# Lesson Notes By Weeks and Term - Senior Secondary 1

Gas laws II

Term: 2nd Term

Week: 7

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject:      Chemistry

Topic:-       Gas laws II

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

1. State Gay-lussac’s law
3. State the ideal gas equation

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

 PRESENTATION TEACHER’S ACTIVITY STUDENT’S ACTIVITY STEP 1 INTRODUCTION The teacher reviews the previous lesson on Boyle’s aw, Charles’ law and the general gas laws Students pay attention STEP 2 EXPLANATION He states and explains Gay-Lussac and Avogadro’s law, stating their formulae repetitively. Students pay attention and participates STEP 3 DEMONSTRATION He outlines the ideal gas equations and solves problems on all the gas laws learnt Students pay attention and participate STEP 4 NOTE TAKING The teacher writes a summarized note on the board The students copy the note in their books

NOTE

GAS LAWS
GAY LUSSAC’S LAW OF COMBINING VOLUME.

Gay-Lussac’s law is a gas law which states that the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas. In other words, the pressure exerted by a gas is proportional to the temperature of the gas when the mass is fixed and the volume is constant.

This law was formulated by the French chemist Joseph Gay-Lussac in the year 1808. The mathematical expression of Gay-Lussac’s law can be written as follows:

P T ; P/T = k

Where:

• P is the pressure exerted by the gas
• T is the absolute temperature of the gas
• k is a constant.

The relationship between the pressure and absolute temperature of a given mass of gas (at constant volume) can be illustrated graphically as follows.

Formula and Derivation

Gay-Lussac’s law implies that the ratio of the initial pressure and temperature is equal to the ratio of the final pressure and temperature for a gas of a fixed mass kept at a constant volume. This formula can be expressed as follows:

(P1/T1) = (P2/T2)

Where:

• P1 is the initial pressure
• T1 is the initial temperature
• P2 is the final pressure
• T2 is the final temperature

This expression can be derived from the pressure-temperature proportionality for gas. Since P ∝ T for gases of fixed mass kept at constant volume:

P1/T1 = k (initial pressure/ initial temperature = constant)

P2/T2 = k (final pressure/ final temperature = constant)

Therefore, P1/T1 = P2/T2 = k

Or, P1T2 = P2T1

Solved Exercises on Gay-Lussac’s Law

Exercise 1

The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the initial temperature of the gas if its initial pressure was 1 atm?

Solution

Given,

Initial pressure, P1 = 1 atm

Final pressure, P2 = 1.5 atm

Final temperature, T2 = 250 K

As per Gay-Lussac’s Law, P1T2 = P2T1

Therefore, T1 = (P1T2)/P2 = (1*250)/(1.5) = 166.66 Kelvin.

Exercise 2

At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K.

Solution

Initial pressure, P1 = 3 atm

Initial temperature, T1 = 300K

Final temperature, T2 = 900 K

Therefore, final pressure (P2) = (P1T2)/T1 = (3 atm*900K)/300K = 9 atm.

It states that equal volume of all gases at the same temperature and pressure contain the same number of molecules irrespective of their physical and chemical properties

IDEAL GAS EQUATIONS

The ideal gas law is derived from the observational work of Robert Boyle, Gay-Lussac and Amedeo Avogadro. Combining their observations into a single expression, we arrive at the Ideal gas equation, which describes all the relationships simultaneously.

The three individual expressions are as follows:

Boyle’s Law

V∝1/P

Charles’s Law

V∝T

V∝n

Combining these three expressions, we get

V∝nT/P

The above equation shows that volume is proportional to the number of moles and the temperature while inversely proportional to the pressure.

This expression can be rewritten as follows:

V=RnT=nRT

P       P

Multiplying both sides of the equation by P to clear off the fraction, we get

PV=nRT

The above equation is known as the ideal gas equation.

Ideal Gas Solved Problems

1. What is the volume occupied by 2.34 grams of carbon dioxide gas at STP?
Solution:
To determine the volume, rearrange the ideal gas law as follows:

V=nRT/P

Substituting the values as follows, we get

V=[(2.34g/44g.mol1)(0.08206L atm mol1 K1) (273.0K)]/1.00atm

V=1.19L

1. Calculate the temperature at which 0.654 moles of neon gas occupies 12.30 litres at 1.95 atmospheres.

Solution:
To determine the temperature, rearrange the ideal gas equation as follows:

T=PV/nR

Substituting the value, we get

T=(1.95atm))(12.30L)/(0.654mol)(0.08206L atm mol1 K1)

T=447K

EVALUATION:    1. State Gay-Lussac’s law

2. 3. A sealed jar whose volume is exactly 1 liter contains exactly 1 mole of air at 20.00 °C. Assuming the air behaves as an ideal gas, what is the pressure inside the jar in pascals?
3. A car tire's pressure is measured to be 200,000 Pa; its volume is 10.00 liters; and its temperature is 30.00 °C.

I) How many moles of gas are in the tire?

II) Assuming air is 80% nitrogen and 20% oxygen, what mass of air is in the tire?

5. A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.

6. A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively