Category: Physic Tutorials

Quarks are the fundamental particles. They are what make up nucleons, i.e. protons and neutrons, as well as other particles. Electron is a fundamental particle, one of a group of fundamental particle called leptons.

The term fundamental means the particles does not have a sub structure; it can’t be split.

They are the fundamental building blocks which build up matter, i.e., they are seen as the “elementary particles”.

There are six quarks, but they are usually talked in terms of three pairs: up/down, charm/strange, and top/bottom.

Up – has a charge of +2/3

Down – has a charge of -1/3

The up and down quarks are different in mass, charge, and spin. These quarks are the most common and least massive.

Four main configurations

(i) Proton

It is made up of two ‘up’ quarks and one ‘down’ quark (UUD)

Proton has a charge of +1 = +2/3 (U) +2/3(U) -1/3(D)

(ii) Neutron

It is made up of 2 ‘down’ quarks and an ‘up’ quark (DDU)

Neutron has a charge of 0 = -1/3 (D) -1/3 (D) + 2/3 (U)

(iii) The Pion (π+)

It is made up of an ‘up’ and an ’anti-down’ quark.

(iv) The Kaon (K+)

It is made up of an ‘up’ and an ‘anti-strange’ quark.

Hadrons

Hadrons are unstable with the exception being the proton-the only stable Hadron.

Hadrons are composed of smaller fundamental particles called Quarks.

Meson have 2 Quarks and Baryons 3. Hence mesons don’t decay to protons or neutrons.

Some carry charge i.e. (p, Kˉ, K+)

Some have no charge i.e. (n, Ko)

Simple quark model of hadrons

Below is the diagram of simple model of hadrons

Difference between Hadrons and Leptons

• Hadrons feel the strong force while lepton do not
• Hadrons have masses much larger than that of leptons

Solution to CIE A level physics questions on fundamental particles

Question 1

Which combination of up (u) and down (d) quarks forms a neutron? (Oct/Nov 2017 P12)
A u u u    B u u d    C u d d    D d d d

Neutron has one u and two d = udd

C is the correct option

Question 2

What is the quark composition of a hydrogen-3 nucleus, 3H1 ? (Oct/Nov 2017 p11)

Solution

Hydrogen – 3 nucleus has one proton and two neutron

one proton =two u and one d= 2u and d

two neutrons = 2(one u and two d) = 2U and 4d

In all, Hydrogen – 3 nucleus has 4 u and 5 d

A is the answer

This article will guide you on how to learn physics and be good at it. Physics is one of the subjects that students do have difficulty to understand. However, physics is easy to learn.

The first thing you need to understand is that understanding the concepts in any of the topics in physics will make you gain mastery of the subject. I came to the understanding of this fact when I started to prepare students for CIE A level.

I discovered that majority of my students fear physics because they think it is weird. The fear was as a result of their inadequacy to understanding the basic concept in physics.

Step to step guide on how you can learn physics and become better at physics are discussed below.

 

How to learn physics for beginners

1. Get different textbooks on the subject

One thing I have discovered about gaining mastery on any subject is to have different textbooks on that subject. Each writer has different ways of explanation on each topic of the subject. This will give you an opportunity to be exposed to different ways of explanation and which is to your benefit.

2. Get a good physics teacher

In this journey, you will need someone who will guide you through. And this is the role the teacher will play. Physics isn’t a literature book or a newspaper that you can study without anybody putting you through.

3. Seek for the concept

Without a proper understanding of physics concept, your effort to understanding physics will be in vain. A concept in physics is an idea behind each topic in physics or the message the topic is trying to pass across in relation to nature, matter, and energy. For example, in physics, acceleration is the rate of change of velocity, and this means the instant at which the velocity of your bicycle or car changes your bicycle has accelerated.

4. Be inquisitive

Always ask questions during class or meet with your teacher before or after school if you need additional assistance.

5. Link every knowledge gain with what happens around you

Physics is the branch of science concerned with the nature and properties of matter and energy. It is what you see around and relate with every day. So your knowledge must go beyond what you have read and it must be seen in the natural standpoint.

For example, if you read about acceleration, try to relate it to the motion of a car or bicycle then you will have a better understanding of acceleration.

6. Know when, where, and how to apply physics formulas

Most of the topics in physics have at least one formula and one interesting thing is that the formulas are linked to each other. It isn’t difficult to know these formulas offhand; you only need to familiarize yourself with them or memorize them. But understanding the concepts in each topic will help you to know the formulas effortlessly and help you to pass an exam.

