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

P1V1 = P2V2 V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

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

n = number of moles

V = Volume gas Unit – dm3 or m3

P = Pressure Unit – Nm-2/Pa/kPa/atm

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

Boyle’s Law Pressure Law

P1 = P2

T1 T2

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

V ∝ n

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

T V ∝ T

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

P1V1 = P2V2 V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

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

Pressure Law

P1 = P2

T1 T2

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

V ∝ n

Boyle’s Law PV = nRT (n, T fix) PV = constant V = constant P • V inversely proportional to P

V ∝ 1 P P1V1 = P2V2

Boyle’s Law Lab Simulator

Video on Boyle’s Law

Boyle’s Law

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

P1V1 = P2V2 V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

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

Pressure Law

P1 = P2

T1 T2

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

V ∝ n

Charles’s Law Lab Simulator

Video on Charles’s Law

Boyle’s Law

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

• V directly proportional to T

V ∝ T V1 = V2

T1 T2

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

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

P1V1 = P2V2 V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

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

Pressure Law

P1 = P2

T1 T2

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

V ∝ n

Pressure Law Lab Simulator

Video on Pressure Law

Boyle’s Law

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

P directly proportional to T P ∝ T

P1 = P2

T1 T2

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

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

P1V1 = P2V2 V1 = V2

T1 T2

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

P ∝ T

V1 = V2

n1 n2

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

Pressure Law

P1 = P2

T1 T2

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

V ∝ n

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 22.4dm3/22400cm3/24L

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 22.4L 22.4L 22.4L

Particles 6.02 x 1023 6.02 x 1023

6.02 x 1023

22.4L

“ 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) =101.325kPa 1m3 = 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” ↓

22.4L

22.4L 22.4L

Video on Avogadro’s Law

1 mole

gas

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

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

P ∝ T

PV = nRT Fix 2 variables → change different gas Laws

Pressure Law

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

V ∝ n

Boyle’s Law

Combined Gas Law

Boyle’s Law Charles’s Law

V ∝ 1 P

V ∝ T

Combined Boyle + Charles Law

PV = constant T PV = 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

2 different variables 3 different variables

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

V = constant/P

V ∝ 1/p

Charles’s Law Avogadro’s Law

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

T V ∝ T

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

P ∝ T

PV = nRT Fix 2 variables → change different gas Laws

Pressure Law

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

V ∝ n

Boyle’s Law

Boyle’s Law Charles’s Law

V ∝ 1 P

V ∝ n Boyle + Charles + Avogadro Law

Proportionality constant Gas constant, R

V ∝ n T P

4 different variables

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

Boyle + Charles + Avogadro Law Ideal Gas Equation

2 different variables

PV = nRT

Ideal Gas Equation

Avogadro’s Law

V ∝ T

PV = n R T

Charles’s Law Pressure Law

PV = nRT

Avogadro’s Law Boyle’s Law

V ∝ 1 P

V ∝ T P ∝ T V ∝ n

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

For 1 mole – PV = RT For n mole – PV = nRT P1V 1 = P2V2

T1 T2

PV = nRT

Ideal Gas Equation Combined Gas Law

+

+

2 different variables

3 different variables

4 different variables

PV = nRT

R = P V n T

n 1 mol

Temp/T oC → 273K

Pressure/P 101 325 Pa(Nm-2)

Volume/V 22.4dm3 → 22.4 x 10-3 m3

R = 101325 x 22.4 x 10-3

1 x 273

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

Find R (Universal Gas Constant)

at molar volume

n = 1 mol

T = 273K

P = 101325Pa/Nm-2

V = 22.4 x 10-3m3

R = ?

