1. CHAPTER:
GAS LAWS

2. 12.1.0: SPECIFIC OBJECTIVES
By the end of this topic, the learner should be able to:
State Boyle’s and Charles’ laws.
Carry out calculations involving the gas laws.
Use combined gas laws in calculations.

3. SPECIFIC OBJECTIVES CONT’
4. Explain diffusion in liquids and gases in terms of kinetic theory.
5. Relate the rate of diffusion to the relative molecular mass of a gas.
6. State Graham’s law of diffusion.

4. 12.2.0: CONTENT
BOYLE’S LAW AND CHARLES’ LAW
Boyle’s law, Charles’ law and the combined gas laws
Explanation of the laws (use graphs to illustrate)
Calculations involving gas laws
NB: Use of SI units should be emphasized

5. 12.2.2: GRAHAM’S LAW OF DIFFUSION
CONTENT CONT’
12.2.2: GRAHAM’S LAW OF DIFFUSION
Graham’s law of diffusion
Relationship between rate of diffusion and density or relative molecular mass of a gas
Calculations on diffusion

6. GAS LAWS
INTRODUCTION
Gas particles are spread a part compared to particles in the solid and liquid states. Gas particles are always in a continuous random motion as explained by kinetic theory of matter. i.e

7. Gases are made up of many small tiny particles known as molecules or atoms.
These particles are in a state of continuous or rapid random movement.
The molecules and atoms collide with one another and with the walls of their containers without losing any energy.
The distance between the particles is much greater than their diameters which means there exist relatively small intermolecular attractions between the particles.

8. SI UNITS IN GAS LAWS
International system of units of measurements abbreviated as SI from a French phrase systeme d’unite’s international.
TEMPERATURE
SI unit for temp is based on the absolute scale (Kelvin (K)). This was suggested by Lord Kelvin in Absolute temp scale is based on the lowest possible temp that can be achieved. There are no negative values on the absolute scale. The lowest value is 0 K known as the absolute zero temp.
The Celsius scale has negative values and is not an absolute scale. A temp on the Celsius scale is converted to absolute scale/Kelvin scale by adding 273 to the Celsius value. The Kelvin units do not have the degree symbol (oc). Similarly temp is converted to Celsius scale by subtracting 273 from the Kelvin value.

9. EXERCISE
Convert the following to Kelvin units
25 oC
0 oC
-273 oC
40 oC
100 oC
46 oC
300 oC
200 OC
Convert the following to Celsius units
373K
73K
315K 0K
500K
700K
100K

10. PRESSURE
Pressure is force acting perpendicularly per unit area. Its units are Newton’s per square meter (N/M2). However the SI units for pressure are Pascal’s (Pa). Named after, Blaise Pascal a 17th Centaury Mathematician cum scientist.
1 Pa = 1 N/M2
Pressure exerted by the air column at sea level is 1 atmosphere. When a mercury barometer is used to measure this pressure, the mercury rises to 760mmHg (read as 760 mm of mercury). Pressure is calculated using the formula:
P=hρg Where:
h= is the height of mercury in meters
ρ= is the density of mercury and
g= is the gravitational field intensity measured in Newton’s per Kilogram

12. PRESSURE CONT’
Thus pressure exerted by 760 mmHg is equal to:
P=hρg
= (760/1000) × (13600Kg/M3) ×10N/Kg
= Pa
Therefore 1 atmosphere = Pa or
×105 Pa =760 mmHg

13. VOLUME
Volume is the amount of space occupied by matter. The SI unit of volume is the cubic metre (M3)
1 Litre = 1000 cm3 = 1 M3
STANDARD CONDITIONS OF A GAS
By convention or international agreement the following are the standard conditions:
Standard temp is 273K
Standard pressure is 1 Pa or 760 mmHg
These conditions are referenced as stp (Standard temperature and pressure

14. BOYLE’S LAW
If a piston is used to push gas particles in a syringe such that they occupy only half of the tube, keeping the temp constant then pressure of the particles on the walls of the tube would be double. This is because of their population in half the original space (or vol) left.

16. The quantitative relationship between vol and pressure of gases is on the basis of Boyle’s Law Published in 1662 by Robert Boyle ( ). The law states that the volume of a fixed mass of gas at constant temp is inversely proportional to its pressure.