For example, V= u +at and V= u-at for motion under gravity, one is for when an object is thrown downward while the other is when an object is thrown upward. So if you don’t understand this simple concept, it will be difficult to apply it even if you know the formula.

7. You must believe in your ability to understand it

Your belief system has a way it has an effect on your outward performance. If you believe it will be difficult for you to understand it, it will difficult indeed. You just have to see physics has been simple to learn and pass easily in an exam.

 

Tips to remember physics concepts

• Link them to your daily activity
• Always try to put them in your daily conversation
• Make it a duty to teach someone the concept when you have the opportunity
• Lay your hands on books or articles that do discuss the concepts you have learned

Cambridge A-level physics is easy to pass with grade A or A*, if you can work hard and be smart. A smart student would have gone through all the available past questions and find solutions to them. It is usual of Cambridge to bring out questions from CIE A level past papers. This means you would have come across some of the questions in the past papers if you have taken your time to study them and find solutions to them.

I have decided to provide explanatory answers to CIE a level physics past papers especially paper 1. And the reason is because CIE has already provided answers to a level physics paper 2 and paper 4 with a degree of explanation. But this is not so for paper 1 (the correct options are only supplied without detailed explanation on how they arrived at the answers).

It is also imortant that you know the key to excel in Cambridge A level exam by reading through some reasons why some students fail the exam. It is wisdom for you to look at the mistakes of others and try everything within your power not to repeat it. Click here to read the major reasons why people have “U/D” in A level.

I want to be uploading solutions to CIE A level physics past papers 1. This will help you to understand some CIE A level physics questions better.

Note: you will need to have the CIE past papers with you for you to relate it with the solutions

To download past papers for physics click here

 

Cambridge International Examination, CIE A Level Physics Past Papers Solutions (Paper 1)

The solution below is Cambridge A level physics past question for Oct/Nov 2017. Click and download

CIE physics 9702 oct

The textbook you are about to download contains all the AS and AL topics in Phyics 9702 and it will assist you to have a good grade in Cambridge A level Physics.

Download for free Preparatory guide Physics textbook for Cambridge A level, UTME, and Post UTME (pdf)

I will upload for other years as soon as possible.

In this article, I will show the kinematic equations for both linear and angular, how they are derived, and graphs. Kinematics in physics is about the motion of a body or an object. And the terms associated with kinematics are displacement, speed, velocity, and acceleration. These are the variables physicists play around in this topic.

Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

The starting point is from the definition of acceleration which is the rate of change of velocity. This is like a car traveling at a certain speed then the driver decides to move faster or slower by pressing the accelerator. The speedometer in the car will change and at that instance, the car has to either accelerate or decelerate.

First Equation

Acceleration = change of velocity / time = (v – u) / t

u is the initial velocity, v is the final velocity, t is the time taken, and a is the acceleration

cross multiply,

Make v the subject of the equation

The first equation above is the Newton’s first equation of motion

Second equation

The next step,

From average velocity,

Average velocity = s / t

Average velocity = (v +u)/2

Equate the two equations together, you get

s / t = (v +u)/2

Cross multiply and rearrange

s = [(v + u)t]/2

substitute v = u + at into  s = [(v + u)t]/2

you have

This is the Newton’s second equation of motion

Recommended: Short notes on motion

Third equation

Final step:

Make t the subject of the Newton’s first equation of motion

t = (v – u) / a

substitute t = (v – u) / a into Newton’s second equation of motion

Cross multiply,

This is the Newton’s third equation of motion

Angular Kinematics Equations

The equation above is the linear kinematic equations but here we will look at the branch which deals with the rotational motion of anybody. In this case,

• Displacement is replaced by a change in angle.
• Initial and final velocities are replaced by initial and final angular velocities.
• Acceleration is replaced by angular
• Time is the only constant.

Kinematic Graphs

The next thing is to look at Kinematics graphs. The graphs are related to the equations that I have given above. The velocity graph is the derivative of the displacement graph while the acceleration-time graph is the derivative of a velocity-time graph in kinematics.

Magnetic resonance imaging, MRI is a diagnostic technique used in medicine. It can provide images of the inside of a patient. It does not rely on exposing patients to ionizing radiation such as x-rays; it relies on the fact that some atomic nuclei behave like tiny magnets in an external magnetic field.

