Worked Examples on Gas Laws and Kinetic Theory

WORKED EXAMPLES

ON

GAS LAWS & KINETIC

THEORY

by

Shefiu S. Zakariyah, PhD

i

PREFACE

In an effort to facilitate learning and teaching of engineering and physical sciences to

potential engineers and scientists amongst others, what follows is a series of solutions to

questions (original and modified) found in standard textbooks in the aforementioned

fields of study.

This booklet presents 45 worked examples on gas laws and kinetic theory, which is

preceded by an introduction to the theory, laws and formulae associated with the topic.

Questions used in this work are drawn from physics and chemistry textbooks designed

for A-level, foundation year and college students or what can be considered as

equivalent. It is however anticipated that students in the early years of degrees in

engineering and related fields will also find this helpful especially if kinetic theory

forms part of an elective module. Additionally, it could be adopted by teaching staff as

a reference guide during classes.

Since this work is part of a series of ‘learn by examples’ undertaken by the author,

pertinent suggestions, feedbacks and queries are highly welcome. This can be directed

to the author at the address below. Coming soon in this series are:

Worked Examples on Mechanics

Worked Example on Complex Number

Worked Examples on Calculus

Worked Examples on Circuit Theorems

And, of course, many more.

Finally, many thanks to my colleagues who have offered help and/or suggestions,

especially Ismail K. Adeboye (Advanced Technovation Ltd), Khadijah O. Olaniyan

(Loughborough University), Abdul Lateef Balogun (Universiti Teknologi PETRONAS &

Advanced Technovation Ltd), Misbahu Ladan Mohammed (London South Bank

ii. i

i

University), K. F. Tamrin (University of Malaya & Advanced Technovation Ltd),

Luqman O. Onadiran (Federal University of Technology Akure), G. A. Ibraheem, Juan

Pablo Casadiego Gonzalez and Nsajigwa Emmanuel Mwangosi.

© Shefiu S. Zakariyah 2013

Email: [email protected]

mailto:[email protected]

iii

Disclaimer

Every effort has been made by the author in ensuring the accuracy of the information,

including questions and associated solutions, presented in this booklet. The author does

not assume and hereby disclaims any liability to any party for any loss, damage, or

disruption caused by errors or omissions, either accidently or otherwise.

iv

CONTENTS

PREFACE ............................................................................................................................................................. I

DISCLAIMER ...................................................................................................................................................... III

CONTENTS ........................................................................................................................................................ IV

INTRODUCTION TO GAS LAWS ........................................................................................................................... 1

WORKED EXAMPLES .......................................................................................................................................... 9

SECTION 1: BOYLE’S LAW ......................................................................................................................................... 9

SECTION 2: CHARLES’ LAW .................................................................................................................................... 12

SECTION 3: GAY-LUSSAC’S LAW ............................................................................................................................ 15

SECTION 4: COMBINED GAS LAW ........................................................................................................................... 17

SECTION 5: IDEAL GAS EQUATION .......................................................................................................................... 26

SECTION 6: ADDITIONAL EXAMPLES ...................................................................................................................... 32

BIBLIOGRAPHY AND FURTHER READING ............................................................................................................ 37

1

INTRODUCTION TO GAS LAWS

1. Gas Laws

1.1. Boyle’s law

This law states that the volume ( ) occupied by a fixed mass of gas is inversely

proportional to the pressure ( ) provided that the temperature remains constant. Since

the temperature is constant (or no heat flow) in this process (Boyle’s law), it is referred

to as isothermal condition. Mathematically, Boyle’s law can be written as:

Or

From Boyle’s law above and for the purpose of calculation it can be stated that if and

are the initial pressure and volume respectively and and are the final pressure

and volume respectively then we can write the law as:

Note: Boyle’s law is not always true for real gases since at high pressures, a real gas can

condense into liquid due to the inter-molecular force of attraction in the gas molecules.

1.2. Charles’ law

This law states that the volume of a given mass of gas at constant pressure is directly

proportional to its temperature ( ) in Kelvin. If the volume is fixed, i.e. the gas is not

allowed to expand, then Charles’ law can no longer hold or be used. This can be written

mathematically as:

Or

As with Boyle’s law, Charles’ law can also be expressed as:

2

1.3. Gay-Lussac’s law

This law states that the pressure of a fixed mass of gas at constant volume is directly

proportional to its absolute temperature. In other words,

Or

Also,

1.4. Avogadro’s law

This law states that at a fixed temperature and pressure, the volume of gas is directly

proportional to the number of moles (or molecules), . This can be expressed as

Or

As with others, we can have

Avogadro’s law implies that at the same conditions of temperature and pressure, equal

volumes of all gasses contain equal numbers of molecules.