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

Different Units Used

Volume/V 22.4dm3 → 22.4 x 10-3 m3

PV = nRT Pressure/P

101 325 Pa(Nm-2)

Temp/T oC → 273K

n 1 mol

R = P V n T

R = 101325 x 22.4 x 10-3

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

PV = nRT

R = P V n T

n 1 mol

Temp/T oC → 273K

Volume/V 22.4L

Pressure/P 1 atm

R = 1 x 22.4

1 x 273 R = 0.0821 atmLmol-1K-1

1 atm ↔ 760 mmHg/Torr ↔ 101 325Pa/Nm-2 ↔ 101.325kPa 1m3 ↔ 103 dm3 ↔ 106cm3

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

cm3 ↔ dm3 ↔ m3

x 10-3 x 10-3

Unit conversion

Different Units Used

Determination of RMM using Ideal Gas Equation

Volatile Liquid (Propanone)

Volatile Gas (Butane)

Syringe Method Direct Weighing Direct Weighing

PV = nRT

Converted to gas

Heated

PV = n x R x T PV = mass x R x T M M = m x R x T P V

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

Density, ρ = m V

PV = n x R x T PV = mass x R x T M M = m x R x T V P M = ρ x R x T P

RMM = M

or n = mass M

Ideal Gas Equation

n = mass M

Determination of RMM (LIQUID) using Ideal Gas Equation

Direct Weighing

PV = nRT

PV = n x R x T 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

Data Collection

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

1. Cover the top with aluminium foil.

Procedure

2. Make a hole on aluminium foil

3. Record mass flask + foil

4. Pour 2 ml volatile liquid into flask

5. Place flask in water, heated to boiling Temp and record pressure

6. Vapour fill flask when heated

7. Cool flask in ice bath – allow vapour to condense to liquid

8. Take mass of flask + foil + condensed liquid

Mass flask 115.15 g

Mass flask + foil +

condensed vapour

115.67 g

Mass condensed vapour 0.52 g

Atmospheric pressure 101325 Pa

Temperature of boiling

water

100 0C →373K

Volume of flask 284 cm3 → 2.84 x 10-4 m3

Data Processing

Vol gas = Vol water = Vol water in flask = Mass water Assume density water = 1g/ml

Click here for lab procedure

Video on RMM determination

Syringe Method

PV = nRT

PV = n x R x T 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

Data Collection

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

1. Set temp furnace to 98C.

Procedure

2. Draw 0.2ml liquid into a syringe Record mass syringe + liquid.

5. Inject liquid into syringe

6. Liquid will vaporise , Record vol of heated vapour + air

4. Record vol of heated air.

Data Processing

Mass syringe + liquid

before injection

15.39 g

Mass syringe + liquid

after injection

15.27 g

Mass of vapour 0.12 g

Atmospheric Pressure 100792Pa

Temp of vapour 371 K

Volume heated air 7 cm3

Volume heated air +

vapour

79 cm3

Volume of vapour 72 – 7 = 72 cm3

72cm3 →7.2 x 10-5 m3

Click here for lab procedure

Video on RMM determination

Determination of RMM (LIQUID) using Ideal Gas Equation

Direct Weighing

PV = nRT

PV = n x R x T PV = mass x R x T M M = m x R x T PV = 0.65 x 8.314 x 294.75 99.17 x 2.76 x 10-4

= 58.17

Data Collection

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

1. Fill a flask with water and invert it .

Procedure

2. Record pressure + temp of water

3. Mass of butane + lighter (initial)

4. Release gas into flask

6. Measure the vol gas

7. Mass of butane + lighter (final)

Mass butane + lighter

(initial)

87.63 g

Mass butane + lighter

(final)

86.98 g

Mass butane 0.65 g

Pressure gas 743.9 mmHg

760 mmHg → 101325 Pa

743.9mmHg → 99.17Pa

Temperature water 21.70C → 294.75K

Volume of gas 276cm3 → 2.76 x 10-4 m3

Data Processing

Click here for lab procedure

Total Pressure (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 Pressures: Total pressure of mix of gases = sum of the partial pressures of all the individual gases

5. Adjust water level in flask until the same as atm pressure

Determination of RMM (GAS) using Ideal Gas Equation

Video on RMM determination

RMM butane RMM butane Collection gas

Density of gas at stp is 2.78gdm-3. Find RMM of gas Answer 1 dm3 gas at stp = 1.78g 22.4 dm3 at stp = 1.78 x 22.4g (Molar Volume) = 39.87g (Molar Mass) RMM gas = 39.87

