SS 2 SECOND TERM PHYSICS

LINEAR MOMEMTUM

CONTENT

1. Newton’s First Law of Motion, Impulse and Momentum
2. Newton’s First Law of Motion
3. Inertia
4. Applications of Newton’s First Law of Motion
5. Momentum
6. Impulse
7. Newton’s Third Law of Motion and Applications

 

Newton’s First Law of Motion, Impulse and Momentum

The influence of unbalanced forces and the laws governing the motion are discussed in this topic.

Newton’s First Law of Motion

The first law states that every object continues in its state of rest or uniform motion in a straight line unless acted upon by an external force.

Inertia

It is the tendency of a body to remain in its state of rest or uniform linear motion. Newton’s law of motion is called law of inertia.

Applications of Newton’s First Law of Motion

1. When a moving vehicle is suddenly brought to rest by the application of the brakes, the passengers suddenly jerk forward as they tend to continue in their straight line motion. That is why it is advisable to use a safety belt.
2. A car driver in a stationary car hit by another car from behind is likely to suffer neck injuries because when the car is hit, his body is pushed forward, but his head stays still and is jerked backward in relation to his body. It is advisable to have a headrest to protect the driver and passengers from injury.
3. A moving body comes to rest due to opposing forces such as air resistance, friction or pull of gravity.

Momentum (p)

The momentum of a body is defined as the product of its mass and its velocity. The S.I unit is kgms^-1 or Ns

P=mv

M = mass in kg,    V = velocity in ms^-1

Impulse

It is the product of the average force acting on a particle and the time during which it acts. It is numerically equal to change in momentum.

I=F×t,Ft=mv−mu.

mv is final momentum,     mu is initial momentum

The unit of impulse is Newton-second (Ns) or kgms ^{– 1}

Newton’s second law of motion: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

Force α change in momemtumtimeF α mv−mutF α m(v−u)t

But v−ut=aF α ma,f=kma,ifk=1f=ma

a is acceleration in ms^-2

f is resultant force acting on the body and the unit is Newton.

Calculations:

1. A net force of magnitude 0.6N acts on a body of mass 40g, initially at rest.

Calculate the magnitude of the resulting  acceleration.

Solution:

f=0.6N, m=40g=0.04kg,u=0, a=?f=ma0.6=0.04×aa=0.60.04=15ms−2

1. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms^-1 and rebounds with the same speed. Determine the impulse experienced by the ball.

f=0.6N m=5.0kgu=2mls,

v=−2m/s (negative because of a change in direction)

Impulse=mv−muImpulse=5(−2−2)=5×(−4)=−20Ns

EVALUATION

1. A ball of mass 0.1kg approaching a tennis player with a velocity of 10ms^-1, is hit back in the opposite direction with a velocity of 15ms^-1. If the time of impact between the racket and the ball is 0.01s, calculate the magnitude of the force with which the ball is hit.
2. A body of mass 20kg is set in motion by two forces 3N and 4N, acting at right angles to each other. Determine the magnitude of its acceleration.

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Newton’s Third Law of Motion and Applications

Newton’s Third Law of Motion

It states that Action and Reaction are equal and opposite. Or to every Action there is an equal and opposite Reaction.

Applications of Newton’s Third Law of Motion

1. Gun and bullet:

When a bullet is shot out of a gun the person firing it experiences the backwards recoil force of the gun. The recoil force of the gun ( reaction) is equal to the propulsive force(action) acting on the bullet.

Force is proportional to change in momentum,

The momentum of bullet is equal and opposite to the momentum of gun.

m_b – mass of bullet

v_b –  velocity of bullet

m_g –  mass of  gun

v_g – velocity of gun

mbvb=−mgvgvg=−mbvbmg

2. Rocket and jet propulsion:

The momentum of the stream of hot gases issuing out of the nozzle behind the jet or rocket impacts an equal and opposite momentum to the rocket or aeroplane which undergoes a forward thrust.

m_r – mass of rocket

v_r –  velocity of rocket

m_g –  mass of hot gas

v_g – velocity of hot gas

mrvr=−mgvgvr=−mbvbmg

 

Newton’s Third Law of Motion

This states that when a body acts on another body with a force F, the second body acts on the first body with an equal amount of force but in the opposite direction.