1.5. Combined gas law

As the name suggests, it combines two or more laws. Generally, it is considered as a combination of Boyle’s and Charles’ laws (so it is Boyle’s-Charles’ gas law) or Boyle’s, Charles’ and Gay-Lussac’s laws (so it is Boyle’s - Charles’ - Gay-Lussac’s gas law).

3

Mathematically, the combined gas law can be expressed as:

And then

The above expression of the combined gas law is because the quantity of the gas (moles

or mass) involved remains constant from initiation to its final state. However, it is

possible that the amount of the sample changes, hence the number of mole (n) is

included. We can therefore write a more encompassing expression of combined gas law

as:

The last expression can be regarded as Boyle’s - Charles’ - Gay-Lussac’s –Avogadro’s gas law since it is a combination of the four laws. It is important to point out here that it is possible to obtain all the previous four laws from the last formula. For example, Boyle’s law can be obtained if the temperature and mass are both constant. In other words, if and in the above formula, the resulting expression will be

Substitute for T2 and n2

Multiply both sides by T1n1

This cancels out as

4

We then have

1.6. Dalton’s law

This law states that the total pressure of a mixture of gases occupying a given volume is equal to the sum of the partial pressures of the individual gases in the mixture. The partial pressure of gas in a mixture is the pressure the gas would exert if it occupied the container alone. It is important to mention that Dalton’s law of partial pressures applies to gases which do not react chemically, i.e. a mixture of gases. For example, a mixture of hydrogen and oxygen gases obeys this law if they have not reacted to form a compound, e.g. water or water vapour.

Mathematically, Dalton’s law can be written as:

where PT is the total pressure of the mixture and , , , … are the pressures of the gases , , , … in the mixture.

2. The Ideal Gas Equation

This equation is a combination of the above laws (excluding Dalton’s law) which

establishes relationship between temperature, pressure, volume and number of

molecules in a given sample of gas. Combining Boyle’s, Charles’ and Avogadro’s laws

will produce the ideal gas equation as follows

By multiplying the right-hand sides of the above expressions, we have

(

) ( )( )

5

Based on the above expression, ideal gas equation can be stated as:

‘The volume of a given mass of an ideal gas is directly proportional to the

temperature in kelvin and number of moles (or molecules) and inversely

proportional to the pressure.’

Removing the (sign of) proportionality, we have

k is the constant of proportionality which, for this case, is denoted by letter and is

referred to as the universal gas constant. The value of R is 8.31 J mol-1K-1 for ideal

gases, but it is different and varies for real gases. Hence, the ideal gas equation is

More commonly represented as:

General Note

The behaviour of gases is approximately equal to the above laws at room temperature and atmospheric pressure, but they deviate from the laws at high pressures and/or low temperatures.

For calculation purpose, temperature is measured in Kelvin. The relationship between temperatures in degree Celsius and Kelvin is: ( ) ( ) .

At standard temperature and pressure, commonly abbreviated as STP, temperature and pressure are taken as 273.15 K (or 0 oC) and 1 atm respectively. It is also common to define the behaviour of gas at what is referred to as standard ambient temperature and pressure (abbreviated as SATP). At SATP, temperature is 298.15 K (or 25 oC) and pressure is 1 atm (or 1 bar).

In this booklet, 273 K is used for absolute zero temperature.

3. Kinetic theory

The movement of ideal gas molecules can be described using the kinetic theory based on three fundamental assumptions.

i. Collision: Gas molecules move constantly but randomly in straight lines and collide with each other and with the walls of their containers. This collision is perfectly elastic, i.e. without any energy loss during the process

6

ii. Space / Volume: The ideal gas molecules are considered to be of negligible dimensions and so do not take up space

iii. Inter-molecular forces: The gas molecules exert forces on each other only on collision.

Real gases do not meet the conditions above. While it is highly likely that the collisions are perfectly elastic, the inter-molecular force of attraction in gases becomes significant when they are close to each other and this explains why gases condense to become liquid and/or gas. Similarly, the space taken up by gas molecules cannot be ignored at high pressures.

3.1. Gas pressure

From the kinetic theory, it is known that the pressure exerted by gas molecules on the walls of the container depends on: (i) the mass of the molecules, (ii) the speed of the molecules, and (iii) the number of molecules in the container.

The relationship is expressed as:

where

is the pressure of the gas (Pa); is the volume of the gas (m3)

is the number of molecules; is the mass of one molecule (kg)

is an average molecular speed (m/s)

Re-arranging the above formula, we can have

Since density is

It follows that pressure

7

where is the density of the gas in kgm-3.

Note that the average molecular speed here is a root-mean-square (r.m.s.) and not simply mean. In all cases, the latter is either less than or equal to the former.

3.2. Energy

Since the molecules of gases are in constant motion, they possess kinetic energy. The value of the kinetic energy can be determined using a formula derived from the ideal gas equation as will be shown in this section.

For the current case, it is assumed that the number of molecules is equal to one mole. One mole of gas has an Avogadro’s number of molecules, denoted by NA. This has a constant value of and is called Avogadro’s constant. Hence, for one mole of gas, we can have

and

Combining these two equations, we have

This can be simplified to give

Re-arranging this, we have

(

)

The left-hand side is a kinetic energy formula which can be written as:

(

)

8

The ratio of the gas constant (R) to Avogadro constant (NA) is also a constant, , which is known as the Boltzmann constant. Its value is or

. The kinetic energy of the gas molecules can therefore be expressed as

Also, the is the (r.m.s.) mean kinetic energy of gas molecule. Based on the above, the ideal gas equation can also be written as:

Derivation of We know that

( ) But

( )

This implies that

( )

Substituting R in equation (1), we have

( )

The number of molecules, N, in gas is related to Avogadro’s constant and number of moles as

( )

Combining equations (4) and (5), we have

9

WORKED EXAMPLES

Section 1: Boyle’s Law

1) In the following questions, calculate the final volume of gas at the specified conditions assuming the temperature and mass remain constant.

(a) V1 = 200 cm3, P1 = 600 mmHg and P2 = 800 mmHg.

Step 1: Choose a formula. In this case, it

is Boyle’s law.

Step 2: Re-arrange the formula to make

the unknown the subject of the formula.

Step 3: Substitute the values of the

known variables to determine the

unknown variable.

The above steps will be used for

questions 1b & 1c.

(b) V1 = 24 m3, P1 = 700 mmHg and P2 = 200 mmHg.

(c) V1 = 110 cm3, P1 = 750 mmHg and P2 = 660 mmHg.

2) In the following questions, calculate the final pressure of gas at the specified conditions assuming the temperature and mass remain constant.

(a) V1 = 1575 cm3, V2 = 1050 cm3 and P1 =

1.0 atm.


is Boyle’s law.

10





unknown variable.

Important

When using gas laws, it essential that appropriate (combination of ) SI units are used. For example, when Pascal (Pa) or Newton per square metre (N/m2) are used for pressure, the volume should be in cubic metre (m3). Similarly, when pressure is in atm, the volume should be in litres.

Therefore, conversion should be carried out prior to calculation. Example for is

760 mmHg = 1 atm = 101.325 kN-2.

For the question (2a) converting the volumes (V1 and V2) to litres would not make any difference to the final answers since it is in ratio and the units can be eliminated as shown below.

( )


questions 2b & 2c.

(b) V1 = 48 litres, V2 = 32 litres and P1 = 1.8 atm.

(c) V1 = 22 m3, V2 = 40 m3 and P1 = 80 kPa.

3) A fixed mass of gas at a constant temperature has a pressure of 2000 Pa and a volume of 0.02 m3. It is compressed until the volume is 0.005 m3. What is its new pressure?

Step 1: List the known (and unknown)

variables.

11

V1 = 0.02 m3, P1 = 2000 Pa, V2 = 0.005 m3 and P2 = ?


is Boyle’s law.



Thus



unknown variable.

4) Initially gas has a volume of 0.14 m3 and a pressure of 300 kPa. What will be its volume when the pressure becomes 60 kPa if the temperature and mass remain unchanged?


variables.

V1 = 0.14 m3, P1= 300 kPa, P2 = 60 kPa and V2 = ?


is Boyle’s law.



Thus



unknown variable.

5) A gas with a volume of 2 m3 is compressed from a pressure of 100 kPa to a pressure of 500 kPa. If the temperature remains unchanged, what is the resulting volume?


variables.

V1 = 2 m3, P1= 100 kPa, P2 = 500 kPa and V2 = ?


is Boyle’s law.



Thus

12



unknown variable.

Section 2: Charles’ Law

6) Calculate the final/initial temperature of gas at the following specified conditions assuming that pressure and mass remain constant.

(a) V1 = 200 cm3, V2 = 300 cm3 and T1 = 27.0 oC.


is Charles’ law.





unknown variable.

( )

= 450 K

( )


questions 6b – 6d. Also remember

to convert the temperature to

Kelvin where appropriate.

(b) V1 = 30 litres, V2 = 25 litres and T1 = 39 oC.

( )

(c) V1 = 56 dm3, V2 = 40 dm3 and T2 = -13 oC.

( )

13

7) Calculate the initial volume of gas at the following specified conditions assuming that pressure and mass remain constant.

(a) V2 = 100 cm3, T1 = 270 K and T2 = 300 K.

(b) V2 = 25 dm3, T1 = 240 K and T2 = 330 K

Similar to number 7b

8) The volume of a fixed mass of gas at

27 oC is 20 cm3. Find its volume at 57 oC if the pressure is kept constant.


variables.

