The document provides information on kinetic theory of gases, gas laws (Boyle's law, Charles' law, Avogadro's law, pressure law), and determination of relative molecular mass (RMM) of gases using the ideal gas equation. It discusses Maxwell-Boltzmann distribution curve, assumptions of kinetic theory, effects of temperature and molecular mass on molecular speeds. It also describes various gas laws and how to derive them from the ideal gas equation by keeping certain variables constant. Methods to determine RMM such as direct weighing of gases, syringe method and using density are presented along with sample calculations. Empirical formula determination from percentage composition and combustion analysis is also summarized.

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
IB Chemistry Kinetic Theory, Ideal Gas Equation and
RMM determination of gases.

2. 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
1
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

3. 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
.
2
1
vmKE =
Xe – mass high ↑ - speed v low ↓ 2
.
2
1
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
2
1
mvKE =

4. 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-1
K-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
VolPressure
TempVol Temp
Pressure Vol
n
PV = nRT
P = constant x T
P ∝ T
P1 = P2
T1 T2

5. 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 constant
Unit - 8.314 Jmol-1
K-1
V = Vol gas
Unit – m3
PV = nRT
Boyle’s Law
n, T fix
P = Pressure
Unit – 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

6. 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 constant
Unit - 8.314 Jmol-1
K-1
V = Vol gas
Unit – m3
PV = nRT
n, P fix
P = Pressure
Unit – 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

7. Ideal Gas Equation
PV = nRT
4 diff variables P, V, n, T→
R = gas constant
Unit - 8.314 Jmol-1
K-1
V = Vol gas
Unit – m3
PV = nRT
n, V fix
P = Pressure
Unit – 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

8. Ideal Gas Equation
PV = nRT
4 diff variables P, V, n, T→
R = gas constant
Unit - 8.314 Jmol-1
K-1
V = Vol gas
Unit – m3
PV = nRT
P, V fix
P = Pressure
Unit – 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

9. 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

10. Unit conversion
1 atm 760 mmHg 100 000 Pa 100 kPa↔ ↔ ↔
1m3
10↔ 3
dm3
10↔ 6
cm3
1 dm3
1000 cm↔ 3
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
nT
Vα
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

11. 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
ρ=
×=
=
=
Density ρ = m (mass)
V (vol)PV
mRT
M
RT
M
m
PV
nRTPV
=
=
=
Molar
mass
RMM using Ideal Gas Eqn
PV = nRT

12. 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 0
C
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

13. 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 0
C
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

14. 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 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

15. 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

16. 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 23.7
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
ρ=
×=
=
=
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

17. 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
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
==
××
=
=
Density gas 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
ρ=
×=
=
=
Density ρ = m (mass)
V (vol)
M = (2.6 x 103
) x 8.31 x (298)
101 x 103
M = 63.7

18. 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
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→
Density gas 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
ρ=
×=
=
=
Density ρ = m (mass)
V (vol)
M = (1.25 x 103
) x 8.31 x (298)
101 x 103
M = 30.6

19. 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

20. Gas occupy at (constant P)
V – 125 cm3 ,
T - 27 C
Find 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 kPa
V1 - 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 325Pa
PV = 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) 2SO→ 3(g)
Find vol of SO2 , dm3
at stp
PV = nRT (at constant P,T)
V n∝
2SO2(g) + 1 O2(g) 2SO→ 3(g)
2 mol 1 mol 2 mol
2 vol 1 vol 2 vol
3dm3
2dm3
?
↓
SO2 is limiting
2dm3
SO2 2dm→ 3
SO3
3dm3
SO2 3dm→ 3
SO3
Boyle, Charles, Avogadro Law
no need to convert to SI units
cancel off at both sides
2 variables involved
n 2.00/28→
= 0.0714 mol

21. A syringe contains gas at
V1 – 50 cm3 ,
P1 – 1 atm, T1 - 293K
What vol , V2, if gas heat to
V2 ? 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 /K
A. Doubled Doubled
B. Halved Halved
C. Doubled Halved
D. Halved Doubled
Initial P1 Final P→ 2 = 1/2P1
Initial T1 Final T→ 2 = 2T1
Initial V1 Final V→ 2 = ?
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 = 2T1
Initial P1 Final P→ 2 = 2P1
Initial T1 Final T→ 2 = 2T1
Initial V1 800 Final V→ 2 = ?
A. 200 cm3
B. 800 cm3
C. 1600 cm3
D. 3200 cm3
3
2
2
2
22
1
11
13
373
5
293
501
cmV
V
T
VP
T
VP
=
×
=
×
=
12
1
21
1
11
2
22
1
11
2
2
VV
T
VP
T
VP
T
VP
T
VP
=
=
=
12
1
21
1
11
2
22
1
11
4
22
VV
T
VP
T
VP
T
VP
T
VP
=
×
=
=
3
2
1
21
1
1
2
22
1
11
800
2
2800
cmV
T
VP
T
P
T
VP
T
VP
=
=
×
=

