http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

IB Chemistry Kinetic Theory, Ideal Gas Equation and RMM determination of gases.

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 lineVol gas – negligible IMF - (negligible)

Low temp

High temp

Kinetic Theory simulation

2

21 mvKE

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 ArNe Xe

Why kinetic energy same for small and large particles?He – mass low ↓ - speed v high ↑ 2.21 vmKE

Xe – mass high ↑ - speed v low ↓ 2.21 vmKE

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

21 mvKE

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 constantUnit - 8.314 Jmol-1K-1

V = Vol gasUnit – 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 = PressureUnit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

VolPressure

TempVol Temp

Pressure Vol

n

PV = nRT P = constant

x T P ∝ T

P1 = P2

T1 T2

Ideal Gas Equation

PV = nRT (n, T fix)PV = constant

V = constant/P V ∝ 1/p

PV = nRT 4 diff variables → P, V, n, T

P1V1 = P2V2

R = gas constantUnit - 8.314 Jmol-1K-1

V = Vol gasUnit – m3

PV = nRT

Boyle’s Law

n, T fix

P = PressureUnit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

Pressure

Vol

Boyle’s Law Lab Simulator

Video on Boyle’s Law

Ideal Gas Equation

Charles’s Law

PV = nRT 4 diff variables → P, V, n, T

PV = nRT V = constant x

T V = constant T

V ∝ T

V1 = V2

T1 T2

R = gas constantUnit - 8.314 Jmol-1K-1

V = Vol gasUnit – m3

PV = nRT

n, P fix

P = PressureUnit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

Vol

Temp

Charles’s Law Lab Simulator

Video on Charles’s Law

Ideal Gas Equation

PV = nRT 4 diff variables → P, V, n, T

R = gas constantUnit - 8.314 Jmol-1K-1

V = Vol gasUnit – m3

PV = nRT

n, V fix

P = PressureUnit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

Pressure

Temp

Pressure Law

PV = nRT (n, V fix) P = constant x T

P ∝ T

P1 = P2

T1 T2

Pressure Law Lab Simulator

Video on Pressure Law

Ideal Gas Equation

PV = nRT 4 diff variables → P, V, n, T

R = gas constantUnit - 8.314 Jmol-1K-1

V = Vol gasUnit – m3

PV = nRT

P, V fix

P = PressureUnit – Nm-2/Pa/kPa

n = number of moles

T = Abs Temp in K

Vol

n

Video on Pressure Law

Avogadro’s Law

PV = nRT V = constant x n

V ∝ n

V1 = V2

n1 n2

Avogadro Law Lab Simulator

Video on Avogadro Law

Avogadro’s Law

Gas Helium Nitrogen Oxygen

Mole/mol 1 1 1Mass/g 4.0 28.0 32.0

Press /atm 1 1 1 Temp/K 273 273 273

Vol/L 22.7L 22.7L 22.7LParticles 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 kPa1m3 ↔ 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 = P2V2V1 = V2

T1 T2

V1 = V2

n1 n2

PV = nRT

Boyle’s Lawn, 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

TVP

TVP

Combined Boyle + Charles

nRTPV

PnTV

Find R at molar vol

n = 1 molT = 273K

P = 100 000 PaV = 22.7 x 10-3 m3

R = ?R = 8.31 JK-1 mol-1

nTPVR

T = 273K

V = 22.7 x 10-3 m3

P = 100 000 Pa

Volatile Liquid (Propanone)

Volatile Gas (Butane)

Syringe MethodDirect WeighingDirect Weighing

Heated – convert to gas

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

n = mass M

PRTM

PRT

VmM

RTMmPV

nRTPV

Density ρ = m (mass) V (vol)

PVmRTM

RTMmPV

nRTPV

Molar mass

RMM using Ideal Gas Eqn

PV = nRT

Direct Weighing

PV = nRT PV = mass x R x T MM = 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 gMass 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 waterAssume density water = 1 g/ml

Click here for lab procedure

Video on RMM determination

RMM (LIQUID) using Ideal Gas Eqn

Procedure

Data Collection

Direct Weighing1. 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 CollectionMass 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 MM = m x R x T PV = 0.65 x 8.314 x 294 99.17 x 2.76 x 10-4

= 58.17

Syringe Method1. Set temp furnace to 98C.

