Circus Charlie: Charles’ Law!

AStrategic Intervention Material

in Chemistry

Presented by:

DON KING EVANGELISTACHRISTOPHER LOUIS ORQUIZA

TABLE OF CONTENTS

I. Cover page

II. Table of Contents

III. Guide Card

IV. Introduction

V. Activity Card1

VI. Activity Card 2

VII. Assessment Card

VIII. Enrichment Card

IX. Answer Card 1

X. Answer Card 2

XI. Reference Card

GUIDE CARD

Understanding the different concepts governing the behavior of gasses (i.e. Ideal Gas) has always been focused by chemistry as it plays a very important role in our society, in the industry and in the different fields of our lives.

In this Strategic Intervention Material, the student is subjected to a deeper understanding of Charles Law. After completing this SIM the learner is expected to:

State and Define the Charles Law. Recognize the key concepts regarding the behavior of ideal

gasses at constant pressure. Identify the applicability and limitations of Charles Law and its

association with other physical concepts (e.g. Ideal Gas Law, Kinetic Theory & Absolute Zero).

Solve practical problems involving Charles Law.

Now you are ready to learn! Let us have the basics of Charles Law!

INTRODUCTIONCharles' law (also known as the law of volumes) is an

experimental gas law which describes how gases tend to expand when heated. It was first published by French natural philosopher Joseph Louis Gay-Lussac in 1802, although he credited the discovery to unpublished work from the 1780s by Jacques Charles. The law was independently discovered by British natural philosopher John Dalton by 1801, although Dalton's description was less thorough than Gay-Lussac's.[2] The basic principles had already been described a century earlier by Guillaume Amontons.

Whatever the priority of the discovery, Gay-Lussac was the first to demonstrate that the law applied generally to all gases, and also to the vapours of volatile liquids if the temperature was more than a few degrees above the boiling point. His statement of the law can be expressed mathematically as: where V100 is the volume occupied by a given sample of gas at 100 °C; V0 is the volume occupied by the same sample of gas at 0 °C; and k is a constant which is the same for all gases at constant pressure. Gay-Lussac's value for k was 1⁄2.6666, remarkably close to the present-day value of 1⁄2.7315.

A modern statement of Charles's law is:

“At constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature on the absolute temperature scale (i.e. the gas expands as the temperature increases).”

which can be written as:

where V is the volume of the gas; and T is the absolute temperature. The law can also be usefully expressed as follows:

The equation shows that, as absolute temperature increases, the volume of the gas also increases in proportion.

ACTIVITY CARDActivity 1Read and analyze each item. Write your answers on the corresponding boxes. Complete the Cipher Code by filling in the number for the corresponding English Alphabet Letter. Decode the mystery phrase.

Silly CIPHER1.2.

3 19 8 16 20 24

3

23 26 174.

12 19 16 13 8 24 12

5.

22 12 24 26 8

At constant (1)_______, the (2)______ of a given mass of an ideal (3)_____ increases or decreases by the same factor as its temperature on the absolute temperature scale(i.e. the gas expands as the temperature increases).”

So at constant pressure, if the temperature (K) is doubled, the volume of gas is also (4)_______.

A hypothetical gas which obeys Charles' Law at all temperatures and pressures is called an (5)_____ gas.

CIPHER CODEA = F = 1

1K = 2

1P = U = Z = 1

B = G = L = Q = 18

V =

C = 25

H = 10

M = R = W = 15

D = I = N = 7 S = X = 2E = J = 9 O = T = 4 Y = 1

4

Decode me!

C H ‘ W25 10 26 5 8 24 17 8 26 15 22 17 2

68 1

719

K N W N T H L A F21 7 19 15 7 26 17 4 10 24 8 26 1 1 1

6524171716524

5 9 1

3 19 8 16 20 24 17

The Cipher Text says that ____________________________________.

ACTIVITY CARDActivity 2In the following items, assume that the gas is hold at constant pressure.

CROSS NUMBERA B

C D

E F G H

I

J K

L M

N

DownA. V1=20L, T1=10°C, V2=__L, T2=52°CB. The volume of a gas at a certain

temperature is 224L. If a drop of 12°C will reduce the volume by 24L, what is the original temperature?

D. V1=____mL, T1=10°C, V2=70L, T2=20°C

E. Find the increase in temperature of a gas whose original volume and temperature is 150L and 100°C respectively, if the new volume is equal to 412.5L?

F. If after the temperature is tripled the new volume of a gas is 1107L, find the original volume.

G. V1=46mL, T1=Φ°C, V2=____mL, T2=9Φ°C

H. V1=___mL, T1=10Φ°C, V2=10mL, T2=Φ°C

L. 147 An ideal gas has a volume of 100L under the temperature of 100°C. If the temperature grows by 47°C, what is the new volume?

M.V1=600L, T1=300°C, V2=__L, T2=315°C

AcrossA. V1=74mL, T1=____°C, V2=200mL,

T2=300°CC. The volume of a given mass of

gas, at 288K is 400 ml. At what temperature, will it occupy a volume of 600 ml?