Principle of operation of MRI

Nuclei of hydrogen (and certain other nuclides) have spin/have a magnetic axis.  In a strong magnetic field, the nuclei precess about the direction of the field with a frequency known as the Larmor frequency.  Radio wave pulse at Larmor frequency causes resonance and nuclei precess in high-energy state.  After pulse has ceased, nuclei relax, emitting an RF signal.

How MRI works

MRI scanning involves radio frequency electromagnetic radiation. The patient lies on a bed in strong magnetic field. RF pulse at Larmor frequency transmitted to patient. RF emissions from patient are detected and processed. Hydrogen nuclei within patient have Larmor frequency dependent on magnetic field strength so that location (and concentration) of hydrogen nuclei can be detected.  Total image built up by varying the non-uniform field to give specific field strength at different positions within the patient. The receiving coils pick up the relaxation signal and pass it to the computer. Therefore, a picture of the patient’s insides is built up by a computer.

Recommended: Uses and operation of Ultrasound

Component of magnetic resonance imaging

(i) Superconducting magnet

(ii) RF transmission coil

(iii) RF receiver coil

(iv) Gradient coils

(v) Computer

MRI uses

It is use in medical diagnosis (spinal/back problems and fine detail provided of soft and hard tissues)

Advantages of MRI

i) It does not use ionizing radiation which causes a hazard to patients and staff

ii) There are no moving mechanisms, just changing currents and magnetic fields

iii) The patient feels nothing during a scan (although the gradient coils are noisy as they are switched), and there are no after-effects.

iv) MRI gives better soft-tissue contrast than a CT scan, although it does not show bone as clearly

v) Computer images can be generated showing any section through the volume scanned, or as a three-dimensional image

Disadvantage of magnetic resonance imaging

Every metallic objects in the patient, such as surgical pins, can become heated. Loose steel objects must not be felt in the room as these will be attracted to the magnet, and the room must be shielded from external radio fields.

Reference

David Sang et al, Cambridge International AS and A level Physics Coursebook

Ultrasound is an oscillating sound pressure wave with a frequency greater than the upper limit of the human hearing range( 20KHz in healthy, young adults). Ultrasound devices operate with frequencies from 20KHz to several gigahertz.

How Ultrasound is generated

i) Like audible sound, ultrasound is produced by a vibrating source. The frequency of the source is the same as the frequency of the waves it produces.

ii) Ultrasound waves may be generated and detected using a piezo-electric transducer.

iii) A transducer converts energy from one form to another.

iv) Electrical energy is converted into ultrasound energy by means of a piezo-electric crystal such as quartz(it is widely used because of its mechanical strength, small dielectric loss, resistance to moisture and the stability of its properties in the face of temperature change).

v) The structure of quartz is made up of a large number of tetrahedral silicate.

vi) An external electric field is applied to the crystal, the ions in each unit cell are displaced by electrostatic forces, resulting in the mechanical deformation of the whole crystal i.e. expand or contract

vii) An alternating voltage with frequency f causes the crystal to contract and expand at the same frequency f. The alternating voltage applied across the crystal is then acts as the vibrating source of the ultrasound waves.

Uses of Ultrasound

• To detect objects and measure distance
• Sonography(ultrasonic imaging) used both in veterinary and human medicine
• For nondestructive testing of products and structures
• To detect invisible flaws
• For cleaning and for mixing, and to accelerate chemical processes

Principle of Ultrasound scan

There are several different types of ultrasound scan which are used in practice, but their principle are the same(i.e. echo sounding)

• Direct the ultrasound waves into the body
• These waves pass through various tissues and are partially reflected at each other boundary where the wave speed changes
• The reflected waves are then detected and used to construct internal image of the body

Based on the principle of an ultrasound scan a diagnostic information can be obtained.

A-scan and B-scan are two techniques commonly use for the display of an ultrasound scan.

In an A-scan a pulse of ultrasound is transmitted into the body through the coupling medium. At each boundary between, some of the energy of the pulse is reflected and some transmitted. The transducer detects the reflected pulses. The signal is amplified and displayed on a cathode ray oscilloscope.

Recommended: Short note on electronic sensor

The page provides all the CIE Physics 9702 past papers.

To have A* in Cambridge A level CIE physics isn’t difficult as you may think, all you need is adequate preparation and productive reading. It requires personal development which is crucial to one’s success in life. It is all about adding the required or needed values that enhance personal academic growth.