V1 = 20 cm3, T1 = 27 oC = 300 K, T2 = 57 oC = 330 K and V2 = ?


is Charles’ law.



Thus



unknown variable.

9) At a temperature of 200 K, the volume of a sample of gas in a cylinder with a smoothly fitting piston is 0.0024 m3. The cylinder is heated while allowing the gas to expand at constant pressure. Calculate the volume of the gas at a temperature of 300 K.


variables.

V1 = 0.0024 m3, T1 = 200 K, T2 = 300 K and V2 = ?


is Charles’ law.

14



Thus



unknown variable.

10) A gas has a volume of 0.010 m3 at 18 °C. What will be its volume at 85°C if the pressure acting on the gas remains unchanged?


variables.

V1 = 0.010 m3, T1 = 18 oC = 291 K, T2 = 85 oC = 358 K and V2 =?


is Charles’ law.



Thus



unknown variable.

11) A gas has a volume of 0.40 m3 at 10 °C and is heated to a temperature of 120 °C. What will be its volume if the pressure remains the same?


variables.

V1 = 0.40 m3, T1 = 10 oC = 283 K, T2 = 120 oC = 393 K and V2 = ?


is Charles’ law.



Thus

15



unknown variable.

Section 3: Gay-Lussac’s Law

12) At a temperature of 200 K, the pressure of air in a flask is 100 kPa. What will the pressure be at a temperature of 300 K? Assuming the volume of the flask is constant.


variables.

T1 = 200 K, P1 = 100 kPa, T2 = 300 K and P2 = ?


is Gay-Lussac’s law.



Thus



unknown variable.

13) A vessel of 356 cm3 capacity contains oxygen at a pressure of 760 mmHg and temperature of 15 oC. Assuming that the volume remains constant, calculate:

(a) the pressure inside the vessel when it was warmed to 100 oC, and

(b) the temperature to which it must be raised to give a pressure of 2 atm.


variables.

P1 = 760 mmHg, T1 = 15 oC = 288 K

a) When T2 = 100 oC = 373 K and P2 = ?

Step 2a: Choose a formula. In this case, it

is Gay-Lussac’s law; re-arrange this to

make the unknown the subject of the

formula.

(

)

(

)

16

b) When P2 = 2 atm = 2(760) mmHg and T2 = ?

Step 2b: Choose a formula. In this case, it



formula.

(

)



unknown variable.

(

)

Step 4: Convert the temperature to

degree Celsius.

Note that since volume is

constant at 356 cm3

, it is not

required in the calculation and

so Gay-Lussac’s law is the

appropriate formula.

14) Two vessels A (100 cm3 capacity) and B (300 cm3 capacity) were separated by a tap. A was filled with gas at 748 kNm-2 pressure and B was evacuated until its inner pressure was negligible.

Assuming that the temperature remains constant, calculate the pressure in the flask when the tap is opened.

Step 1: Work out what the variables are

from the information given in the

question.

This is because vessel B was evaluated or emptied at this state.

( )

This is because the tap between the vessels is now open and the gas can fill the two vessels.

Step 2: Now List the known (and

unknown) variables.

For this case, the initial variables, V1 and P1, are the value of pressure and volume of gas in vessel A before the tap was opened. The final variables, V2 and P2, are the value of pressure and volume of gas in the combined vessels (A and B).

Therefore, we have the following state variables:

17

V1 = 100 cm3, P1 = 748 kNm-2, V2 = 400 cm3 and P2 = ?




formula.

(

)



unknown variable.

(

)

15) In the morning, when the temperature is 286 K, a bicyclist finds that the absolute pressure in his tires is 501 kPa. That afternoon he finds that the pressure in the tires has increased to 554 kPa. Ignoring expansion of the tires, find the afternoon temperature.


variables.

T1 = 286 K, P1 = 501 kPa, P2 = 554 kPa and T2 = ?


is Gay-Lussac’s law.



Thus



unknown variable.

Section 4: Combined gas law

16) Calculate the final temperature of gas at the following specified conditions assuming that mass remains constant.

(a) V1 = 466 dm3, V2 = 500 dm3, P1 = 300 mmHg, P2 = 450 mmHg and T1 = 47 oC.


is combined gas law.



(

) (

)

18



unknown variable.

( ) (

) (

)


questions 16b & 16c. Again

remember to convert

temperature to Kelvin scale.

(b) V1 = 700 m3, V2 = 550 m3, P1 = 39 kPa, P2 = 77 kPa and T1 = 57 oC.

( ) (

) (

)

(c) V1 = 5 litres, V2 = 15 litres, P1 = 3.6 atm, P2 = 1.8 atm and T1 = -11 oC.