Mass of 1 dm3 gas is 1.96g at stp. Find RMM of gas Answer Mass of 1dm3 at stp = 1.96g Mass of 22.4dm3 at stp = 22.4 x 1.96g = 43.9g RMM = 43.9

IB Questions on Ideal Gas

1 2

3 4 Density of gas is 2.6gdm-3 , T- 25C and P - 101kPa Calculate RMM of gas Answer PV = n R T PV = m x R x T M M = m x R x T V P M = ρ x RT P Density = 2.6 gdm-3 → 2.6 x 103 gm-3

P = 101kPa → 101 x 103 Pa M = ρ x RT P M = (2.6 x 103) x 8.31 x (298) 101 x 103

= 63.7

Density ρ = m (mass) V(volume)

Density gas is 1.25gdm-3 at T- 25C and P- 101kPa. Calculate RMM of gas Answer Using PV = n R T n = m M PV = m x RT M M = m x RT V P M = ρ x RT = 1.25 x 103 x 8.31 x 298 P 1.01 x 103 = 30.6gmol-1

m = density, ρ = 1.25gdm-3

V

Calculate RMM of gas Mass empty flask 25.385g Mass flask filled gas = 26.017 Mass flask filled water = 231.985g Temp = 32C Pressure = 101kPa Answer Mass gas = (26.017 – 25.385) = 0.632g Vol flask = (231.985 – 25.385) = 206.6 x 10-6 m3

P = 101kNm-2 → 101 x 103 Nm-2

M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206.6 x 10-6 = 76.8

IB Questions on Ideal Gas

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

Vol gas = 804.303cm3 → 804.303 x 10-6m3

Pressure = 97.7kPa → 97700Pa Temp = 23C → (273 + 23) = 296K Assume density water = 1g/cm3

Answer 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

= 146.7g/ml

5 6

3.376g gas occupies 2.368dm3 at T- 17.6C, P - 96.73kPa. Find molar mass Answer 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 = 35.61g/mol

Unit conversion

T – 17.6C → (273 = 17.6 = 290.6) P – 96.73kPa → 96730Pa

7

6.32 g gas occupy 2200cm3, T- 100C P -101kPa. Calculate RMM of gas

Answer PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x (100+273) n = 7.17 x 10-2 mol n = mass M RMM = mass n RMM = 6.32 7.17 x 10-2

= 88.15

8

Find empirical formula for composition by mass. S 23.7%, O 23.7%, CI 52.6% Density of its vapour at T- 70C and P- 98kNm-2 = 4.67g/dm3 What molecular formula?

Empirical formula - SO2CI2

PV = nRT PV = m x R x T M M = m x R x T V P Density ρ = m (mass) V (volume) M = ρ x RT P Density = 4.67gdm-3 → 4.67 x 103 gm-3

P = 98kN-2 → 9.8 x 104 Nm-2

M = (4.67 x 103) x 8.31 x (273 +70) 9.8 x 104

= 135.8 135.8 = n [ 32 + (2x16)+(2 x 35.5) ] 135.8 = n [ 135.8] n = 1 MF = SO2CI2

IB Questions on Ideal Gas

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 = 1 0.738

1

1.48 = 2 0.738

2

1.48 = 2 0.738

2

9

Answer Answer

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

• T - 27C Calculate its volume at 35C

Answer: (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 = 250kPa V1 - 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

IB Questions on Ideal Gas

T → (20 + 273) = 293K n → 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 ∝ 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 sides

• 2 variables involved

10 11 12

13 14 15

IB Questions on Ideal Gas

Combined gas Law

• no need to convert to SI units

• cancel off at both sides

• 3 variables involved

16 17 3

18 19

A syringe contains gas at V1 - 50cm3

P1 – 1atm T1 - 20C → 293K What volume , V2, if gas heated to V2 - ? T2 - 100C → 373K P2 - 5 atm

Answer: (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 = V1 Volume 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 T1 → Final T2 = 2T1

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 = 2T1

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

Initial T1 → Final T2 = 2T1

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 halved T double

Ideal Gas Law Lab simulation

Ideal Gas Law Videos