That is, for every action, there is an equal but opposite reaction.

 

Applications of the Newton Third Law of Motion

1. Garden sprinkler:

This has a Z-shaped tube mounted on a pivot and through which water flows into the tube.As water is forced out of the open end of the tube, the tube is pushed backward with a equal but opposite reaction. This way, the tube is able to spin round and round and water the field all around it.

2. Weightlessness or weight loss in a lift:

A person standing on a weighing machine in a lift descending with a certain acceleration will experience weight loss and if the downward acceleration of the lift is equal to the prevailing acceleration due to gravity at that location, the person becomes weightless and float around in the lift.

i. When the lift is ascending with acceleration a;

(Note: the value of R represents the reading on the weighing machine on which the person stands). In this case, the reading on the machine is greater than the usual weight W. The person will feel heavier as the lift ascends.

ii. When the lift is ascending but at constant velocity.

R−mg=NetforceR−mg=0R=mg

(Note that in this case, the reading recorded on the weighing machine is the exact weight of the body at that location)

iii. When the lift is descending with acceleration a

Mg−R=maR=mg−maR=m(g−a)

In this case, the person will feel lighter as the lift descend.

However, if the lift is descending with acceleration a  =  g

R=m(g−g)R=m(0)R=0

In this case, the person will experience weightlessness (free fall). In fact, he will float around in the lift

 

Worked Problems:

1. A tight rope walker of mass 60 kg stands in the middle of a rope and such that at his feet, the rope makes angle 5⁰ to the horizontal. Calculate the tension in the rope.[mediator_tech]

mg=Tsin5+Tsin560×10=T(2×0.087)6000.174=TT=3448N

Worked Example 2:

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player is 90.0 kg and he is accelerating at 1.2 m/s². What is the force of friction between the losing player’s feet and the grass?

ma=800−fr90×1.2=800−frfr=800−108fr=692N

Worked Example 3:

A woman of mass 57 kg stands on a weighing machine inside a lift ascending at 0.2 m/s². What is the reading on the machine?

R−mg=maR=ma+mgR=m(a+g)R=57(0.2+10)R=57×10.2R=581.4N

 

GENERAL EVALUATION

1. A bullet of mass 0.045kg is fired from a gun of mass 9kg, the bullet moving with an initial velocity of 200m/s. Find the initial backward velocity of the gun.

MECHANICAL ENERGY

CONTENT

1. The Principle of Conservation of Linear Momentum
2. Types of Collisions

 

The Principle of Conservation of Linear Momentum

In a closed system of colliding bodies, the total momentum after the collision is equal to the total momentum before the collision provided there is no net external force acting on the system.

Case 1: Consider two bodies A and B of masses m₁ and m₂ moving in the same direction with velocities u₁ and u₂ respectively. After collision their velocities were V₁ and V₂ as shown below.

Applying the principle of conservation of linear momentum,

M1U1+M2U2=M1V1+M2V2

Case 2: Consider two bodies A and B of masses m₁ and m₂ moving towards each other with velocities u₁ and u₂ respectively. After collision there velocities were V₁ and V_{2 .}

M1U1+M2U2=M1V1+M2V2

Assuming M1U1>M2U2,

Applying the principle of conservation of linear momentum,

M1U1+M2U2=(M1+M2)V

 

CALCULATIONS:

Example 1:

A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0ms^-1 collides perfectly with a stationary trolley of the same mass on the same platform. Calculate the total momentum of the two trolleys immediately after the collision.