( ) (

) (

)

17) Calculate the initial pressure at the following specified conditions assuming that mass remains constant.

(a) V1 = 100 dm3, V2 = 1.0 dm3, P2 = 400 kNm-2, T1 = 250 K and T2 = 200 K.





(

) (

)



unknown variable.

(

) (

)

(b) V1 = 2.8 litres, V1 = 1.8 litres, P2 = 2.0 atm, T1 = 280 K and T2 = 350 K.

(

) (

)

(c) V1 = 1.50 m3, V2 = 2.40 m3, P2 = 90.0 kNm-2, T1 = 300 K and T2 = 360 K.

19

(

) (

)

18) A given mass of gas has a volume of 480 cm3 at STP, calculate its volume under the following conditions:

(a) 39 oC and 800 mmHg.

In the current case,V1

= 480 cm3

,

P1

= 760 mmHg,T1

= 273K and V2

will be determined under the

known conditions.


variables.

V1 = 480 cm3, P1 = 760 mmHg, T1 = 273 K, P2 = 800 mmHg, T2 = 39 oC = 312 K and V2 = ?


is combined gas law; re-arrange this to


formula.

(

) (

)



unknown variable.

(

) (

)


questions 18b & 18c.

(b) -13 oC and 960 mmHg.

V1 = 480 cm3, P1 = 760 mmHg, T1 = 273 K, P2 = 960 mmHg, T2 = -13 oC = 260 K and V2 = ?

(

) (

)

V1 = 480 cm3, P1 = 760 mmHg, T1 = 273 K, P2 = 120 kNm-2, T2 = 240 K and V2 = ?

(

) (

)

(c) 546 K and 50 kNm-2.

V1 = 480 cm3, P1 = 100 kNm-2, T1 = 273 K, P2 = 50 kNm-2, T2 = 546 K and V2 = ?

(

) (

)

20

19) What volume would the following gases occupy at STP?

(a) 76 cm3 measured at 14 oC and 820 mmHg.

This is similar to the previous

question (No 15) except that the

values at STP are the final

values. In other words, we have

P2

= 760 mmHg, T2

= 273K and V2

is determined using the known

values as V1

, P1

and T1

.

V1 = 76 cm3, T1 = 14 oC = 287 K, P1= 820 mmHg, T2 = 273 K, P2 = 760 mmHg and V2 = ?

(

) (

)

(

) (

)

(b) 64.4 litres measured at 49 oC and 1.5 atm.

V1 = 64.4 litres, T1 = 49 oC = 322 K, P1 = 1.5 atm, T2 = 273 K, P2 = 1 atm and V2 = ?

(

) (

)

(c) 20 dm3 measured at 400 K and 80 kNm-2.

V1 = 20 dm3, T1 = 400 K, P1 = 80 kNm-2, T2 = 273 K, P2 = 100 kNm-2 and V2 = ?

(

) (

)

20) Calculate the final pressure of the following gases.

(a) V1 = 160 cm3, T1 = 47 oC, P1 = 700 mmHg, V2= 140 cm3 and T2 = 79 oC.





(

) (

)

21



unknown variable.

(

) (

)

(b) V1 = 7.5 litres, T1 = -23 oC= 250 K , P1 = 20 atm, V2 = 15 litres, T2 = 27 oC = 300 K and P2 = ?

(

) (

)

(c) V1 = 2 dm3, T1 = 300 K, P1 = 150 kNm-

2, V2 = 3 dm3, T2 = 150 K and P2 = ?.

(

) (

)

21) Calculate the final temperature of the following gases in oC:

(a) V1 = 750 cm3, T1 = 2 oC = 275 K, P1 = 350 mmHg, V2 = 630 cm3, P2 = 400 mmHg and T2 = ?





This implies

Thus,

(

) (

)



unknown variable.

(

) (

)

Step 4: Convert the temperature to

degree Celsius.


questions 21b & 21c.

(b) V1 = 150 cm3, T1 = 42 oC = 315 K, P1 = 1.0 atm, V2 = 200 cm3, P2 = 2.0 atm and T2 = ?

22

(

) (

)

(c) V1 = 8 dm3, T1 = 240 K, P1 = 70 kNm-2, V2 = 7 dm3, P2 = 100 kNm-2 and T2 = ?

(

) (

)

22) A gas has a volume of 0.100 m3 at a temperature of 25°C and a pressure of 140 kPa. What will be its temperature at a pressure of 700 kPa if the new volume is 0.200 m3?


variables.

V1 = 1.00 m3, V2 = 0.200 m3 , T1 = 25 oC = 298 K, P1= 140 kPa, P2 = 700 kPa and T2 = ?


is combined gas law; re-arrange it to


formula.

(

) (

)



unknown variable.

