV

nRTPV

Density ρ = m (mass) V (vol)

M = (2.6 x 103) x 8.31 x (298) 101 x 103

M = 63.7

Sodium azide, undergoes decomposition rxn to produce N2 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

PVn

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

Density gas is 1.25g dm-3 at T- 25C ,P- 101 kPa. Find RMM of gas

P

RTM

P

RT

V

mM

RTM

mPV

nRTPV

Density ρ = m (mass) V (vol)

M = (1.25 x 103) x 8.31 x (298) 101 x 103

M = 30.6

PV

mRTM

RTM

mPV

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

29331.8088.06

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

PVn

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