I. What is the new volume of a gas after the temperature is tripled if the original volume is 21667mL?

J. The temperature is reduced by 2/3. If the original volume is 1737mL, what is the new volume?

K. The temperature is squared. If the original volume is 20, find the new volume.

L. The volume of a gas is 9L under a temperature of 90°C, if after applying heat the volume changes to 10.6L, find the new temperature?

N. V1=1080mL, T1=300°C, V2=__mL, T2=200°C

ASSESSMENT CARDActivity 1Read and analyze each Problem. Show your solution.

PROBLEM SOLVING

1. A sample of gas at 101.3kPa had a volume of 1.2L at 100oC. What would its volume be at 0oC at the same pressure?Vi = 1.2L                               Vf = ? Ti = 100oC = 100 + 273 = 373K   Tf = 0oC = 0 + 273 =273K

2. A balloon had a volume of 75L at 25oC. To what does the temperature need to raised in order for the balloon to have a volume of 100L at the same pressure?Vi = 75L                                 Vf = 100L Ti = 25oC = 25 + 273 = 298K     Tf = ? (K)

ENRICHMENT CARDCharles’ Law and its effect on Real Gasses

A Real Gas is one which approaches Charles' Law as the

temperature is raised or the pressure lowered.

As a Real Gas is cooled at constant pressure from a point

well above its condensation point, its volume begins to increase

linearly. As the temperature approaches the gases

condensation point, the line begins to curve (usually downward)

so there is a marked deviation from Ideal Gas behaviour close to

the condensation point. Once the gas condenses to a liquid it is

no longer a gas and so does not obey Charles' Law at all.

Absolute zero (0K, -273oC approximately) is the temperature at

which the volume of a gas would become zero if it did not

condense and if it behaved ideally down to that temperature.

ANSWER CARD1.2.

V

O L U M E

3 19 8 16 20 243

G A S23 26 17

4.D O U B L E D12 19 16 13 8 24 12

5.I D E A L

22 12 24 26 8

At constant (1)_______, the (2)______ of a given mass of an ideal (3)_____ increases or decreases by the same factor as its temperature on the absolute temperature scale(i.e. the gas expands as the temperature increases).”

So at constant pressure, if the temperature (K) is doubled, the volume of gas is also (4)_______.

A hypothetical gas which obeys Charles' Law at all temperatures and pressures is called an (5)_____ gas.

Activity 1

The Cipher Text states that Charles’ Law is also known as the Law of Volumes.

Activity 2

A1 1

B1

0 1C

4 D

3 2 E

1 F

3 5 G

4 H

1

7 I

6 5 0 0 1 0 J

5 7 9 0K

4 0 0 L

1 0 M

6

4 3 N

7 2 0

PRESSURE6524171716524

ANSWER CARDAssessment 1

a. A sample of gas at 101.3kPa had a volume of 1.2L at 100oC. What would its volume be at 0oC at the same pressure?

Vi = 1.2L                                   Vf = ?

Ti = 100oC = 100 + 273 = 373K     Tf = 0oC = 0 + 273 =273K

1.2/373 = Vf/273

3.22 x 10-3 = Vf/273

Vf = 3.22 x 10-3 x 273 = 0.88L (880mL)

b. A balloon had a volume of 75L at 25oC. To what does the temperature need to raised in order for the balloon to have a volume of 100L at the same pressure?

Vi = 75L                                 Vf = 100L

Ti = 25oC = 25 + 273 = 298K     Tf = ? (K)

Vi/Ti = V

f/Tf

75/298 = 100/Tf

0.2517 = 100/Tf

Tf = 100/0.2517 = 397K (397-273 = 124oC)

REFERENCE CARD

1. Gay-Lussac, J. L. (L'An X – 1802), "Recherches sur la dilatation des gaz et des vapeurs", Annales de chimie XLIII: 137. English translation.

2. http://www.chemistryexplained.com/Fe-Ge/Gay-Lussac-Joseph- Louis.html

3. Fullick, P. (1994), Physics, Heinemann, pp. 141–42, ISBN 0435570781.

4. Clapeyron, E. (1834), "Mémoire sur la puissance motrice de la chaleur", Journal de l'École Polytechnique XIV: 153–90. Facsimile at the Bibliothèque nationale de France (pp. 153–90).

5. Thomson, William (1848), "On an Absolute Thermometric Scale founded on Carnot's Theory of the Motive Power of Heat, and calculated from Regnault's Observations", Philosophical Magazine: 100–6, http://zapatopi.net/kelvin/papers/on_an_absolute_thermometric_scale.html.

6. Thomson, William (1852), "On the Dynamical Theory of Heat, with numerical results deduced from Mr Joule's equivalent of a Thermal Unit, and M. Regnault's Observations on Steam", Philosophical Magazine 4.

7. http://www.ausetute.com.au/charslaw.html