GAS LAWS

 

Week:   FIVE                         Date:                                                   Time:

Period:                                   Duration:                                           Average age of learners:

Subject:                               CHEMISTRY                                                         Class:  SS ONE

Topic:                                                    KINETIC THEORY OF MATTER AND GAS LAW

Sub – Topic: Gas laws

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA[ 4^th Edition]  pg58-68

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: stone,  and water

Entry behavior: The students have been familiar with state of matter

Behavioural objective: At the end of the lesson the students should be able to:

1^st PERIOD

State Boyle’s law and used kinetic theory of gas to explain it

Solved mathematical problem on Boyle’s law

2^nd  PERIOD

State Charle’s law and used kinetic theory of gas to explain it

Solved mathematical problem on Charle’s law

 3^rd PERIOD

Explain general and ideal gas equation

Solved mathematical problem on general gas law

BOYLE’S LAW

1^st PERIOD

STEP I: The teacher States Boyle’s law and used kinetic theory of gas to explain it

Boyle's law developed by Robert Boyle in 1662, states that if we keep the temperature of a gas constant in a sealed container. Its pressure (P) varies inversely with its volume (V). In other words at any given temperature if we pressurize a gas its volume will be reduced proportionately to the pressure change. If we increase the volume of a gas its pressure will increase. 
"For a fixed mass of ideal gas at fixed temperature the product of pressure and volume is constant".
Boyle's Law Formula is expressed symbolically as

Another way to express Boyle's Law Formula is

Where:  P is pressure of the gas  ; V is volume of the gas ; k is a constant, and has units of force times distance

Explanation of Boyle’s Law by the Kinetic Theory

Boyle’s law gives an inverse relationship between the volume of a fixed mass of gas and its pressure at constant temperature. I.e., V α 1/P. The kinetic theory explains the law as follows: From the assumptions of the kinetic theory of gases, the pressure exerted by a gas is due to the collisions between the gas molecules and the walls of the vessel. Hence, the greater the rate of collision, the greater the pressure.

At lower volumes, molecules are closer to one another, thus they collide more frequently with the walls of the container, leading to increase in pressure. The reverse would occur if volumes were increased. Therefore, the relationship between volumes and pressures of a gas at constant temperature is an inverse one as given by Boyle’s law.

STEP II: Solved mathematical problem on Boyle’s law

Solved Examples

Question 1: A sample of gaseous nitrogen in a 65.0 cm³ automobile air bag has a pressure of 745 mm Hg. If this sample is transferred to a 25.0 cm³  bag at the same temperature. what is the pressure of the gas in the 25.0 cm³   bag?
Solution:

It is often useful to make a table of the information provided.

Initial conditions	Final conditions
P1 = 745 mm Hg	P2 = ?
V1 = 65.0 cm3	V2 = 25.0 cm3

We know that P₁V₁ = P₂V₂
Therefore,
745 x 65 = P2 x 25;  P2 = 48425/25 = 1937
2.A sealed syringe contains 10 x10^-6 m3 of air at 1 × 10⁵ Pa. The plunger is pushed until the volume of trapped air is 4 x 10^-6 m³. If there is no change in temperature what is the new pressure of the gas?

Solution

P1 = 1x10⁵; V1= 10 x 10⁶;  V2= 4x 10⁶; P2 = ?

P1V1 = P2V2 ;  1x10⁵  x 10x10⁶ = P2 x 4x10⁶ ; P2 = 10x10¹¹/ 4x10⁶ = 2.5x10⁵

The new pressure in the syringe is 2.5 × 10⁵ Pa

2^nd  PERIOD

CHARLE’S LAW

STEP I: The teacher States Charle’s law and used kinetic theory of gas to explain it

Charles’s law states that if a given quantity of gas is held at a constant pressure, its volume is directly proportional to the absolute temperature. 

 

Explanation of Charles’ Law by the Kinetic Theory

The kinetic theory of gases explain Charles’ law thus:

I. Gases, due to their molecules been very far apart, do not have appreciable volumes, but occupy the volume of the vessel in which they are kept - the greater or higher their molecules are able to move in the vessel, the more volume they occupy, and vice versa.