Solution:

M_{1 =} 4kg,  U₁ = 10m/s,   U₂ = 0,   M_{2 =} 4kg

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=M1V1+M2V2SinceU2=0,M2U2=0

Momentum after collision (M1V1+M2V2)=M1U1=4×1=4.0kgms−1

Example 2:

A ball of mass 0.5kg moving at 10ms^-1 collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

M_{1 =} 0.5kg,  U₁ = 10m/s,   U₂ = 0,   M_{2 =} kg

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=(M1+M2)V0.5×10+0.5×0=(0.5+0.5)V5=1.0VV=5m/s

 

EVALUATION

1. State the following:

(a) Newton’s third law of motion[mediator_tech]

(b) Principle of conservation of linear momentum.

1. A ball P of mass 0.25kg loses one-third of its velocity when it makes a head-on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2m/s in the original direction of P. Calculate the initial velocity of P.

 

Types of Collisions

There are two major types of collisions, elastic and inelastic collisions.

Elastic Collision:

In an elastic collision both momentum and kinetic energy are conserved. This means that for two colliding bodies with masses m₁ and m₂  and initial velocities u₁ and u₂ and final velocities after collision V₁ and V₂ ,

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22=12M1V21+12M2V22

An example of perfectly elastic collision is a ball which bounces off the ground back to its original height.

Inelastic Collision:

In this case momentum is conserved but kinetic energy is not conserved. The energy lost is usually converted to heat, sound or elastic potential energy.

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22≠12M1V21+12M2V22

Example 3:

A body of mass 5kg moving with a velocity of 20m/s due south hits a stationary body of mass 3kg. If they move together after collision with a velocity v due south, find the value of v.

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision = total momentum after the collision.

M1U1+M2U2=(M1+M2)V5×20+3×0=(5+3)V100=8VV=12.5m/s

 

EVALUATION

1. Distinguish between perfectly elastic collision and perfectly in elastic collision.
2. A tractor of mass kg is used to tow a car of mass kg. The tractor moves with a speed of 3.0m/s just before the towing rope becomes taut. Calculate the

(i) speed of the tractor immediately the rope becomes taut.

(ii) loss in kinetic energy of the system just after the car starts moving.

MECHANICAL ENERGY: SIMPLE MACHINES

CONTENT

1. Definition of Machine and Simple Machine
2. Mechanical Advantage/Force Ratio
3. Velocity Ratio
4. Efficiency of a Machine
5. Lever
6. Pulley
7. Inclined Plane
8. Hydraulic Press
9. Screw
10. Wheel and Axle
11. Gear

[mediator_tech]

Definition of Machine and Simple Machine

An arrangement by which work can be done conveniently on a load or against a resistance is known as machine 

A simple machine is a machine in its simplest form. They are devices that use energy to do work. The work is being done by the machine when a small effort is used to overcome a large resistance.

A machine can be defined as a tool or devices that allows a force (or effort) applied to one point to overcome a resisting force (or load) at another point.

Simple machine can be classified into different categories namely the lever, the pulley, the inclined plane, the wedge, the wheel and axle, the screw, the hydraulic press.

Some common examples of simple machines are, scissors, drill brace, the shovel (a form of lever), the pulley at the top of a flagpole, the steering wheel of an automobile (a form of wheel and axle), and the wheelchair ramp (a form of inclined plane). An everyday example of a complex machine is the can opener, which combines a lever (the hinged handle), a wheel and axle (the turning knob), and a wedge (the sharpened cutting disk).

 

Mechanical Advantage/Force Ratio

Mechanical advantage/Force ratio of a machine is defined as the ratio of the load to the effort.

Mechanical Advantage =LoadEffort=Le

If the laod is bigger than the effort, the mechanical Advantage is greater than one.

In pratice,all machines have some friction in them and this reduces the efficiency. Part of the work put into a machine is thus always wasted in overcoming friction and in moving some parts of the machine. Thus, no machine is hundred percent (100%) efficient.

 

Velocity Ratio

The velocity ratio of a machine is defined as the distance moved by the effort to the distance moved by the load.

Velocity Ratio =Distance moved by EffortDistance moved by Load=dedL

 

Efficiency of a Machine

The efficiency of a machine is defined as the ratio of work obtained from the machine to work put into the machine expressed in percentage.

It also defined as the ratio of work output  of the machine to the total work input expressed in percentage.