Tips to excel in CIE AS and A level

(1)  Be inquisitive – Ask the right questions in class. It is when you ask that you get answers, clarity, and understanding

(2)  Relate with your colleagues that are intelligent – Get around people that are ahead of you academically. Whenever important topics are been discussed by your colleagues, always find your place around them. From my personal experience, sometimes studying alone may be boring or not interesting but when you get around friends discussing a particular topic, a level of interest rise up in you. There is a level of challenge you get that will propel you to read. It is that fuel that will make that particular topic interesting. In some cases, the reason why some of us read at times is that we heard some people discussing a topic that we know little or nothing about. This leads to a hunger to know more about the topic. In summary being close to people that are ahead of you academically can be of help to your personal growth.

(3)  Get enough materials on physics expecially the ones recommended by CIE – Different materials have a way in which they discuss different topics. And the more the available materials you have the better your understanding in that area. It is possible that you may not get a satisfactory explanation from a textbook but having more than one textbook will give you an option to check other’s views.

(4)  Determination and consistency – People who are successful aren’t people who have only motivation but are consistent and determined in their motivation.

The free download of Cambridge A level Physics 9702 past papers and markscheme are below:

CIE Physics 9702: Free download Cambridge A Level Physics 9702 past papers

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Markschemes

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Cambridge International Examination Physics may/june 2017 free download

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May/June 2016
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Past physics question papers Cambridge A level may/june 2016
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How to solve Questions on Nuclear physics for UTME

Question 1

A piece of radioactive material contains 1000 atoms. If its half-life is 20 seconds, the time taken for 125
atoms to remain is {UTME 2012}
A. 20 seconds            B. 40 seconds         C. 60 seconds        D. 80 seconds

Solution

N = N0 exp(-ℷt)

ℷ = 0.693/T

ℷ = 0.693/20

125 = 1000 exp(-0.03465t)

125/1000 = exp(-0.03465t)

1000/125 = exp(0.03465t)

8 = exp(0.03465t)

0.03465t = In 8

0.03465t = 2.0794415

t = 2.0794415 /0.03465 = 60 seconds

C is the correct option

Question 2

A radioactive isotope has a decay constant of 10-5s-1. Calculate its half life. {UTME 2011}
A. 6.93 x 10-6s      B. 6.93 x 10-5s     C. 6.93 x 105s      D. 6.93 x 104s

Solution

T = 0.693/ℷ

T = 0.693 / 0.00001

T = 6.93 x 10^4 s

D is the correct option

Question 3

The radioisotope 235U92 decays by emitting two alpha particles, three beta particles and a gamma ray. What is the mass and atomic numbers of the resulting daughter element? {UTME 2010}
A. 91 and 227       B. 92 and 238      C. 227 and 91       D. 215 and 88.

Solution

You should note that in nuclear reaction, proton number and nucleon number must be conserved

235U92 – 2(4He2) + 3(0e-1) +ϒ + 227X91

227X91 means nucleon number(mass number) is 227 and proton number(atomic number) is 91.

C is the correct option

Question 4

A piece of radioactive material contains 10^20 atoms. If the half life of the material is 20 seconds, the number of
disintegrations in the first second is {UTME 2009}
A. 3.47 x 1018       B. 6.93 x 1020       C. 3.47 x 1020          D. 6.93 x 1018

Solution

N = N0 exp(-ℷt)

ℷ = 0.693/T

ℷ = 0.693/20

N = 10^20 exp(-1*0.03465)

N = 10^20/1.03536 = 9.658 x 10^19

the number of disintegrations in the first second is = N0 – N = 10^20 – 9.658 x 10^19

the number = 0.342 x 10^19 = 3.42 x 10^18

A is the correct answer

Question 5

If the decay constant of a radioactive substance is 0.231s-1, the half-life is {UTME 2009}
A. 3.00s      B. 0.12s       C. 0.33s     D. 1.50s

Solution

T = 0.693/ℷ

T = 0.693 / 0.23

T = 3.01 s

A is the correct option

Question 6

In the equation above, the particle X is {UTME 2008}
A. a proton         B. a neutron      C. an α-particle       D. a β-particle

Solution

In nuclear reaction, proton number and nucleon number must be conserved

On the reactant side, the total number of proton = 7 + 2 = 9, the total nucleon number = 14 + 4 = 18

For the product side to be conserved, the nucleon number on X must be 1 and proton number must be 1

That is, 1X1 = proton

A is the correct option.

