Gas Laws – Boyle’s Law

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Welcome to class! 

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

GAS LAWS – BOYLE’S LAW

It states that the volume of a fixed mass of gas is inversely proportional to the pressure provided the temperature remains constant.

Mathematically,

V α 1/P

V = k/P

PV = k

Hence,               P₁V₁ = P₂V₂

Boyle’s law can be represented graphically as shown below.

 

The graph shows that if the pressure is doubled, the volume is reduced to half its former value and if it is halved, the volume is doubled.

 

 

EXPLANATION OF BOYLE’S LAW USING THE KINETIC THEORY

When the volume of a fixed mass of gas is decreased, the molecules of the gas will collide with each other more rapidly. This gives rise to an increase in pressure. However, if molecules are farther apart the number of collisions for unit time decreases, resulting in a decrease in pressure.

CHARLES’ LAW

Charles’ law states that the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in the Kelvin scale.

Mathematically,

V α T

V = k/T

V = kT

Hence,           V₁ / T₁=  V₂ / T₂   

The graphical representation of Charles’ law is as shown below:

EXPLANATION OF CHARLES’ LAW USING THE KINETIC THEORY

When a given gas is heated at constant pressure, the molecules acquire more kinetic energy and move faster. They collide with one another and with the walls of the container more frequently. To maintain the same number of collisions on the walls of the container (i.e. keep the pressure constant) the volume of the gas increases.

CALCULATIONS BASED ON BOYLE’S AND CHARLES’ LAW

1. 200cm3 of a gas has a pressure of 510mmHg. What will be its volume if the pressure is increased to 780mmHg, assuming there is no change in temperature?

Solution:

V₁ = 200cm³, P₁ = 510mmHg, P₂ = 780mmHg V₂ = ?

Using the expression for Boyle’s law:

P₁V₁ = P₂V₂

V₂ = (P₁V₁ ) / P₂ = (510mmHg x 200cm³ ) / 780mmHg= 130.769 = 131 cm³

2. A certain mass of a gas occupies 300cm³ at 35^oC. At what temperature will it have its volume reduced by half assuming its pressure remains constant?

Solution:

V₁ = 300cm³, T₁ = 35^oC = (35 + 273)K = 308K, V₂ = V₁/2 = 300/2 = 150cm³, T₂ = ?

Using the formula for Charles’ law

V₁ = V₂
T₁    T₂

T₂ = (V₂T₁ ) / V₁ = 150cm³ x 308K ) / 300cm³= 154K

EVALUATION

1. State Boyle’s law
2. Explain Charles’ law using the kinetic theory

GENERAL GAS EQUATION

Boyle’s and Charles’ laws are combined into a single expression known as the general gas equation which can be expressed mathematically as

(P₁V₁ ) / T1= (P₂V2) / T₂

IDEAL GAS EQUATION

This equation states that for an ideal gas PV/T is a constant.

That is, (PV) / T = R    (R = molar gas constant)

PV = RT

That is, for n mole of a gas, the equation becomes

PV = nRT

CALCULATIONS

1. What is the volume at s.t.p of a fixed mass of a gas that occupies 700cm³ at 25^oC and 0.84 x 10⁵ Nm^-2pressure?

Solution:

T₁ = 273K, P₁ = 1.01 x 10⁵Nm^-2, T₂ = 25^oC = (25 + 273) = 298K, P₂ = 0.84 x 10⁵Nm^-2,

V₂ = 700cm³, V₁ =?

Using the general gas equation

P₁V₁ = P₂V₂
   T₁        T₂

V₁ = P₂V₂T₁ = 0.84 x 10⁵Nm^-2 x 700cm³ x 273K = 533.337 =533cm^{3
.              P₁T₂                                1.01 x 105Nm-2 x 298K}

2. Calculate the number of moles present in a certain mass of gas occupying 6.5dm³ at     3atm and 15^oC (R = 0.082atmdm³K^-1mol^-1)

Solution:

V = 6.5dm3, P = 3atm, T = 15^oC = (15 + 273)K = 288K, n =?