17. When a graph of pressure against volume at constant temp is drawn it takes the shape below

18. A graph of pressure against 1/volume has the below shape
A graph of pressure against 1/volume has the below shape. Ie pressure variation against reciprocal of volume of a fixed mass of gas at constant temperature.
Pressure
1/volume

19. The mathematical expression of Boyle’s law is
VOLUME (V) α
PRESSURE (P)
V = CONSTANT
PRESSURE Or
PRESSURE × VOLUME = CONSTANT
When pressure of a gas changes from P1 to P2 its volume also changes from V1 to V2. This leads to a relationship P1 V1 = P2 V2

20. WORKED OUT EXAMPLE Soln
A vol of 375 cm3 of a gas has a pressure of 20 atmos. What will be its vol if pressure is reduced to 15 atmos.
P1 v1= p2 v2
P1= 20 atmos V1=375 cm3 V2 =?
V = p1v1 P2
= × 375 15
= cm3

21. QUESTIONS
At a pressure of 5 atmos a given mass of gas occupies a vol of 200 cm3. At what pressure will the gas have a vol of 800 cm3
A certain mass of gas occupies 250 cm3 at 25oc and 750 mmHg. Calculate its volume at 25oc if pressure changes to 760 mmHg in SI units.
At constant temp a gas at 540 mmHg pressure occupies a vol of 300 litres. The gas is expanded to occupy a vol of 600 litres. What is the new pressure of the gas?

22. CHARLES’ LAW
This law deals with how a gas’ volume relates to temp. The following set up is used to investigate Charles’ law

23. When the capillarity tube was immersed in a trough of warm water the mercury drop moves up however when the capillarity tube immersed in a trough of ice cold water the mercury drop in the capillarity tube moves down.
When temp of a given mass of gas is increased and pressure kept constant the volume of the gas increases. This is because the kinetic energy of the gas molecules increases hence the molecules move faster and hit the walls of the container harder. The reverse is true
The relationship between temp and volume of fixed mass of gas is summarized by Charles’ law that states: The volume of a fixed mass of gas is directly proportional to its absolute temp at constant pressure.

24. This implies that as vol of a fixed mass of gas increases in the same proportion as the absolute temp provided the pressure is kept constant. i.e

25. Mathematically Charles law can be expressed as
Volume α absolute temp
V α T therefore
V = K x T (K is some constant)
V = K T
V1 = V2 T T2

26. A graphical representation of Charles’ law is as below
-273 oc oc

27. This extrapolates to zero volume at a temperature of zero Kelvin
This extrapolates to zero volume at a temperature of zero Kelvin. This leads to unreasonable case where if temp is -273oc the vol of the gas will be zero (0 cm3). In practice this is not possible since the gas would have solidified and therefore Charles’ law does not apply.
The temp -273oc at which a gas occupies zero volume theoretically is called absolute zero and is from this that the absolute temp scale is derived. The units for the temp on the absolute scale are Kelvin (K).
WORKED OUT EXAMPLE
A gas occupied 450 cm3 at 27oc. What vol will it occupy at 177oc if pressure is kept constant?
Soln
V1 = V2 T T2
V1=450cm T1= =300K T2= =450K V2=?
V2 = V1T = 450× =675cm3 T

28. QUESTIONS
At a temp of 57oc nitrogen gas occupies a volume of 750cm3. At what temp will the gas occupy 100cm3.
A gas occupies 200cm3 at 0oc and 740 mmHg pressure. What volume will it occupy at 47oc and 740mmHgf pressure?
What is the temperature of a gas that is expanded from 2.5 L at 25ºC to 4.1L at constant pressure?
What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC?

29. COMBINED GAS LAW PV = Constant.
The two equations representing each of the laws (Boyle’s & Charles’) give the combined gas law equation or ideal gas equation or general gas equation. When the two laws are combined they give a general rule for any gas behaving ideally. Ideal gas is expected to obey both Boyle’s and Charles’ law. Such a gas is called a perfect gas or ideal gas.
The mathematical expression of Charles’ law is V α T. It can be combined with Boyles’ law V α 1/P to obtain this expression V α T/P therefore PV α T and hence
PV = Constant. T

30. If a fixed mass of gas of vol V1 exerts pressure P1 at absolute temp T1 then the expression may be written as P1V1 = Constant. T1
Suppose the same mass of gas has a vol V2 and exerts a pressure P2 at Absolute temp T2 then the expression becomes
P2V2 = Constant. T2
Therefore P1V1 = P2V2
T T2
This equation enables the vol of any mass of a gas to be obtained under any conditions of temp and pressure provided its vol under some other conditions of temp and pressure are known i.e ideal gas equation.

