IB Chemistry on Kinetic Theory Gas, Deviation from Ideal Gas and Ideal Gas Equation

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Prepared by Lawrence Kok

Tutorial on Kinetic Theory, Ideal/Real Gas and Deviation from Ideal gas behaviour.

Kinetic Theory of Gases

Kinetic Theory of Gases5 assumptions •Continuous random motion, in straight lines •Perfectly elastic collision•Average kinetic energy directly proportional to absolute temp ( E α T )•Gas consists of tiny negligible volume•Intermolecular forces attraction does not exist bet gases

Straight line Curve line

Volume gas – small (negligible)

Intermolecular forces - (negligible)


Kinetic Theory of Gases5 assumptions •Continuous random motion, in straight lines •Perfectly elastic collision•Average kinetic energy directly proportional to absolute temp ( E α T )•Gas consists of tiny negligible volume•Intermolecular forces attraction does not exist bet gases

Maxwell Boltzmann Distribution Curve•Molecular speed/energies at constant temperature •Molecule at low, most probable, root mean square speed•Higher temperature –greater spread of energies to the right (total area under curve the same)

Straight line Curve line

Volume gas – small (negligible)

Intermolecular forces - (negligible)

Low temp

High temp

Kinetic Theory simulation


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



Distribution of molecular speed, Xe, Ar, Ne, He at same temp

At same temperature•Xe, Ar, Ne and He have same average kinetic energy •Mass He lowest – speed fastest •Mass Xe highest – speed slowest

He ArNe Xe

Why kinetic energy same for small and large particles?



Distribution of molecular speed, Xe, Ar, Ne, He at same temp

At same temperature•Xe, Ar, Ne and He have same average kinetic energy •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 ↑ 2.2

1vmKE

Xe – mass high ↑ - speed v low ↓ 2.2

1vmKE

Kinetic energy SAME

Real Gas vs Ideal Gas

Ideal Gas•No volume•No intermolecular forces attraction

Real Gas•Volume•Intermolecular forces attraction

No forces attractionForces attraction




Deviation of Real gas from Ideal behaviour


1RT

PVconstant1

RT

PV1

RT

PVor


Deviation from ideal increase as



• Molecule close together • Forces of attraction exist

• Kinetic energy low• Molecule closer together – condense• Intermolecular forces stronger



1RT

PVconstant1

RT

PV1

RT

PVor

Pressure increase ↑

Pressure exerton wall less ↓

1RT

PV

Temp decrease ↓

Temp ↓

Positive deviation•Volume of molecules

Negative deviation•Presence of intermolecular attraction


At Low temperature•Molecules close together – condense•Presence of intermolecular attraction•Molecule will exert lower pressure at wall


1RT

PV



At High Pressure•Molecule close together •Forces of attraction exist

At Very High Pressure•Volume gas is significant

Volume gas significant athigh pressure

1RT

PV

At Low temperature•Molecules close together – condense•Presence of intermolecular attraction•Molecule will exert lower pressure at wall


1RT

PV



At High Pressure•Molecule close together •Forces of attraction exist

At Very High Pressure•Volume gas is significant

Vol available for molecule to move abt is less than observed volbecause molecule occupy space. As pressure increase, free space formolecule to move become smaller.

