CONTENTS:

1. Gay- Lussac’s Law and Avogadro’s Law.
2. Graham’s law of diffusion.
3. Molar volume of gases- Avogadro number and the mole concept.
4. Dalton’s law of partial pressure.

 

PERIOD 1: GAY- LUSSAC’S LAW AND Avogadro’s LAW

Gay- Lussac’s law describes the combining volumes of gases that react together. In his experiment, all temperatures and pressures were kept constant:

1. STEAM: Gay- Lussac’s observed that two volumes of hydrogen reacted with one volume of oxygen to yield two volumes of steam
2. B. HYDROGEN CHLORIDE GAS: One volume of hydrogen combined with one volume of chloride to yield two volumes of hydrogen.

Hydrogen + Chlorine → Hydrogen Chloride

 

Gay- Lussac’s noticed that the combining volumes as well as the volumes of the products, if gaseous, were related by simple ratios of whole numbers. He proposed the law of combining volume or gaseous volumes.

Hence; Gay- Lussac’s law combining volumes states that when gases react, they do so in volumes which are in simple ratios to one another and to the volumes of the products, if gaseous provide that the temperature and the pressure remain constant.

EXAMPLES

1. What is the volume of oxygen required to burn completely 45cm3 of methane (CH4)?

By Gay- Lussac’s Law:

1 volume of methane required 2 volumes of oxygen i.e.

1cm3 of methane requires 2cm3 of oxygen

∴ 45cm3 of methane require 90cm3 of oxygen

2.20cm3 of carbon (I) oxide are sparked with 20cm3 of oxygen. If all the volumes of gases are measured at a S.T.P, calculate the volume of the residual gases after sparking?

Equation of reaction                  2COg+ O2g                2Co2g

Combining volume                     2  :  1       : 2

Volumes before sparking           20cm3 10cm3,      20cm3

Volumes after sparking                 –10            20

Residual gases = un-reacted oxygen + carbon (IV) oxide formed

Volume of residual gas = 10cm3+ 20cm3= 30cm3

 

 

AVOGADRO’S LAW

Avogadro’s Law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

This law means that for all of gases e.g. oxygen, hydrogen, Chlorine etc. if their volumes are the same, they will have the same number of molecules.

Avogadro’s Law is easily applied to convert volume of gases to the number of molecules. Avogadro’s Law can be used to solve problem under Gay –Lussac’s law of combining volumes.

The formation of steam from reaction of Hydrogen and Oxygen is given below:

Reaction: Hydrogen + Oxygen → Steam

Volume: 2   1 2

Gay –Lussac’s:  2    :   1       :   2

Avogadro’s Law: 2   :   1       : 2

This agrees with the equation below:

2H2(g)   +   O2(s)       2H2O(s)

i.e. 2 molecules of hydrogen combine with 1 molecule of oxygen to produce 2 molecules of steam

 

Example:

1. 60cm3of hydrogen are sparked with 20cm3 of oxygen at 1000C and 1 atmosphere. What is the volume of the steam produced?

Solution

2H2 + O2 2H2O

From the equation, 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of steam.

2H2  + O2 2H2O

2 vol      1vol   2 vol (combining volumes)

i.e.  2cm3   1cm3           2cm3

From the above information, when 2cm3(2 vol) of H2 react, 1cm3(1 vol) of O2will react i.e. half of H2vol, to give 2cm3(2 vol) of H2O.

Thus, 10cm3 of H2 will react with 5 cm3 of O2 to produce 10cm3 of H2O and so on.

From the question, we have 60cm3of  H2 and 20cm3 of O2, thus, when all the 20cm3 of O2react, only 40cm3 ofH2 will react to give 40cm3 of H2O, because the volume of H2 is the same as that of H2O i.e.

2H2  + O2 2H2O

2 vol     1 vol   2 vol

2cm3   1cm3           2cm3

40cm3 20cm3        40cm3

Thus, the volume of steam (H2O) formed is 40cm3

2. What volume of propane is left unreacted when 80cm3 of oxygen and 20cm3 of propane react according to the equation below?

C3H8(g)   + 5O2(g) 3CO2(g) + 4H2O

Solution 

C3H8(g)   + 5O2(g)   3CO2(g) + 4H2O

1vols          5vols

1cm3        5cm3

4cm3            20cm3

Volume of the propane before the reaction =20cm3

The volume that reacted =4cm3

Volume that did not react= volume before

Reaction – volume that reacted i.e. 20 – 4 =16cm3

 

EVALUATION

1. State Gay –Lussac’s law
2. State Avogadro’s law.
3. 50cm3of methane were burnt completely in oxygen according to the equation below.

CH4 + 2O2 Co2 + 2H2O

Calculate: (a) volume of oxygen used (b) Volume of carbon(Iv) oxide produced

(c) Volume of steam produced.