31. WORKED OUT EXAMPLE
What will be the vol of a given mass of oxygen at 25oc if it occupies 100 cm3 at 15oc (pressure remains constant)
Soln
V1=100cm3 , T1= =288K, P1 = P2, V2 ?
P1V1 = P2V2
T T2
Therefore V2= V1T2 T1
=100 × 298
288
=103.5cm3

32. QUESTIONS
A given mass of gas occupies 20cm3 at 25oc and 670mmHg pressure. Find the volume it will occupy at:
a) 10oc and 335 mmHg
b) 0oc and 760 mmHg
2. A sample of neon gas used in a neon sign has a volume of 15 litres at stp. What is the volume of the neon gas at 2.0 atm and -25oc?
3. A sample of helium gas a volume of 0.18 litres, a pressure of 0.8 atm and a temp of 29oc. What is the new temp of the gas at volume of 90 milliliters’ and pressure of 3.2 atm?

33. DIFFUSION AND GRAHAM’S LAW
When substances are placed at a one point they are capable of spreading out from one area to another. Since substances do not spread as a whole then they must be broken up into smaller particles for them to spread out. Examples to this spreads are as below:
What happens when crystals of potassium manganate (VII) are placed in water?
A crystal of potassium manganate (VII) is introduced in a beaker containing clean water carefully using a straw and the set up left undisturbed for at least 40 minutes and the observations made are recorded.

34. OBSERVATION AND EXPLANATION
The purple colour of the permanganate spread throughout the clean water in the beaker. Spreading out of the purple colour of KMnO4 in water is evidence for the movement of the solid particles. Movement of KMnO4 particles in water is due to the collisions between KMnO4 particles and the water molecules as a result of both having kinetic energy.

35. What happens when bromine water is left in a gas jar?
A bout 4 drops of bromine liquid are put in a gas jar and then an empty gas jar is inverted over the gas jar containing bromine liquid drops and the set up is left undisturbed for 30 minutes.
OBSERVATION AND EXPLANATION
The brown fumes of bromine liquid are observed in the jar that was inverted over the jar containing the bromine liquid. The spread of bromine vapours into the two jars indicates movements of liquid particles in air.

36. What happens when ammonia gas fumes are left in air?
When cotton wool soaked in a concentrated solution of ammonia is placed at one end of a combustion tube and at the other end wet red litmus paper is placed there and the set up is left undisturbed for 30 minutes what happens?
OBSERVATION AND EXPLANATION
The red litmus paper turns blue, indicating that the ammonia molecules must have spread along the combustion tube. This is evidence of movement of gas particles.

37. N/B
Spreading out of gas particles in air takes a short time than liquid particles in water because gas particles are far apart and have more kinetic energy than the liquid particles.
Spreading out of particles of matter from a region of high concentration to region of low concentration through a medium is known as DIFFUSION.
Diffusion occurs faster in gases than in liquids and solids.

38. BROWNIAN MOTION
Brownian motion is the random movement of microscopic particles suspended in a fluid. A fluid is a substance that can flow and has no shape such as gas or liquid. Brownian motion can further be verified by the use of a smoke cell.
GRAHAM’S LAW
It is about the relationship between the rate of diffusion of a gas and its density. It was first investigated in 1833 by Thomas Graham an English scientist.

39. DO MOLECULES OF DIFFERENT GASES DIFFUSE AT THE SAME RATE?
The set up below is used to investigate the above question. Two pieces of cotton wool soaked in conc. ammonia solution and conc. hydrochloric acid separately are plugged at opposite ends of a glass tube simultaneously and the glass tube is then Stoppard. A stop watch is then started immediately and the time taken to for a change to occur in the glass tube to occur is recorded. A mark is made on the glass tube where the change occurred and the distance covered by each gas is measured.