Volume gas significant athigh pressure

1RT

PV

Vol used for calculation

Actual vol is lessfor gas to move

1RT

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

Click here for notes from chemguide

At low temp- greater deviation

Ideal gas equation

At Low pressure + High Temp

Real gas/Van Der Waals equation

Ideal Gas vs Real Gas

Ideal gas Real gas

Ideal gas equation

At High pressure + Low Temp

Ideal gas equation




Ideal gas Real gas

Ideal gas equation



Correction for pressure due tointermolecular forces bet molecules

Correction for volume due tovol occupied by gas molecule

Boiling point CO2 = - 57oCCH4 = - 1640CN2 = - 195oCH2 = 252oC

Deviation greatest forCO2 > CH4 > N2 > H2

Ideal gas equation




Ideal gas Real gas

Ideal gas equation



Correction for pressure due tointermolecular forces bet molecules

Correction for volume due tovol occupied by gas molecule

Deviation of real gases

Temperature decrease ↓

Higher boiling point

Easier to condense

Intermolecular force increase

Negative deviation

Ideal Gas Equation

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

V = constant/P V ∝ 1/p

Charles’s Law

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

Avogadro’s Law

T = Absolute Temperature in K

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

V = constant T

V ∝ T

P1V1 = P2V2V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

R = universal gas constantUnit - 8.314Jmol-1K-1 or 0.0821 atm L mol-1 K-1

n = number of moles

V = Volume gasUnit – dm3 or m3

P = PressureUnit – Nm-2/Pa/kPa/atm

PV = nRTFix 2 variables → change to different gas Laws

Boyle’s Law Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2

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

V ∝ n

P, T fix



Charles’s Law Avogadro’s Law

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

P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2

PV = nRT Fix 2 variables → change to different gas Laws

Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Boyle’s Law





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Boyle’s LawPV = nRT (n, T fix)PV = constant V = constant P• V inversely proportional to P V ∝ 1 PP1V1 = P2V2

Boyle’s Law Lab Simulator

Video on Boyle’s Law

Boyle’s Law





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Boyle’s Law





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Charles’s Law Lab Simulator

Video on Charles’s Law

Boyle’s Law

Charles’s LawPV = nRT (n, P fix) V = constant x T

• V directly proportional to TV ∝ T

V1 = V2

T1 T2

Temp increase ↑ → kinetic energy increase ↑ → collision bet particles with container increase ↑ → volume increase ↑





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Boyle’s Law





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Pressure Law Lab Simulator

Video on Pressure Law

Boyle’s Law

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

P directly proportional to TP ∝ T

P1 = P2

T1 T2

Temp increase ↑ → kinetic energy increase ↑ → collision bet particles with container increase ↑ → pressure increase ↑





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Boyle’s Law





P1V1 = P2V2V1 = V2

T1 T2


P ∝ T

V1 = V2

n1 n2


Pressure Law

n, T fix

n, P fix n, V fix

P1 = P2

T1 T2


V ∝ n

P, T fix

Avogadro Law Lab Simulator

Video on Avogadro Law

Boyle’s Law

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

V directly proportional to n V ∝ n

V1 = V2

n1 n2

• 1 mole of any gas at fix STP (Std Temp/Pressure)• occupies a volume of 22.7 dm3/22700cm3

Avogadro’s Law

http://leifchemistry.blogspot.kr/2011/01/molar-volume-at-stp.html

Gas Helium Nitrogen Oxygen

Mole/mol 1 1 1

Mass/g 4.0 28.0 32.0

Pressure/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 temperature/pressure contain equal numbers of molecules”

T – 0C (273.15K)Unit conversion

1 atm = 760 mmHg/Torr = 101 325Pa(Nm-2) =101.325kPa1m3 = 103 dm3 = 106cm3

1dm3 = 1 litre

P - 1 atm = 760 mmHg = 101 325Pa (Nm-2) = 101.325kPa

Standard Molar Volume

Standard Temp/Pressure

“molar volume of all gases the same at given T and P” ↓ 22.7L

22.7L 22.7L

Video on Avogadro’s Law

1 mole gas






P ∝ T

PV = nRT Fix 2 variables → change different gas Laws

Pressure Law

n, T fix

n, P fix n, V fix


V ∝ n

P, T fix

Boyle’s Law

Combined Gas Law

Boyle’s Law Charles’s Law

V ∝ 1 P

V ∝ T

Combined Boyle + Charles Law

PV = constant TPV = R T

Gas constant, R

V ∝ T P

P1V 1 = P2V2

T1 T2

3 different variables

Charles’s Law Boyle’s Law Pressure Law Avogadro’s Law

Combined Boyle Law + Charles Law Combined Gas Law


2 different variables 3 different variables






P ∝ T

PV = nRT Fix 2 variables → change different gas Laws

Pressure Law

n, T fix

n, P fix n, V fix


V ∝ n

P, T fix

Boyle’s Law

Boyle’s Law Charles’s Law

V ∝ 1 P

V ∝ n Boyle + Charles + Avogadro Law

Proportionality constantGas constant, R

V ∝ n T P


Charles’s Law Boyle’s Law Pressure Law Avogadro’s Law

Boyle + Charles + Avogadro Law Ideal Gas Equation


PV = nRT

Ideal Gas Equation

Avogadro’s Law

V ∝ T

PV = n R T

Charles’s Law Pressure Law

PV = nRT

Avogadro’s Law

n, T fix

n, P fix P, T fixP, T fix

Boyle’s Law

V ∝ 1 P

V ∝ T P ∝ TV ∝ n

When n = 1 mol – Gas constant, R is 8.31 JK-1mol-1or NmK-1

For 1 mole – PV = RTFor n mole – PV = nRTP1V 1 = P2V2

T1 T2

PV = nRT

Ideal Gas EquationCombined Gas Law

++




PV = nRT

R = P V n T

n 1 mol

Temp/T oC → 273K

Pressure/P 101 325 Pa(Nm-2)