II. Gaseous molecules are able to move or expand because they possess kinetic energy. Their average kinetic energy is directly proportional to the absolute temperature. I.e., the higher the temperature, the greater the average kinetic energy, and the more volume they occupy, and vice versa. In general, we can sum-up the explanation this way: since the volume which gas molecules occupy is directly dependent upon their movement, which in turn is directly dependent upon their kinetic energy, and which in turn is directly dependent on the absolute temperature, it goes to prove that the volume is directly proportional to its absolute temperature at constant pressure – Charles’ law (V α T).

STEP II: The teacher leads the students to Solved mathematical problem on Charle’s law

1. Calculate the final volume of a gas that occupies 400 cm³ at 20°C and is subsequently heated to 300°C. Begin by converting both temperatures to the absolute scale:

T ₁ = 20°C = 293.15 K

T ₂ = 300°C = 573.15 K

Then substitute them into the constant ratio of Charles' law:

  

2.) At 27.00 °C a gas has a volume of 6.00 dm³. What will the volume be at 150.0 °C

Answer: The Initial Volume is V₁ = 6.00 dm³. The Initial Temperature is T₁ = 27.00 + 273 = 300 K. The Final Temperature is T₂ = 150.0 + 273 = 423 K. The Final Volume (V₂) is what we are trying to find in the problem.

Plug into the Charles' Law Equation

  ; ;     V₂ = 8.46 

The volume of the container would be 8.46 dm³ .

GENERAL GAS LAW AND THE IDEAL GAS

3^rd PERIOD

Explain general and ideal gas equation

Solved mathematical problem on general gas law

The ideal gas law is the most important gas law for you to know: it combines all of the laws you learned about in this chapter thus far, under a set of standard conditions. The four conditions used to describe a gas—pressure, volume, temperature, and number of moles (quantity)—are all related, along with R, the universal gas law constant, in the following formula:

PV = nRT

where P = pressure (atm), V = volume (L), n = number of moles (mol), R = 0.08206 L · atm/mol · K, and T = temperature (K).

 

 

 

GAY LUSSAC’S LAW OF COMBINING VOLUMES 

Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

The law explains experimental facts about how gaseous atoms combine. Example:

For the reactions:

(i) N_2(g) + 3H_2(g) → 2NH_3(g)

   1 vol.    3 vols.    2 vols.

1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia.

(ii) 2H_2(g) + O_2(g) → 2H₂O_(g)

     2 vols.   1 vol.     2 vols.

2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of steam.

(iii) Cl_2(g) + H_2(g) → 2HCl_(g)

     1 vol.    1 vol.    2 vols.

1 volume of chlorine gas combines with 1 volume of hydrogen to form 2 volumes of hydrochloric acid.

Question: Consider the reaction: 2H_2(g) + O_2(g) → 2H₂O_(g)

(a). What volume of steam is formed from 20 cm³ of hydrogen and 20 cm³ of oxygen mixed together?

(b). What gas(s) is in excess, and by what amount?
 Solution: (a). The ratio of their volumes is 2 vols. : 1 vol. → 2 vols.

20 vols. : 10 vols.  → 20 vols.

That means, 20 cm³ of hydrogen will combine with 10 cm³ of oxygen to form 20 cm³ of steam.

(b). Oxygen is in excess by 10 cm³.

Daltons Law of Partial Pressure

The law states that in a mixture of gases which do not react chemically, the total pressure exerted by the mixture is the sum of partial pressures of the individual gasses that make up the mixture.

P_T = P_A + P_B + P_C ;   P_T = Total pressure of the mixture; P_A, P_B, and P_C = Pressure of gases A, B and C that make up the mixture.

A gas is likely to be saturated with water vapour if it is collected over water and the total pressure becomes.

P_T - P _gas + P _{water vapour}

The correct pressure of the dry gas is obtained by subtracting the vapour pressure from the total pressure.

Calculation Based on Dalton's Law of Partial Pressure

2 moles of oxygen gas, 3 moles of nitrogen gas and 1 mole of carbon(IV) oxide gas are put in a gas balloon whose pressure is maintained at 3.019 x 10⁵ N M^-2 at 28 ^oC. What is the partial pressure of nitrogen?

Solution:

  

Explanation of Graham’s Law of Diffusion by the Kinetic Theory

Graham’s law of diffusion can be explained from the assumption which states that the average kinetic energy of gas molecules is directly proportional to the absolute temperature - molecules of different gases at the same temperature have the same kinetic energy. This means that molecules of one kind, A, must have velocities (or rate of diffusion) that are different from velocities (or rate of diffusion) of molecules of another gas, B, unless they have the same masses.