40. OBSERVATION AND EXPLANATION
Conc. ammonia soln generates ammonia gas while conc. hydrochloric acid generates hydrochloric acid gas. NH3 (g) and HCl (g) diffuse in air in the tube and when their molecules meet they react to form white fumes of NH4Cl i.e
NH3 (g) + HCl (g) NH4Cl (s)
In this experiment it was observed that NH3 (g) whose molecular mass is 17g (H=1, N=14, therefore 14+(1×3)= 17) covered a longer distance before meeting with HCl (g) molecules of molecular mass 36.5g (H=1, Cl= 35.5 therefore = 36.5). The distance covered by NH3 (g) was 12 cm while that of HCl (g) was 8 cm within the same time interval of 5 minutes.

41. Using this data the rate of diffusion of
NH3(g) is 12cm = 2.4cm/min
5 min
While that of HCl (g) is 8cm = 1.6cm/min
The relative rate of diffusion of NH3 (g) compared to HCl (g) in air is
Rate of diffusion of NH3 = 2.4cm/min = 1.5
Rate of diffusion of NH cm/min
i.e Ammonia gas diffuses 1.5 times faster than HCl
Behaviour of gases when they diffuse is summarized by Graham’s law of diffusion which states that: Under the same conditions of temp and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density.

42. MATHEMATICAL EXPRESSION OF GRAHAM’S LAW
Rate (R) α Density (d) d
R= CONSTANT
WHEN THE RATE OF DIFFUSION OF TWO GASES A AND B ARE COMPARED THEN THE EQUATIONS ARE:
I) RA= CONSTANT II) RB= CONSTANT
dA dB

43. In equation (I) R A dA =CONSTANT
In equation (II) R B dB =CONSTANT
Therefore R A dA = R B dB
Thus RA = dB
RB dA

44. THEREFORE THE RATE OF DIFFUSION OF TWO GASES A AND B COMPARED WILL BE
SINCE DENSITY IS DIRECTLY PROPORTIONAL TO MOLECULAR MASS GRAHAM’S LAW CAN ALSO BE EXPRESSED AS
R= CONSTANT
MOLECULAR MASS (M)
THEREFORE THE RATE OF DIFFUSION OF TWO GASES A AND B COMPARED WILL BE
Thus RA = MB
RB MA

45. Since rate is inversely proportional to time (Rα1/Time) thus it is possible to compare the time taken for equal volumes of two gases to diffuse under similar conditions
RA = constant RB = constant
Time A Time B
i.e RATA = Constant, RB TB = Constant
Therefore RA TA = RB TB

46. RA = TB
RB TA
RA = dB
= MB
BUT
RB dA MA
THEREFORE TB = dB = MB
TA dA MA

47. Rco2 Mco 28 =1.254 WORKED EXAMPLE
Equal volumes of carbon (II) oxide and carbon (IV) oxide are allowed to diffuse through the same medium. Calculate the relative rate of diffusion of carbon (II) oxide. (C=12.0, O= 16.0)
Soln
RMM of CO=12+16=28
RMM of CO2=12+32=44
Rco = Mco2
Rco Mco =
=1.254

48. QUESTIONS
Determine the molecular mass of gas Y which diffuses one and half times faster than oxygen.( O=16)
If it takes 30 secs for 100 cm3 of carbon (IV) oxide to diffuse across a porous plate how long will it take 150cm3 of nitrogen(IV) oxide to diffuse across the same plate under similar conditions (C=12, N=14, O=16)
Hydrogen gas diffuses seven times faster than a gas X under the same conditions of temp and pressure. Calculate the relative molecular mass of X if that of hydrogen is two
A volume of 120cm3 of nitrogen gas diffuses through a membrane in 40 secs, how long will it take 240 cm3 of carbon (IV) oxide to diffuse through the same membrane?
If it takes 20 secs for 200cm3 of oxygen gas to diffuse across a porous plug. How long will it take an equal volume of sulphur (IV) oxide to diffuse across the same plug?

49. APPLICATIONS OF DIFFUSION
Uranium separation by diffusion. The heavier isotope diffuses less rapidly than the lighter one
Extraction of helium is a lighter element found in natural gas. It is extracted by passing the natural gas through a very thin sheet of Pyrex glass (EFFUSION –Movement of gas through a small opening)