Using PV = nRT

n = PV            =           3atm x 6.5dm3                     = 0.8257
.      RT                  0.082atmdm3K-1mol-1 x 288K

Number of moles = 0.83 mole

 

DALTON’S LAW OF PARTIAL PRESSURE

This law state that in a mixture of gases which do not react chemically together, the total pressure exerted by the mixture of gases is equal to the sum of the partial pressure of the individual gases that make up the mixture.

Mathematically, the law can be expressed as:

P_total = P_A + P_B +P_C.…….P_n

Where P_total is the total pressure of the mixture and P_A, P_B, P_C is the partial pressure exerted separately by the individual gases A, B, C that make up the mixture.

The pressure each constituent gas exerts is called partial pressure and is expressed as

The partial pressure of gas A (P_A) =         number of moles of gas A   XP_{total
.                                                                            Total number of moles of gas in the mixture}

That is, P_A = n_A x P_total

n_A + n_B + n_C

If the gas is collected over water, it is likely to be saturated with water vapour and the total pressure becomes

P_total = P_gas + P_{water vapour}

P_gas = P_total – P_{water vapour}

CALCULATION ON THE LAW

A gaseous mixture containing 64g of O₂ and 70g of N₂ exerts a total pressure of 1.8oatm. What is the partial pressure exerted by oxygen in the mixture?

Solution:

Molar mass of O₂ = 16 x 2 = 32gmol^-1

Molar mass of N₂ = 14 x 2 = 28gmol^-1

Number of mole of O₂ = 64g    = 2.0mole
.                                  32gmol-1

Number of mole of O₂ = 70g      = 2.5mole
.                                    28gmol-1

Total number of moles of gases in mixture = 2.0 + 2.5 = 4.5 mole

Partial pressure of O₂ = 2.0 x 1.80 = 0.80atm
.                                      4.5

GENERAL EVALUATION/REVISION

1. State Dalton’s of partial pressure.
2. Calculate the pressure at 27^oC of 16.0g O₂ gas occupying 2.50dm³
3. A certain mass of hydrogen gas collected over water at 10^oC and 760mmHg pressure has a volume of 37cm³. Calculate the volume when it is dry at s.t.p (Saturated vapour     pressure of water at 10^oC =1.2mmHg)
4. Determine the number of electrons, protons and neutrons in each of the following: ³⁹K_19,       ⁵Cu₂₉.
5. If an element R has isotopes 60% of ¹²R₆ and 40% ^xR₆ and the relative atomic mass is      12.4, find x.

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y. Ababio, Pg 78-85.

WEEKEND ASSIGNMENT

1. Kelvin temperature can be converted into temperature by A. ^oC = K-273       K + 273 C.     ^oC + 273/K    D.      K + 273/^oC
2. The pressure exerted by a gas is a result of the A. continuous random motion of its particle.
3. bombardment of the walls of the container by its molecules. C. expansion of the gas molecules
4. collision between the gas molecules.
5. From the ideal gas equation, PV = nRT, the unit of n is A. atmdm³ atmdm³/K C. mole D. K/mole
6. What will be the new volume (V) if the new pressure is halved and the initial pressure remains the same A. 2P₁V₁ = P₂V₂ P₁V₁ = 2P₂V₂C. P₁V₁/2= P₂V₂/2 D. P₁V₁ = P₂V₂/2
7. A fixed mass of gas of volume 546cm3 at 0^oC is heated at constant pressure. What is the volume of the gas at 2^oC? A. 550cm³ 560cm³C. 570cm³D. 580cm³ 

THEORY

1. A given mass of nitrogen is 0.12dm³ at 60^oC and 1.01 x 10⁵Nm^-2. Find its pressure at the same temperature if its volume is changed to 0.24dm³; 272cm3 of CO2 was collected over water at 15^oC and 782mmHg pressure. Calculate the volume of dry gas at s.t.p (saturated vapour pressure of water at 15^oC is 12mmHg).

 

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

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