Question 7

A radioactive substance has a half-life of 20 days. What fraction of the original radioactive nuclei will remain after 80
days? {UTME 2007}
A. 1/16      B. 1/8        C. 1/4       D. 1/32

Solution

let the original value be N

N – N/2 …20 days

N/2 – N/4 … 40days

N/4 – N/8 ….60 days

N/8 – N/16 …. 80 days

after 80 days 1/16 0f the original radioactive nuclei will remain

A is the correct option

Question 8

The time it will take a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original
number is {UTME 2005}
A. 150 days       B. 200 days        C. 250 days         D. 300 days.

Solution

let the original value be N

N – N/2…50 days

N/2 – N/4 …100 days

N/4 – N/8 …150 days

N/8 – N/16 … 200 days

N/16 – N/32 … 250 days

It will take 250 days for a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original
number

C is the correct option

Question 9

In the reaction above, X is {UTME 2005}
A. proton          B. neutron          C. electron             D. neutrino

Solution

In nuclear reaction, proton number and nucleon number must be conserved

On the reactant side, the total number of proton = 92 + 0 = 92, the total nucleon number = 235 + 1 = 236

On the product side, the total number of proton = 56 + 36 + A = 92, A = 0,  the total nucleon number = 144 + 90 + 2Z = 236, Z =(236-234)/2 = 2/2 = 1

For the product side to be conserved, the nucleon number on X is 1 and proton number is 0

1X0 = neutron

B is the correct option

Recommended: Short notes on Nuclear physics

Recommended: Question and answer on Nuclear for A level

How to solve questions on gas law for UTME

Question 1

The pressure of a given mass of a gas changes from 300Nm-2 to 120Nm-2 while the temperature drops from 127oC to –73oC. The ratio of the final volume to the initial volume is{UTME 2001}
A. 2 : 5              B. 4 : 5                     C. 5 : 2                     D. 5 : 4

Solution

P1V1/T1  = P2V2/T2

127 degree = 273+127 = 400 k

-73 degree = 273-73 = 200 k

300*V1/400 = 120*V2/200

300*200*V1 = 120*400*V2

V1/V2 = 120*400/300*200 = 24/30 = 4:5

V2/V1 = 5:4

D is the correct option

Question 2

The pressure of 3 moles of an ideal gas at a temperature of 27oC having a volume of 10-3m3 is{UTME 2002}
A. 2.49 x 105Nm-2       B. 7.47 x 105Nm-2      C. 2.49 x 106Nm-2       D. 7.47 x 106Nm-2
[R = 8.3J mol-1K-1]

Solution

PV = nRT

P = nRT/V

P = 3*8.3*(273+27)/10-3

P = 3*8.3*300/0.001

P = 7470000Nm-2 = 7.47 x 106Nm-2

D is correct option

Question 3

Which of the following gas laws is equivalent to the work done?{UTME 2007}
A. Pressure Law       B. Van der Waal’s Law        C. Boyle’s Law        D. Charles’ Law

Solution

PV = K(k=constant), this is Boyle’s Law and is equivalent to workdone

C is the correct option

Question 4

A sealed flask contains 600cm3 of air at 27oC and is heated to 35oC at constant pressure. The new volume is{UTME 2008}
A. 508cm3             B. 516cm3            C. 608cm3              D. 616cm3

Solution

V1/T1 = V2/T2

600/(273+27) = V2/(273+35)

600/300 = V2/308

2/1 = V2/308

V2 = 308*2 = 616cm3

D is the correct option

Question 5

At 40C, the volume of a fixed mass of water is{UTME 2009}
A. constant           B. minimum           C. maximum         D. zero.