Mathematically, if  ½ m_Au²_A = ½ m_Bu²_B and u²_A/u²_B = m_B/m_A

So that u_A/u_B = √m_B/m_A or R_A/R_B = √m_B/m_A

 For example, a methane molecule (molecular weight, 16), A, has a mass one fourth as great as the mass of a sulphur(IV) oxide molecule (molecular weight, 64), B. If m_B/m_A = 4  ; u_A/u_B = √4 = 2  or u_A = 2 x u_B

The methane molecules have an average velocity twice that of the sulphur(IV) oxide molecules. Thus, the kinetic theory shows why methane diffuses twice as fast as sulphur(IV) oxide, and why , in general, different gases diffuse at rates that vary inversely with the square roots of their molecular weights or densities. 

GRAHAM’S LAW OF DIFFUSION

Graham’s law states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure:

 30 cm³ of a gas, the empirical formula of which was CH₃, diffuses through a porous partition in 45.2 s. 30 cm³ of hydrogen diffused in 11.7 s under the same conditions. Calculate (i) the vapour density of the gas (ii) the molecular formula of the gas?

Solution:

(i). The two gases are of equal volumes,

t_x/t_H = √m_x /√m_H  ; 45.2/11.7 = √(m_x/2) ;  2(45.2/11.7)² = m_x ; m_x = 2 x 14.92 = 29.84 (g)

r.m.m. = 2 x v.d.   ; v.d. = r.m.m./2  ;v.d. = 29.84/2 = 14.92

(ii). xCH₃ = 30;  15x = 30;  x = 2

Therefore, molecular formula = C₂H₆

Explanation of Avogadro’s Law by the Kinetic Theory

Avogadro’s law can be explained from the assumptions:

1. The average kinetic energy of gas molecules is proportional to the absolute temperature. Hence, at the same temperature, the average kinetic energy of two different gases are the same.

2. The pressure exerted by a gas confined within a fixed volume is proportional to nE_k , i.e., the number of molecules per unit volume times their average kinetic energy. Therefore, if two gases are at the same temperature, then E_k (the average kinetic energy per molecule) is the same for both gases, and if the two gases are also exerting the same pressure, then nE_k is the same for the two gases.

Thus, if two gases are at the same temperature and pressure, n must be the same for both. The number of molecules per unit volume is the same - i.e., equal volumes of different gases contain the same number of molecules if they are at the same temperature and pressure.

AVAGADRO'S LAW

The law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. It explains then 10cm³ of N2 ; 10cm3 of O2 and 10cm3 of H2

Must all contain the same number of molecules at the same temperature and pressure.

Mole and Molar Volume

1 mole is the atomic mass or molecular mass of a substance expressed in grammes. The unit for the mole is the gram. The molar mass of oxygen molecules is obtained thus:

O + O        O2      (16+ 16 = 32). Therefore one mole of oxygen gas weighs 32g.

One mole of Hydrogen, H2 weighs

H+H          H2        (1.008 + 1.008 = 2.016g)

One mole of sodium atom, Na weighs 23g ; One mole of oxygen atom weighs 16g

The number of particles in one mole of a substance is 6.02 x 10²³ .This number is called Avogadro's number or constant.

Example

How many moles are there in 54g of Aluminum (Al = 27)3

Solution

Mole =  =  = 2 mol

Example

Convert 250 grammes of CaCO3 to mole (Ca = 4O, C = 12, O = 16)

Solution

CaCO3 = 40+12+(16x3) = 40+12+48 =100 ;   250 /100= 2.5 moles

To convert mole to gram

Multiply the mole by the atomic or the molecular mass.

Example

Convert 0.5 mole of Al to gram (Al = 27)

Solution

0.5 x 27 = 13.5 = 13. 5g

Molar Volume is the volume occupied by one mole of a gas at S.T.P. and numerical equal to 22.4 dm3

One mole of these gases (O2, H2, N2, CO2, CO etc) occupy 22.4 dm3 at S.T.P

Question 7

Calculate the volume of carbon (iv) oxide produced at S.T.P when 50g of calcimtrioxocarbonate (iv) is completely decomposed by heat.

CaCO3                    CaO + CO2

40+12+ (16x3)

4O+12+48

100g                                     22.4 dm3

50g                                      22.4 x 50 / 100 = 11.2dm3