 

WEEKEND ACTIVITY: 

 (a) Define Graham’s law of diffusion.(b) What is a mole and mole concept?

 

 

2: GRAHAM’S LAW OF DIFFUSION.

This law states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass or square root of its vapour density. Mathematically, Graham’s law of diffusion can be represented as:

R1R2∝√p2p1  Where R1 and R2 are the rates of diffusion and P1 and P2, the densities of the two gases.

The density is directly proportional to its molecular mass.

EXAMPLES

1. 100cm3of oxygen diffuse through an office in 60 seconds while it takes 120seconds for the same office. Calculate the molecular mass of the unknown gas [0=16]

Solution

RRxO2 = MxMO2

Since the rate of diffusion is inversely proportional to the time taken:

RRxO2 = =txtO2 = √MxMO2

(txtO2) = MxMO2

Mx= MO2 (txtO2)2= 32 ×(12060)2= 32 22

Mx = 32 × 4 = 128g

 

2. 200cm3of hydrogen diffused through a porous pot in 40 seconds. How long will it take 300cm3of chlorine to diffuse through the same pot?

Solution

200cm3of hydrogen diffused in 40secs

∴ 300cm3 of chlorine will diffuse in

300cm3200cm3 × 4020

(3 × 20) = 60seconds

Now, using the equation

t1t2 =√M1M2

Where t= 60s, M1 = molecular mass of hydrogen

i.e H2= (2 ×1) =2

M2= molecular mass of chlorine = cl2 =2× 35.5 = 71

T2 = t1√M1M2= 60√712= 60 35.5 = 60 × 5.96

= 357.5sec

Time of diffusion of chlorine = 358s.

3. How many times the rate of diffusion of hydrogen is faster than that of oxygen and what law do you use to get the answer? [vapour density] of [H=1, O=16]

Solution

Rate (R+) of diffusion of H2= 

Density of O2Density of H2

R1R2=161= R1R2= 41

∴Hydrogen diffuses four times faster. The law used is Graham’s law of diffusion.

 

RELATIVE VAPOUR DENSITY OF A GASE

The vapour density of a gas or vapour is the number of times a given volume of gas (or vapour) is heavier than the same volume of hydrogen measured and weighed under the same temperature and pressure 

Vapour density = mass of 1 vol of a gas or vapourmass of equal volume of hydrogen

Applying Avogadro’s law, it is possible to show that the vapour density of a gas is related to the relative molecular mass of the gas.

V.D = mass of 1 mole of a gas or vapourmass of 1 molecule  of hydrogen

V.D = mass of 1 vol of a gas mass of 2 atoms of hydrogen

∴2 x V.D =relative molecular mass

The density of hydrogen at S.T.P is 0.09dm3

 

Example

Calculate the vapour densities of the following gases from the given data.

560cm3 of oxygen at S.T.P weighs 0.8g

1,400cm3 of sulphur (iv) oxide weighs 4g

Solution

1000m3 of hydrogen at S.T.P weighs 0.09g

∴ 560cm3 of hydrogen at 560cm3100cm3× 0.09

= 0.05g

V.D= massofagivenvolumeofgasmassofeequalvolumeofhydrogen

∴Vapour density of oxygen=

mass of  560 of oxygenmass of 560 of hydrogen

1000cm3 of hydrogen at S.T.P weighs 0.09g.

∴ 1400 of hydrogen will weighs

1400 0.091000= 0.126g

Vapour density= massofagivenvolumeofgasmassofeequalvolumeofhydrogen

∴Vapour density of SO2= mass of  1400cm3 of  SO2mass of 1400cm3of H2

= 4g0.126= 31.74= 32

 

EVALUATION

Deduce the relationship between relative molecular mass and vapour density of a substance.

Define vapour density of a gas.

 

3: MOLAR VOLUME OF GASES- AVOGADRO NUMBER AND THE MOLE CONCEPT

The molar volume of any gas is the volume occupied by one mole of that gas at s.t.p. and is  numerically equally to 22.4dm3 i.e. one mole of any gas at s.t.p. occupies the same volume the value of which is 22.4dm3. This value is called molecular mass or molar mass.

From Avogadro’s law, the molar volume for all gases contains the same number of molecules. This number is called the Avogadro’s number or constant and the value is 6.02 1023 at s.t.p

MOLE: The mole can be defined as the amount of substance which contain as many elementary particles or entities e.g. ions, molecules, atoms, electrons as the number of atoms in exactly 12 grams of carbon -12. 

The mole of any substance represents 6.02 1023 particles of any substance. Therefore, a mole refers to Avogadro’s number of particles of any substance.

In summary, the molar mass of a gas contains Avogadro’s number of molecules which is 6.02 1023 and occupies a volume of 22.4dm3 at s.t.p.