22. Fix mass ideal gas has a V1 = 1dm3, P1, T1
Find V2 ,when T doubled (x2), P tripled (x3)
V2 = ?, P2 = 3P1, T2 = 2T1
Initial P1 Final P→ 2 = 3P1
Initial T1 Final T→ 2 = 2T1
Initial V1 = 1 dm3
Final V→ 2 =?
Fix mass ideal gas has a V1 = 2 dm3 , P1, T1
Find V2 ,when T double (x2), P quadruple (x4)
V2 = ?, P2 = 4P1, T2 = 2T1
Initial P1 Final P→ 2 = 4P1
Initial T1 Final T→ 2 = 2T1
Initial V1 2dm3
Final V→ 2 = ?
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 V→ 2 = 2V1
Initial T1 Final T→ 2 = 2T1
Initial P1 =40 Final P→ 2 = ?
3/2
2
31
2
1
21
1
1
2
22
1
11
=
=
×
=
V
T
VP
T
P
T
VP
T
VP
3
2
1
21
1
1
2
22
1
11
1
2
42
dmV
T
VP
T
P
T
VP
T
VP
=
=
×
=
40
2
240
2
1
12
1
1
2
22
1
11
=
×
=
×
=
P
T
VP
T
V
T
VP
T
VP
What conditions would one mole
CH4, occupy the smallest vol?
A. 273 K and 1.01×105 Pa
B. 273 K and 2.02×105 Pa
C. 546 K and 1.01×105 Pa
D. 546 K and 2.02×105 Pa
PV = nRT
V = nRT
P
= low T, high P

23. Find total vol and composition of remaining gas
10cm3
ethyne react with 50cm3
hydrogen
C2H2(g) + 2H2 (g) C→ 2H6 (g) at stp
PV = nRT (at constant P,T)
V n∝
C2H2 (g) + 2H2(g) C→ 2H6 (g)
1 mol 2 mol 1 mol
1 vol 2 vol 1 vol
10cm3
20cm3
10cm3
C2H6 = 10cm3
produced
H2 = 50-20 =30 cm3
remain (excess)
Which conditions does a fix mass
of an ideal gas have greatest vol?
Temperature Pressure
A. low low
B. low high
C. high high
D. high low
PV = nRT
V = nRT
P
= high T, low P
Find vol of 6 g of chlorine at 27o
C and 101 kPa.
pV = nRT
T = 300K,
n = 6/71 = 0.08451 mol CI2, Mr(Cl2)= 71
p = 101000 Pa.
3
002087.0
101000
30031.808451.0
mV
V
P
nRT
V
=
××
=
5 litre container contain 0.5 kg butane gas (C4H10).
Cal press at 25o
C.
Mr(C4H10)= 58, 0.5kg = 500g
moles n = 500/58 = 8.621, T = 298K
5 litre = 5 dm3
= 5 x 10-3
m3
kPaP
P
V
nRT
P
4272
105
29831.8621.8
3
=
×
××
=
=
−

24. P = 202600 Pa
n = 0.050 mol
T = 400K
V = ? 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 20o
C?
3
00082.0
202600
40031.8050.0
mV
V
P
nRT
V
nRTPV
=
××
=
=
=
kPaP
P
V
nRT
P
nRTPV
3248
105.7
29331.810
3
=
×
××
=
=
=
−
50 L fill with argon to a press of 101 kPa at
30o
C.
How many moles of argon ?
P = 101000Pa
V = 50 x 10-3
m3
n = ? mol
T = 303 K
moln
n
RT
PV
n
nRTPV
2
30331.8
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
nR
PV
T
nRTPV
15.76
31.81.0
10250.0253250 3
=
×
××
=
=
=
−

25. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
http://kwokthechemteacher.blogspot.hk/2011/07/ideal-gas-assumptions.html
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com