2.Put 0.2ml liq into a syringe3. 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 + liqbef injection

15.39 g

Mass syringe + liqafter injection

15.27 g

Mass of vapour 0.12 gPressure 100792Pa

Temp of vapour 371 KVol 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 MM = 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 gMass flask fill gas = 26.017

gMass flask fill water =

231.985 gTemp = 32C, P = 101 kPa

Find molar mass gas by direct weighing, T-23C , P- 97.7 kPaMass empty flask = 183.257 gMass flask + gas = 187.942 gMass flask + water = 987.560 gMass gas = (187.942 – 183.257) = 4.685 gVol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3

RMM determination

PV = nRTPV = mass x R x T MM = 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

0.003930.000393

= 10

0.007970.00039

3= 20

0.0003930.000393

= 1

Conservation of massMass C atom before = Mass C atom afterMass 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)

Ideal Gas Equation

Element S O CI

Composition 23.7 23.7 52.6

Moles 23.732.1

= 0.738

23.716.0

= 1.48

52.635.5

= 1.48

Mole ratio 0.738 0.738

1

1.48 0.738

2

1.48 0.738

2PRTM

PRT

VmM

RTMmPV

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 = 135.8135.8 = n [ 32 + (2 x 16)+(2 x 35.5) ]135.8 = n [ 135.8]n = 1MF = 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 massPV = nRTPV = mass x RT MM = 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

nRMM = 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

RMMmassMol

00.102.600.65

PnRTV

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 1.340341.0109080

15.29831.850.1

dmmV

V

PnRTV

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

PRTM

PRT

VmM

RTMmPV

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

RTPVn

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

7.115.29931.81036116150 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

PRTM

PRT

VmM

RTMmPV

nRTPV

Density ρ = m (mass) V (vol)

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

M = 30.6

PVmRTM

RTMmPV

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

101.3810130029331.8088.0

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 0.250 101.32KCIO3(s) → 2KCI(s) + 3O2 (g)

RTPVn

nRTPV

2

3

.010.029931.8

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

Gas occupy at (constant P)

V – 125 cm3 , T - 27 CFind its vol at 35 C

V1 = V2 (constant P) T1 T2

↓ 125 = V2

(27+273) (35 + 273)

↓V2 = 128 cm3

Find final vol, V2, at (constant T) compressed to P2 = 250

kPaV1 - 100 cm3 , P1 – 100 kPa V2 - ? P2 – 250 kPa

p1V1 = p2V2 (constant T)↓

100 x 100 = 250 x V2

↓V2 = 40 cm3

What vol (dm3) of 1 mol gas at

P - 101325 Pa, T - 25C pV = nRT V = nRT P

V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3

Find vol (m3) of 1 mol of gas at

T - 298K, P - 101 325PaPV = nRT V = nRT P

V = 1 x 8.314 x 298 101325 = 0.0244 m3

Find vol (dm3) of 2.00g CO at

T → 20C, P → 6250Nm-2

PV = nRT V = nRT P

= 0.0714 x 8.314 x 293 6250

= 0.0278 m3 = 27.8dm3

IB Questions on Ideal Gas

T → 293K

3.0 dm3 of SO2 react with 2.0 dm3 of O2

2SO2(g) + O2(g) → 2SO3(g)Find vol of SO2 , dm3 at stp

PV = nRT (at constant P,T)V ∝ n

2SO2(g) + 1 O2(g) → 2SO3(g)

2 mol 1 mol 2 mol

2 vol 1 vol 2 vol

3dm3 2dm3 ?

↓SO2 is limiting

2dm3 SO2 → 2dm3 SO3

3dm3 SO2 → 3dm3 SO3

Boyle, Charles, Avogadro Law

no need to convert to SI units

cancel off at both sides2 variables involved

n → 2.00/28= 0.0714 mol

A syringe contains gas atV1 – 50 cm3 , P1 – 1 atm, T1 -

293KWhat vol , V2, if gas heat toV2 ? T2 - 373 K, P2 - 5 atm

Find vol of gas when its press and temp are

double ?

Volume no change

Which change in conditions would increase

vol by x4 of a fix mass of gas?

Pressure /kPa Temperature /KA. Doubled DoubledB. Halved HalvedC. Doubled HalvedD. Halved Doubled

Initial P1 → Final P2 = 1/2P1

Initial T1 → Final T2 = 2T1Initial V1 → Final V2 = ?

Vol increase by x4

Fix mass ideal gas has a V1 = 800cm3 , P1, T1

Find vol, V2 when P and T doubled.P2 = 2P1

T2 = 2T1Initial P1 → Final P2 = 2P1

Initial T1 → Final T2 = 2T1

Initial V1 800 → Final V2 = ?