(

) (

)

23) A compressed-air tank holds 0.500 m3 of air at a temperature of 285 K and a pressure of 880 kPa. What volume would the air occupy if it were released into the atmosphere, where the pressure is 101 kPa and the temperature is 303 K?


variables.

V1 = 0.500 m3, T1 = 285 K, P1= 880 kPa, T2 = 303 K, P2 = 101 kPa and V2 = ?




formula.

(

) (

)



unknown variable.

(

) (

)

23

24) A gas cylinder contains 0.11 m3 of gas at an absolute pressure of 1000 kPa and a temperature of 15 °C. What will be the volume of the gas at the atmospheric pressure of 101 kPa and a room temperature of 25 oC?


variables.

V1 = 0.11 m3, T1 = 15 oC = 288 K, P1= 1000 kPa, T2 = 25 oC = 298 K, P2 = 101 kPa and V2 = ?




formula.

(

) (

)



unknown variable.

(

) (

)

25) A gas has a volume of 0.40 m3 at a pressure of 90 kPa and a temperature of 30 °C. What will be its volume at STP?


variables.





formula.

(

) (

)



unknown variable.

(

) (

)

26) A cylinder of compressed air has a volume of 2.12 litres and a pressure of 11.0 atm at 15 oC. What volume would the air from the cylinder occupy at STP?


variables.

V1 = 2.12 litres, P1= 11.0 atm, T1 = 15 oC = 288 K, P2 = 760 mmHg = 1.00 atm, T2 = 273 K and V2 = ?



24


formula.

(

) (

)



unknown variable.

(

) (

)

27) 1.12 litres of oxygen measured at 30 oC and 754 mmHg has a mass of 1.43 g. Calculate the volume occupied by 32.0 g of oxygen at STP.


variables.

V1 = 1.12 litres, P1 = 754 mmHg, T1 = 30 oC = 303 K, P2 = 760 mmHg, T2 = 273 K and V2 = ?




formula.

(

) (

)



unknown variable.

(

) (

)

Step 4: Determine the volume occupied

by 32.0 g of oxygen gas (using

Avogadro’s law).

That is to say, 1.12 litres of O2 at 30 oC and 754 mmHg is equivalent to 1.001 litres of the same gas (i.e. O2) at STP. Thus, 1.001 litres of O2 at STP has a mass of 1.43 g.

This implies that 32 g would have a

volume of

28) A gas with a volume of 0.40 m3 at STP is heated until it occupies a volume of 0.46 m3 at a pressure of 115 kPa. What will be its temperature?


variables.

V1 = 0.40 m3, T1 = 0 oC = 273 K, P1= 101 kPa, V2 = 0.46 m3, P2 = 115 kPa and T2 = ?




formula.

This implies that

25

Thus,

(

) (

)



unknown variable.

(

) (

)

29) A gas at a pressure of 250 kPa and temperature 20 °C has a volume of 0.050 m3. What will be its volume at a pressure of 400 kPa and a temperature of 100 °C?


variables.





formula.

(

) (

)



unknown variable.

(

) (

)

30) 1 litre of hydrogen at STP weighs 0.090 g. Calculate the mass of 2.56 litres of hydrogen measured at 25 oC and 736 mmHg.


variables.

V1 = 2.56 litres, P1 = 736 mmHg, T1 = 20 oC = 293 K, P2 = 760 mmHg, T2 = 273 K and V2 = ?




formula.

(

) (

)



unknown variable.

(

) (

)

Step 4: Convert the volume to equivalent

weight.

26

But, 1 litre of H2 at STP weighs 0.09 g, which implies that 2.31 litres would weigh ( ) .

31) A balloon contains 3.7 litres of nitrogen gas at a temperature of 87 K and a pressure of 101 kPa. If the temperature of the gas is allowed to increase to 24 oC and the pressure remains constant, what volume will the gas occupy?


variables.

V1 = 3.7 litres, P1 = P2 =101 kPa, T1 = 87 K, T2 = 24 oC = 297 K and V2 = ?




formula.

(

) (

)



unknown variable.

(

) (

)

Since the pressure remains

constant, the question can

simply be solved using Charles’

law. As it is apparent in the

above that

.

Section 5: Ideal gas equation

32) An air cylinder for a scuba diver contains 150 moles of air at 15 oC. The cylinder has a volume of 0.012 m3. (The gas constant R = 8.3 J K-1 mol-1.)

a. Calculate the pressure in the air cylinder.

b. Calculate the gas volume when it is all released at atmospheric pressure of 100 kPa and a temperature of 25 oC.


variables.

n = 150 mol, R = 8.3 J K-1 mol-1, V1 = 0.012 m3, T1 = 288 K, T2 = 298 K, P2 = 100 kPa, P1 =? and V2 = ?

a) Pressure in the air cylinder


is ideal gas equation; re-arrange this to


formula.