The atomic mass of every element also contains Avogadro’s number of atoms.

The mole concept– This says that one mole of any substance contains the same number of particles; which can be atoms, molecules or ions. This number is 6.023 1023dm3 (the Avogadro’s number)

Examples

1.158g of a gas at s.t.p. occupies a volume of 5000dm3. What is the relative molecular masss of the gas? (Molar volume at s.t.p= 22.4dm3 mol-1

Solution

Volume of gas: V = 50.00dm3

Molar volume of gas; V = 22.4dm3 mol-1

N= amount in moles

=VV

N= 5022.4dm3mol-1= 2.23mol

Molar mass M of the gas = Mn = 158g22.4dm3 mol-1 = 70.8

Molar mass = 71 gmol-1

2. 2. What is the mass of 3 moles of oxygen gas O2? (O = 16)

Mass of 1 mole of O2= (2 16)g =32g

Mass of 3moles of O2= (3 32)g = 96g

3. 3. How Many moles are there in 20g of CaCO3? [CaCO3 =100]

Molar mass of CaCO3= 100g

100g of CaCO3= 1 mole

20g of CaCO3 =20 1001mole = 0.2moles

 

EVALUATION

1. Using the relationship between mole and Avogadro’s number. Define mole in six ways.

 

 

4: DALTON’S LAW OF PARTIAL PRESSURE

Dalton’s law of partial pressure states that for a mixture of gases that do not react chemically, the total pressure exerted by the mixture of gases is equal to the sum of the partial pressures of the individual gases. 

Mathematically, Dalton’s law of partial pressure for a mixture of n gases can be expressed as:

Ptotal = P1 +P2+P3 +………..+ Pn where Ptotal is the total pressure exerted by the mixture of gases that dot not react, P1, P2, P3……Pn are partial pressure of the individual gases.

Example:

If 20.0dm3 of hydrogen were collected over water at 17oC and 79.7kNm-2 pressure; Calculate the 

(a) Pressure of dry hydrogen at this temperature.

(b) Volume of dry hydrogen at s.t.p.

( vapour pressure of water is 1.90kNm-2 at 17OC)

Solution:

(a) PH2  =Ptotal – Pwater vapour

=  79.7 – 1.90

=  77.8kNm-2

(b) P1V1T1  =  P2V2T2

77.9X20290  =  101.3V2273

V2=  14.5dm3

 

GENERAL EVALUATION

OBJECTIVE TEST

1. A liquid begins to boil when (a) Its vapour pressure is equal to the vapour pressure of its solid at the given temperature (b) Molecules start escaping from its surface (c) Its vapour pressure equals the atmosphere pressure (d) Its volume is slightly increased.
2. Hydrogen diffuses through a porous plug (a) At the same rate as oxygen (b) Twice as fast as oxygen (c) Three times as fast as oxygen. (d) Four times as fast as oxygen
3. When pollen grains are suspended in water and viewed through a microscope, they appear to be in a state of constant but erratic motion. This is due to: (a) Convention current (b) small change in temperature (c) a chemical reaction between the pollen grains and the water (d) the bombardment of the pollen grain by molecules of water.
4. If the quantity of oxygen occupying a 2.76litre container at pressure of 0.825 atmosphere and 300k is reduced by one-half, what is the pressure exerted by the remaining gas? (a) 1.650atm (b) 0.825atm (c) 0.413atm (d) 0.275atm
5. 200cm3of oxygen diffused through a porous plug in 50secs. How long will 80cm3 of methane (CH4) take to diffuse through the same porous plug under the same conditions (C= 12, O= 16, H=1) (a) 40sec  (b) 20sec  (c) 14sec (d) 7sec

ESSAY QUESTIONS

1(a) State Graham’s law of diffusion

Arrange the following gases in decreasing order of diffusion rate: Chlorine, hydrogen chloride, hydrogen sulphide and Carbon (IV) oxide

[H=1, C= 12, O=16, S= 32, Cl=35.5]

2. (a)  What do you understand by s.t.p?

(b) If the volume of a given mass of gas at 298k and pressure of 205.2 103 Nm-2 is 2.12dm3, what is the volume at S.T.P? Standard pressure= 101.3103 Nm. Standard temperature= 273k

3. (a) Calculate the number of moles of the following at s.t.p
4. 16g of oxygen
5. 67.2dm3 of nitrogen gas, and

iii. 1.14dm3 of hydrogen chloride gas.

O=16, H=14, N=1. Molar volume of gas at S.T.P = 22.4dm3

(b) (i) Convert 33℃ and -41℃ to Kelvin scale 

(ii) Convert 270k and 315k to 0℃

 

WEEKEND ACTIVITY: 

List all the separation techniques that you know.