Gas Laws II – Avogadro’s Law

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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:

2CH_4(g) + 3O_2(g) → 2CO_2(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 100^oC 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.

READING ASSIGNMENT

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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