Solution

Anomalous behaviour of water is between 0 – 4 degree. At this temperature water contract i.e volume of water decreases. So the volume will be minimum

B is the correct option

Question 6

The pressure of two moles of an ideal gas at a temperature of 270C and volume 10-2m3 is{UTME 2009}
A. 4.99 x 105 Nm-2       B. 9.80 x 103 Nm-2      C. 4.98 x 103 Nm-2       D. 9.80 x 105 Nm-2
[R = 8.313 J mol-1 K-1]

Solution

PV = nRT

P = nRT/V

P = 2*8.313*(273+27)/10-2

P = 2*8.313*300/0.01

P = 498780Nm-2 = 4.99 x 105 Nm-2

A is correct option

Question 7

The pressure of one mole of an ideal gas of volume 10-2m3 at a temperature of 270C is {UTME 2010}
A. 2.24 x 104 Nm-2        B. 2.24 x 105 Nm-2     C. 2.49 x 105 Nm-2      D. 2.49 x 104 Nm-2.
[Molar gas constant = 8.3 Jmol-1K-1]

Solution

PV = nRT

P = nRT/V

P = 1*8.3*(273+27)/10-2

P = 1*8.313*300/0.01

P = 249000Nm-2 = 2.49 x 105 Nm-2

C is correct option

Question 8

2000cm3 of a gas is collected at 27oC and 700mmHg. What is the volume of the gas at standard temperature and
pressure?{UTME 2012}
A. 1896.5cm3        B. 1767.3cm3       C. 1676.3cm3        D. 1456.5cm3

Solution

At s.t.p pressure = 760mmHg, temperature = 273K

P1V1/T1 = P2V2/T2

700*200/300 = 760*V2/273

V2 = 700*2000*273/(300*760) =1676.3cm3

C is the correct option

Question 9

A gas at a pressure of 105Nm-2 expands from 0.6m3 to 1.2m3 at constant temperature, the work done is{UTME 2013}
A. 6.0 x 104J          B. 7.0 x 107J            C. 6.0 x 106J         D. 6.0 x 105J

Solution

workdone = P(V2 – V1)

workdone = 100000(1.2 – 0.6) = 100000(0.6)

workdone = 60000 J = 6.0 x 104J

A is the correct option

short notes on gas law for UTME

Short notes on gas law for UTME

Boyle’s law

Boyle’s law states that the volume of a fixed mass of gas is inversely proportional to the pressure provided that the temperature remains constant.

If p is the pressure, v the volume and t the temperature in the Kelvin, Then the law can be stated as follows:

P α 1/v   i.e p = k/v (k is constant)

PV = constant (k is constant).

If P1 and V1 are the initial pressure and volume of the gas respectably, and P2 and V2 are the final pressure and volume after the change, then Boyle’s law can be written thus:

P1V1 = P2V2

Charles’ law

Charles’ law state that the volume of a fixed mass of gas is directly proportional to its absolute temperature (T) provided the pressure remains constant. Mathematically,

V α T  where V is the volume, T = absolute temp

V = kT(k is constant)

V/T  =k(k= constant)

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature and V2 and T2 are the final volume and temperature respectively.

Pressure law

Pressure law or gay Lussa’s law states that the pressure of a fixed mass of gas at constant volume is proportional to its absolute temperature.

Mathematically

P α T

P/T = k(k = constant)

P1/T1 = P2/T2

Where P1 and T1 are initial pressure and Temperature while P2 and T2 are final pressure and temperature respectively

The ideal gas equation (general gas equation)

The ideal gas equation is a combination of the three gas laws i.e. Boyle’s law, Charley’s law and the pressure law.

From Boyle’s law we have PV = constant at constant temperature

From Charles’ law we have V/T = constant at constant temperature

From pressure law we have P/T = constant at constant volume combining these three laws

Therefore PV/T = constant

P1V1/T1 = P2V2/T2

Example

An ideal gas has a pressure of 100cmHg at a temperature of 27 degree when its volume is 120cm3. When the pressure and temperature are increased to 150cmHg and 127 degree respectively. Calculate the new volume of the gas.

Solution

The temperature must be converted to kelvin. 1.e

The first temperature 27 degree = 273+27 = 300 kelvin

The second temperature 127 degree = 273 + 127 = 400 kelvin

P1V1/T1 = P2V2/T2

100*120/300 = 150*V2/400

V2 = (100*120*400)/(300*150) = 106.67cm3

The standard and temperature is 0 degree or 273k while the standard pressure is 760mmHg. The standard volume is 22.4 dm3.

The kinetic theory and its explanation of the gas laws

The kinetic theory of gas is made up of larger number of molecules. These molecules move about in their container randomly with different velocity, colliding with one another and the container walls. As the gas molecules hit the walls of the container and their velocity change as well as their momentum. The wall experience some force as a result of change in momentum of the gas molecule. Therefore, some pressure is exerted on walls of the vessel used by the collision of the molecules, since pressure is force per unit area.

Questions and answers on gas law for UTME