# Gas Laws II – Avogadro’s Law

Welcome to class!

In today’s class, we will be talking more about gas laws – Avogadro’s law. Enjoy the class!

### GAS LAWS II – AVOGADRO’S LAW

This law states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. This means that 1 mole of any gas at s.t.p has a volume of 22.4dm3.

#### GAY LUSSAC’S LAW OF COMBINING VOLUMES

It states that when gases react they do so in volumes which are simple ratios to one another and the volumes of the products if gaseous, provided that the temperature and pressure remain constant.

##### CALCULATION ON THE LAW

Calculate the volume of oxygen required to burn 500cm3 of methane completely.

Solution:

The equation for the reaction is:

2CH4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

By Gay Lussac’s law,

2 volumes of CH4 requires 3 volumes of O2 for complete combustion

Therefore, 2cm3 of CH4 requires 3cm3 of O2

500cm3 of CH4 will require Xcm3 of O2

Xcm3 of O2 = 500cm3 x 3cm3 ) / 2cm3 = 750cm3

Thus, 750cm3 of O2 is required.

EVALUATION

1. State the Gay Lussac’s law of combining volumes
2. 40cm3 of hydrogen was sparked with 160cm3 of oxygen at 100oC and 1atm. Determine the volume of oxygen left after the reaction.

#### GRAHAM’S LAW OF DIFFUSION

It states that the rate of diffusion of a gas is inversely proportional to the square root of its density at constant temperature and pressure.

Mathematically,

R α 1/√d

R = k/√d where k is a constant

Comparing the rate of diffusion of two gases:

R1 = √d2
R2   √d1

In terms of relative molecular mass, M

R α 1/√M

For two gases,

R1 = √M2
R2    √M1

But the rate of diffusion is reciprocal of time, R =1/t

That is,

R1 = t2
R2    t1

From the inverse relationship, we can deduce that the less dense a gas is, the higher the rate of diffusion and vice versa.

##### CALCULATION
1. A given volume of SO2 diffuses in 60 seconds. How long will it take the same volume of  CH4 to diffuse under the same condition (SO2 = 64, CH4 = 16)

Solution:

Using the expression:

t1= √M2
t2   √M1

t2 = √M2 x t1 = √16 x 60seconds = 30seconds
√M1               √64

###### GENERAL EVALUATION/REVISION
1. State Graham’s law of diffusion
2. Under the same condition of temperature and pressure, hydrogen diffuses 8 times as fast as gas Y. Calculate the relative molecular mass of Y.
3. State the following rule/principle: (a) Hund’s rule of maximum multiplicity (b) Aufbau principle
4. Write the electronic configuration of (a) oxide ion, (b) Aluminium ion, (c) potassium (d) phosphorus.

New School Chemistry for Senior Secondary School by O.Y. Ababio, Pg 86-92

###### WEEKEND ASSIGNMENT
1. 400cm3 of a gas X diffuses through a porous pot in 2 minutes. Calculate the rate at which X diffuses. A. 6.3cm3s-1 20cm3s-1 C. 200cm3s-1D. 3.33cm3s-1#
2. The relationship between the density (d) of gas and the rate at which the gas diffuses is A. R = kd
3. R= k/√d C. R = k√d D. k/d*
4. Calculate the minimum volume of oxygen required for the complete combustion of a mixture of 20cm3 CO and 20cm3 of H2. A. 10cm3 20cm3C. 40cm3D. 60cm3
5. If sulphur (iv) oxide and methane (CH4) are released at the same time at opposite ends of a tube, the rate of diffusion will be in the ratio A. 2:1 B. 4:1 C. 1:4 D. 1:2
6. ‘Equal volume of all gases at the same temperature and pressure contain the same number of molecules’ is a state of which law A. Avogadro’s law B. Boyle’s law Charles’ law      D. Chemical law

THEORY

1. Arrange the following gases in order of increasing rate of diffusion: CO, SO2, H2S, NO2 and O2.
2. The vapour densities of O2 and Cl2 are 16 and 36 respectively. If 60cm3 of O2 diffuses through a porous partition in 14 seconds, how long will it take 1000cm3 of Cl2 to diffuse through the same partition?

In our next class, we will be talking about Identification and Types of Alloys.  We hope you enjoyed the class.

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