Thus

27

For this case, V = V1 and T = T1

Step 3a: Substitute the values of the


unknown variable.

It follows that

Mega is one of the many suffixes

used in science and

engineering, which is

equivalent to 106

. For example

b) Volume.




formula.

(

) (

)

Step 3b: Substitute the values of the


unknown variable.

For this case, P1 = 2.988 x 107 Pa

(

) (

)

33) What is the mass of gas at a pressure of 500 kPa and a temperature of 50 °C if it occupies a volume of 0.10 m3? The gas has a characteristic gas constant of 189 J kg-1 K-1.


variables.

R = 189 J kg-1 K-1, V = 0.10 m3, T = 50 oC = 323 K, P = 500 kPa and m = ?




formula.

Thus



unknown variable.

28

Since the universal gas constant

is measured in J/kg/K, the

number of moles (n) in the

original equation is replaced by

m, the mass of the gas. This is

true since mass of gas is directly

proportional to mole since the

molecular mass remains

constant.

34) A rigid gas container of an internal volume of 1.2 m3 holds 1.8 kg of gas at 18 °C. What is the pressure of the gas and to what value will it change if a further 0.8 kg of gas is added at the same temperature? The gas has a characteristic gas constant of 287 J kg-1 K-1.


variables.

m = 1.8 kg, R = 287 J kg-1 K-1, V = 1.2 m3, T = 18 oC = 291 K and P = ?




formula.

Thus



unknown variable.

Step 4: Now calculate the pressure when

1.7 kg of the gas is added.

Let the new pressure be pn , and

m = (1.8 + 0.8) kg = 2.6 kg.

Thus

35) What is the mass of gas at a pressure of 350 kPa and a temperature of 35 °C

29

if it occupies a volume of 0.030 m3. The gas has a characteristic gas constant of 290 J kg-1 K-1?


variables.





formula.

Thus



unknown variable.

36) The Goodyear blimp spirit of Akron is 62.6 m long and contains 7023 m3 of helium. When the temperature of the helium is 285 K, its absolute pressure is 112 kPa. Find the mass of the helium in the blimp.


variables.

R = 8.31 J mol-1 K-1, V = 7023 m3, T = 285 K, P = 112 kPa and m = ?


is the ideal gas equation; re-arrange this

to make the unknown the subject of the

formula.

Thus



unknown variable.

Step 4: Determine the mass of helium.

37) A typical region of intersteller space may contain 106 atoms per cubic meter (primarily hydrogen) at a temperature of 100 K. What is the pressure of this gas?

30


variables.

N = 106 atoms, k = 287 J kg-1 K-1, V = 1 m3, T = 100 K and P = ?

Note that volume is taken as 1

m3

and number of particle as 106

atoms. Alternatively, it is also

possible to consider 106

atoms

per cubic meter as the N / V in

the formula.




formula.

Thus

Or

(

)



unknown variable.

( ) ( )

Alternatively, we can consider the first approach, i.e. with V = 1 m3 and N =106, we can have

( )

It is important to note here that

the final answer in this question

is given in one significant

figure (s.f.) since it is the least

s.f. in the values of the

quantities given in the question.

This applies to all answers.

38) A rigid container with an internal volume of 0.85 m3 contains gas at a pressure of 275 kPa and temperature of 15 °C. What will be the pressure of the gas in the container if an additional 1.7 kg of the gas is pumped

K N / V T

31

into the container at the same temperature? The gas has a characteristic gas constant of 290 Jkg-1 K-1.


variables.





formula.

Thus



unknown variable.

Step 4: Now calculate the pressure when

1.7 kg of the gas is added.

When additional 1.7 kg is added, the new mass of the gas is 4.5 kg.

Therefore, the pressure of the gas in the container can be calculated as follows

39) A gas has a temperature of 310 K and a pressure of 101 kPa. (a) Find the volume occupied by 1.25 mol of this gas, assuming it is ideal. (b) Assuming the gas molecules can be approximated as small spheres of diameter 2.5 x 10-10 m, determine the fraction of the volume found in part (a) that is occupied by the molecules.


variables.

n = 1.25 mol, R = 8.31 J mol-1 K-1, T = 310 K, P = 101 kPa, diameter = 2.5 x 10-10 m and V = ?

a) Volume




formula.

Thus

32



unknown variable.

b) Fraction of the volume occupied by the molecules.

Step 4: Determine the volume occupied

by a molecule.

Volume occupied by a molecule = volume of a sphere

(

)

But the number of molecules, N, in the gas is

Therefore, the total volume of the molecules

Thus, the fraction occupied by the molecules

Section 6: Additional examples

40) The air pressure of a constant volume

gas thermometer is 1.010 x 105 Pa at

ice point and 1.600 x 105 Pa at steam

point. Determine the Celsius

temperature when the pressure of the

gas is 1.250 x 105 Pa.


variables.