Volume/V22.7dm3 → 22.7 x 10-3 m3

R = 101325 x 22.7 x 10-3

1 x 273

R = 8.31 JK-1mol-1or NmK-1

Find R (Universal Gas Constant) at molar volumen = 1 molT = 273KP = 101325Pa/Nm-2

V = 22.7 x 10-3m3

R = ?

Value of gas constant, R (Universal Gas Constant) at molar volume

DifferentUnits Used

Volume/V22.7dm3 → 22.7 x 10-3 m3

PV = nRTPressure/P

101 325 Pa(Nm-2)

Temp/T oC → 273K

n 1 mol R = P V

n T

R = 101325 x 22.7 x 10-3

1 x 273R = 8.31 JK-1mol-1 or NmK-1

PV = nRT

R = P V n T

n 1 mol

Temp/T oC → 273K

Volume/V22.7L

Pressure/P 1 atm

R = 1 x 22.7 1 x 273

R = 0.0821 atmLmol-1K-1

1 atm ↔ 760 mmHg/Torr ↔ 101325Pa/Nm-2↔ 101.325kPa1m3 ↔ 103 dm3 ↔ 106cm3

1dm3 ↔ 1000cm3 ↔ 1000ml ↔ 1 litre x 103 x 103

cm3 dm3 m3

x 10-3 x 10-3

Unit conversion

DifferentUnits Used

A gas occupy at (constant P)• V - 125cm3

• T - 27CCalculate its volume at 35C

Find final vol, V2, gas at (constant T) compressed to P2 = 250kPaV1 - 100cm3

P1 - 100kPa V2 - ?P2 – 250kPa

What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C

Find volume (m3) of 1 mol of gas at• T - 298K• P - 101 325Pa

Find volume (dm3) of 2.00g CO at• T → 20C• P → 6250Nm-2

IB Questions on Ideal Gas

3.0 dm3 of SO2 reacted with 2.0 dm3 of O2

2SO2(g) + O2(g) → 2SO3(g)Find volume of SO2 in dm3 at stp

1 2 3

4 5 6

A gas occupy at (constant P)• V - 125cm3

• T - 27CCalculate its volume at 35CAnswer: (Charles

Law)V1 = V2 (constant P)T1 T2

125 = V2(27+273) (35 + 273)V2 = 128cm3

Find final vol, V2, gas at (constant T) compressed to P2 = 250kPaV1 - 100cm3

P1 - 100kPa V2 - ?P2 – 250kPa

Answer: (Boyle Law)p1V1 = p2V2 (constant T)100 x 100 = 250 x V2

V2 = 40cm3

What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C

Answer: (Ideal gas eqn)pV = nRT V = nRT P V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3

Find volume (m3) of 1 mol of gas at• T - 298K• P - 101 325Pa

Answer: (Ideal gas eqn)PV = nRT V = nRT P V = 1 x 8.314 x 298 101325 = 0.0244m3

Find volume (dm3) of 2.00g CO at• T → 20C• P → 6250Nm-2

Answer: (Ideal Gas Eqn)PV = nRT V = nRT P = 0.0714 x 8.314 x 293 6250 =0.0278m3 = 27.8dm3


T → (20 + 273) = 293Kn → 2.00/28 = 0.0714 mol

Using PV = nRT (Ideal gas eqn)• Need to convert to SI units• 4 variables involved

3.0 dm3 of SO2 reacted with 2.0 dm3 of O2

2SO2(g) + O2(g) → 2SO3(g)Find volume of SO2 in dm3 at stp

Answer: (Avogadro Law)PV = nRT (at constant P,T)V ∝ n2SO2(g) + 1 O2(g) → 2SO3(g)2 mol 1 mol 2 mol2 vol 1 vol 2 vol3dm3 2dm3 ?