A. 200 cm3

B. 800 cm3

C. 1600 cm3

D. 3200 cm3

32

2

2

22

1

11

13373

5293

501

cmV

VTVP

TVP

12

1

21

1

11

2

22

1

11

22

VVTVP

TVP

TVP

TVP

12

1

21

1

11

2

22

1

11

422

VVTVP

TVP

TVP

TVP

3

2

1

21

1

1

2

22

1

11

800

22800

cmV

TVP

TP

TVP

TVP

Fix mass ideal gas has a V1 = 1dm3, P1, T1

Find V2 ,when T doubled (x2), P tripled (x3)

V2 = ?, P2 = 3P1, T2 = 2T1Initial P1 → Final P2 = 3P1

Initial T1 → Final T2 = 2T1

Initial V1 = 1 dm3 → Final V2 =?

Fix mass ideal gas has a V1 = 2 dm3 , P1, T1

Find V2 ,when T double (x2), P quadruple (x4)

V2 = ?, P2 = 4P1, T2 = 2T1Initial P1 → Final P2 = 4P1

Initial T1 → Final T2 = 2T1

Initial V1 2dm3 → Final V2 = ?

Fix mass ideal gas has a P1 = 40 kPa , V1, T1

Find P2 of gas when V and T doubled.P2 = ?, V2 = 2V1, T2 = 2T1

Initial V1 → Final V2 = 2V1

Initial T1 → Final T2 = 2T1

Initial P1 =40 → Final P2 = ?

3/22

31

2

1

21

1

1

2

22

1

11

VTVP

TP

TVP

TVP

32

1

21

1

1

2

22

1

11

1

242

dmV

TVP

TP

TVP

TVP

402

240

2

1

12

1

1

2

22

1

11

PTVP

TVTVP

TVP

What conditions would one mole

CH4, occupy the smallest vol?A. 273 K and 1.01×105 PaB. 273 K and 2.02×105 PaC. 546 K and 1.01×105 PaD. 546 K and 2.02×105 Pa

PV = nRT V = nRT P = low T, high P

Find total vol and composition of remaining gas

10cm3 ethyne react with 50cm3 hydrogen C2H2(g) + 2H2 (g) → C2H6 (g) at stp

PV = nRT (at constant P,T) V ∝ nC2H2 (g) + 2H2(g) → C2H6 (g)

1 mol 2 mol 1 mol1 vol 2 vol 1 vol10cm3 20cm3 10cm3

C2H6 = 10cm3 producedH2 = 50-20 =30 cm3 remain (excess)

Which conditions does a fix mass

of an ideal gas have greatest vol?

Temperature PressureA. low lowB. low highC. high highD. high low

PV = nRT V = nRT P = high T, low P

Find vol of 6 g of chlorine at 27oC and 101 kPa.

pV = nRTT  = 300K,n = 6/71 = 0.08451 mol CI2, Mr(Cl2)= 71p = 101000 Pa.

3002087.0101000

30031.808451.0

mV

V

PnRTV

5 litre container contain 0.5 kg butane gas (C4H10).

Cal press at 25oC.Mr(C4H10)= 58, 0.5kg = 500gmoles n = 500/58 = 8.621, T = 298K5 litre = 5 dm3 = 5 x 10-3 m3

kPaP

P

VnRTP

4272105

29831.8621.83

P = 202600 Pan = 0.050 mol T = 400KV = ? m3

What vol need to store 0.050 moles

of helium, P= 202.6 kPa , T= 400 K V = 7.5 x 10-3 m3

molar mass (H2) =2.016 g mol-1 n = 20.16 ÷ 2.016 = 10 mol T  = 293 K 

What press exerted by 20.16 g hydrogen

in 7.5 L cylinder at 20oC?

300082.0202600

40031.8050.0

mV

V

PnRTV

nRTPV

kPaP

P

VnRTP

nRTPV

3248105.7

29331.8103

50 L fill with argon to a press of 101 kPa at 30oC.

How many moles of argon ?P = 101000PaV = 50 x 10-3 m3

n = ? mol T = 303 K

moln

n

RTPVn

nRTPV

230331.8

1050101000 3

What temp does a 250 mL cylinder containing

0.40 g helium need to be cool to a press of 253.25 kPa?P = 253250 Pa

V = 0.250 x 10-3 m3

mass = 0.40 g , molar mass (He) = 4.003 n = 0.40 ÷ 4.003 = 0.1 mol 

KT

T

nRPVT

nRTPV

15.7631.81.0

10250.0253250 3

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/http://kwokthechemteacher.blogspot.hk/2011/07/ideal-gas-assumptions.html

Prepared by Lawrence Kok

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