T1 = 0 oC = 273 K, P1 = 1.010 x 105 Pa, T2 =100 oC = 373 K, P2 = 1.600 x 105 Pa, P3 = 1.250 x 105 Pa and T3 =?

Step 2: Work out the pressure and

temperature difference.

( ) ( )

This is proportional to temperature difference

33

Also,

( ) ( )

This is proportional to temperature difference

( )


is a ‘modified’ Gay-Lussac’s law and re-

arrange accordingly.

Using the Gay-Lussac’s law, we can have



unknown variable.

41) Helium at atmospheric pressure of

100 kPa has a density of 0.17 kg m-3 at

273 K. Calculate the r.m.s. speed of the

molecules.


variables.

P = 100 kPa, ρ= 0.17 kg m-3, T = 273 K, c = ?

Step 2: Choose a formula and re-arrange

this to make the unknown the subject of

the formula.

√



unknown variable.

√

42) Compute (a) the number of moles and

(b) the number of molecules in 1.00

cm3 of an ideal gas at a pressure of 100

Pa and a temperature of 220 K.

34


variables.

P = 100 Pa, V = 1.00 cm3, T = 220 K, NA =

6.02 x 1023 mol-1, R = 8.31 J mol-1 K-1, k =

1.38 x 10-23 J K-1, N= ? and n= ?

a) the number of moles, n




formula.

Thus

Step 3a: Substitute the values of the


unknown variable.

b) the number of molecules




formula.

Thus

Step 3b: Substitute the values of the


unknown variable.

Alternatively, since we have found the value of n in (a), we can use the formula below to find the number of molecules in 1.00 cm3 of the gas as follows:

( ) ( )

( )

43) A mole of hydrogen molecules, each

of mass 3.3 x 10-27 kg, is contained in a

cylinder of volume 0.050 m3. The

molecules have an r.m.s. speed of 800

m s-1. Calculate the pressure of the

gas. (Avogadro constant NA = 6.0 x

1023 mol-1).


variables.

m = 3.3 x 10-27 kg, V = 0.05 m3, c = 800 m/s,

and P?



the formula.

35

For one mole of hydrogen, we have the number of molecules, N, equal to Avogadro’s constant, NA



unknown variable.

44) A metal can of volume 2000.0 cm3 contains 2.0 g of nitrogen gas and 6.0 g of hydrogen gas at exactly 200 oC. What is the total pressure of gas in the can?


variables.

V = 2000.0 cm3, T = 200 oC = 473 K, mass of nitrogen gas = 2.0 g, mass of hydrogen gas = 6 g




formula.

Note that the partial pressures

for nitrogen and hydrogen must

be calculated.

Let and be the partial pressures for nitrogen and hydrogen gases respectively. Also, let and be the number of moles for nitrogen and hydrogen gases respectively.

Step 3: Determine the number of moles for nitrogen and hydrogen gases

Therefore,

Step 3: Determine the partial pressures of the gases.

For nitrogen gas, we have

36

For hydrogen gas, we have

Step 3: Determine the total pressure in the can using Dalton’s law of partial pressures.

Let be the total pressure in the can. Therefore,

( ) ( )

45) A gas expands at a constant pressure of 1.0 x 105 Pa. Its volume increases

from 0.1 m3 to 0.15 m3. How much work is done in the expansion?


variables.

V1 = 0.1m3, V2 = 0.15 m3, P = 1.0 x 105 Pa and W = ?



the formula.



unknown variable.

( ) ( )

37

Bibliography and Further Reading

1) Bolton, W., 2006. Engineering Science. 5th ed. Oxford: Elsevier Newnes.

2) Brown, C. and Ford, M., 2009. Higher Level Chemistry developed specifically for the IB

Diploma. Harlow: Pearson Education.

3) Akusoba, E.U. and Ewelukwu, G. O., 1989. Calculations in Chemistry for Senior

Secondary School. Onitsha: Africana First Publishers.

4) Halliday, D., Resnick, R. and Walker, J., 2001. Fundamental of Physics. 6th ed. New

York: John Wiley & Sons.

5) Johnson, K., Hewett, S., Holt, S. and Miller, J., 2000 Advanced Physics for You.

Cheltenham: Nelson Thornes.

6) Lewis, R. and Evans, W., 2011. Chemistry. 4th ed. London: Palgrave Macmillan.

7) Norris, R., Ryan, L. and Acaster, D., 2011. Cambridge International AS and A Level

Chemistry Coursework Cambridge: Cambridge University Press.

8) Walker, J. S., 20010. Physics. 4th ed. San Francisco: Addison-Wesley.

.

Documents

Worked Examples on Gas Laws and Kinetic Theory