SO2 is limiting2dm3 SO2 → 2dm3 SO3

3dm3 SO2 → 3dm3 SO3

Boyle, Charles, Avogadro Law• no need to convert to SI units• cancel off at both sides• 2 variables involved

1 2 3

4 5 6


7 8 3

9 10

A syringe contains gas atV1 - 50cm3

P1 – 1atmT1 - 20C → 293KWhat volume , V2, if gas heated toV2 - ?T2 - 100C → 373KP2 - 5 atm

Find volume fixed mass gas when its pressure and temp are double ?

Which change in conditions would increase the volume by x4 of a fix mass of gas?

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

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

T2 = 2T1


Combined gas Law• no need to convert to SI units• cancel off at both sides• 3 variables involved

7 8 3

9 10

A syringe contains gas atV1 - 50cm3

P1 – 1atmT1 - 20C → 293KWhat volume , V2, if gas heated toV2 - ?T2 - 100C → 373KP2 - 5 atmAnswer: (Combine Gas

Law)P1V1 = P2V2 T1 T2

1 x 50 = 5 x V2 293 373 V2 = 13cm3

Find volume fixed mass gas when its pressure and temp are double ?

Answer: (Combine Gas Law)Initial P1 → Final P2 = 2P1

Initial T1 → Final T2 = 2T1

Initial V1 → Final V2 = ?P1V1 = P2V2 T1 T2

P1V1 = 2P1V2 T1 2T1

V2 = V1Volume no change

P and T double

Which change in conditions would increase the volume by x4 of a fix mass of gas?

Pressure /kPa Temperature /K

A. Doubled Doubled

B. Halved Halved

C. Doubled Halved

D. Halved Doubled

Answer: (Combine Gas Law)Initial P1 → Final P2 = 1/2P1


Initial V1 → Final V2 = ?P1V1 = P2V2 T1 T2

P1V1 = P1V2 T1 2 x 2T1

V2 = 4V1

Volume increase by x4

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

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

T2 = 2T1Answer: (Combine Gas Law)Initial P1 → Final P2 = 2P1


Initial V1 800 → Final V2 = ?P1V1 = P2V2 T1 T2

P1 x 800 = 2P1V2 T1 2T1

V2 = 800

A. 200 cm3

B. 800 cm3

C. 1600 cm3

D. 3200 cm3

P halvedT double


Fix mass ideal gas has a V1 = 1dm3

P1

T1

Find V2 ,when T doubled (x2), P tripled (x3)V2 = ?P2 = 3P1

T2 = 2T1

11 Fix mass ideal gas has a V1 = 2dm3 P1

T1

Find V2 ,when T double (x2), P quadruple (x4)V2 = ?P2 = 4P1

T2 = 2T1

12

Fix mass ideal gas has a P1 = 40kPa V1

T1

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

T2 = 2T1

13

A. 1 dm3

B. 2 dm3

C. 3 dm3

D. 4 dm3


Fix mass ideal gas has a V1 = 1dm3

P1

T1

Find V2 ,when T doubled (x2), P tripled (x3)V2 = ?P2 = 3P1



Initial V1 = 1dm3 → Final V2 =?P1V1 = P2V2 T1 T2

P1 x 1 = 3P1 x V2 T1 2T1

V2 = 2/3

A. 1/3B. 2/3C. 3/2D. 1/6

11 Fix mass ideal gas has a V1 = 2dm3 P1

T1

Find V2 ,when T double (x2), P quadruple (x4)V2 = ?P2 = 4P1



Initial V1 2dm3 → Final V2 = ?P1V1 = P2V2 T1 T2

P1 x 2 = 4P1V2 T1 2T1

V2 = 1 dm3

12

Fix mass ideal gas has a P1 = 40kPa V1

T1

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

T2 = 2T1 Answer: (Combine Gas Law)Initial V1 → Final V2 = 2V1


Initial P1 =40 → Final P2 = ?P1V1 = P2V2 T1 T2

40 x V1 = P2 x 2V1 T1 2T1

P2 = 40

A. 10kPaB. 20kPaC. 40kPaD. 480kPa

13

Combined gas Law• no need to convert to SI units• cancel off at both sides• 3 variables involved


Calculate total volume and composition of remaining gas10cm3 ethyne react with 50cm3 hydrogen to produce ethaneC2H2(g) + 2H2(g) → C2H6(g) at stp

Which conditions does a fix mass of an ideal gas have greatest volume?

What conditions would one mole of, CH4, occupy the smallest volume?

14 15

16


Calculate total volume and composition of remaining gas10cm3 ethyne react with 50cm3 hydrogen to produce ethaneC2H2(g) + 2H2(g) → C2H6(g) at stpAnswer: (Avogadro Law)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 =30cm3 remains(excess)

Which conditions does a fix mass of an ideal gas have greatest volume? Temperature Pressure

A. low lowB. low highC. high highD. high low

Answer: (Ideal Gas Eqn)PV = nRT V = nRT P = high T, low P

What conditions would one mole of, CH4, occupy the smallest volume?

Answer: (Ideal Gas Eqn)PV = nRT V = nRT P = low T, high P

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

14 15

16


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

5 litre container contain 0.5kg butane gas (C4H10). Cal pressure at 25oC.

17 18

What vol needed to store 0.050 moles of helium at 202.6 kPa and 400 K?

19 What pressure exerted by 20.16 g hydrogen in a 7.5 L cylinder at 20oC?

20


AnswerpV = nRTT = 273 + 27 = 300K,n = 6/71 = 0.08451 mol chlorine, Mr(Cl2) = 2 x 35.5 = 71p = 101 x 1000 = 101000 Pa.V = nRT/pV = 0.08451 x 8.314 x 300/101000 = 0.002087 m3

AnswerMr(C4H10) = (4 x 12) + 10 = 58, 0.5kg = 500gmoles n = 500/58 = 8.621, T = 273 + 25 = 298KPV = nRT, P = nRT/V, 5 litre = 5 dm3 = 5 x 10-3 m3

P = 8.621 x 8.314 x 298/(5 x 10-3)P = 4271830 Pa = 4272 kPa

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

5 litre container contain 0.5kg butane gas (C4H10). Cal pressure at 25oC.

17 18

AnswerPV = nRTP = 202.6 kPa - 202600 Pan = 0.050 mol T = 400KV = ? L R = 8.314 J K-1 mol-1

202600 x V = 0.050 x 8.314 x 400 202600 x V = 166.28 V = 166.28 ÷ 202600V = 0.00082 m3

What vol needed to store 0.050 moles of helium at 202.6 kPa and 400 K?

19

AnswerPV = nRTV = 7.5 L → 7.5 x 10-3 m3

molar mass (H2) = 2 x 1.008 = 2.016 g mol-1 n = 20.16 ÷ 2.016 = 10 mol T = 20 + 273 = 293 K PV = nRTP x 7.5 x 10-3 = 10 x 8.314 x 293 P x 7.5 x 10-3 = 24360.02 P = 3248 kPa

What pressure exerted by 20.16 g hydrogen in a 7.5 L cylinder at 20oC?

20

50 L cylinder fill with argon to a pressure of 101 kPa at 30oC. How many moles of argon ?


21

To what temperature does a 250 mL cylinder containing 0.40 g helium need to be cooled to a pressure of 253.25 kPa?

22

AnswerP = 101 kPa → 101000PaV = 50 L → 50 x 10-3m3

n = ? mol T = 30 + 273 = 303 KPV = nRT101000 x 50 x 10-3 = n x 8.314 x 303 5050 = n x 2519.142 n = 5050 ÷ 2519.142 = 2.00 mol

50 L cylinder fill with argon to a pressure of 101 kPa at 30oC. How many moles of argon ?


21

AnswerP = 253.25 kPa → 253250PaV = 250 mL → 250 ÷ 1000 = 0.250 L → 0.250 x 10-3 m3

mass = 0.40 g molar mass (He) = 4.003 g mol-1 n = 0.40 ÷ 4.003 = 0.10 mol T = ? KPV = nRT25325o x 0.250 x 10-3 = 0.10 x 8.314 x T 63.3125 = 0.8314 x T T = 63.3125 ÷ 0.8314 = 76.15 K

To what temperature does a 250 mL cylinder containing 0.40 g helium need to be cooled to a pressure of 253.25 kPa?

22

Ideal Gas Law Lab simulation

Ideal Gas Law Videos

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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Education

IB Chemistry on Kinetic Theory Gas, Deviation from Ideal Gas and Ideal Gas Equation