Efficiency =Work outputWork input×100

Derivation of the Formula

Efficiency E=MAVR×100

The efficiency of a machine can be determined by taking ratio of the work output to work input of the machine or its velocity ratio and mechanical advantage.

Therefore:

Efficiency (E)=Work outputWork input×100

Efficiency (E)=Work done in loadWork done in effort×100

Since Work=force×distance

Efficiency

(E)=Load(L)×distance moved by load(dL)Effort(E)×distance moved by effort(de)×100

Therefore, E=L×dLe×(de)×100

But MA=Leand1VR=dLde

Therefore, E=M.AV.R×100

Mechanical Advantage (M.A)=Le

Efficiency (E)=MechanicalAdvantageVelocityRatio×100

Example 1:

A system of lever with velocity ratio 30 overcomes resistance of 2500 Newton when an effort of 125 Newton is applied to it, calculate

• The mechanical advantage of the system.
• It’s efficiency.

Solution:

Mechanical Advantage =LoadEffort=2500125=20N

Efficiency =M.AV.R×100

Efficiency =2030×100=2003=66.75[mediator_tech]

GENERAL EVALUATION

1. Explain what is meant by a machine.
2. Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine.
3. Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation E=M.AV.A×100
4. Explain why the efficiency of a machine is usually less than 100%.

 

Lever

In a lever the relative positions of Force(F), Effort(e) and Load(L) may vary and this leads to different types of lever. The lever operates on the principle of moment.

First Order Lever

In first order lever, the fulcrum is between the load and the effort e.g crowbar, claw hammer,  pliers, scissors, see-saw e.t.c

Taking moment about F gives;

Clockwise moment = anticlockwise moment

Y×L=X×ELE=XY=M.A=V.R

Second Order Lever

In second order lever, the load is between the effort and the fulcrum. E.g wheelbarrow, bottle opener, nut cracker

Third Order Lever

In third order lever,  the effort is between the load and fulcrum. E.g Forearm of a human being, tongs etc.

 

Pulleys

Definition of Pulley

A simple pulley is a fixed wheel with a rope passing round a groove in its rim. A load is attached to one end of the rope while effort is applied at the other end. If there is no friction, load is equal to the effort which is equal to the tension in the rope.

L = T = E, therefore  M.A = V.R

For a block and tackle or systems of pulley of n pulleys, the velocity ratio is equal to the number of pulley.

Velocity ratio = number of pulleys = n

A system of pulley is used for lifting loads. They are used by builders for hauling heavy loads to high floors or in loading and unloading ships.

CALCULATIONS:

Example 1:

The velocity ratio and efficiency of a system of pulleys are 6 and 80% respectively. How much effort is required to lift a load of mass 120kg  with this system? [g = 10ms^-1]

Solution:

V.R = 6,  Efficiency = 80%,  Load = 1200N, e = ?

Calculating M.A,

Efficiency =M.AV.R×10080=M.A6×100M.A=80×6100=4.8M.A=LoadEffortEffort=LoadM.A=12004.8=250N

EVALUATION

A pulley system with a velocity ratio of 6 is used to raise a load of 80N through a vertical height of 16m,[mediator_tech]

1. Draw a diagram of this arrangement
2. Calculate the effort required in the system, if its efficiency is 70%.
3. Calculate the work done.

 

Inclined Plane

An inclined plane is a sloping surface used for pulling or pushing a load up, rather than lifting them vertically.

Velocity Ratio =Distance moved by EffortDistance moved by Load=xh

Velocity Ratio =1sinθ

 

Hydraulic Press

Definition of Hydraulic Press

Hydraulic press is a device used to produce a very large force to compress or lift up a heavy load. It is used in the printing press where a large force presses the type with ink on it against the paper. Pressure is transmitted equally to all parts of a liquid at the same level.

 

If A₁ is the area of small piston and A₂ area of the large piston, pressure(P) is transmitted equally at the same level, then

P=EA1,E=P×A1P=LA2,L=P×A2

Therefore a small effort lifts a large load.

M.A=LE=A2A1=R2r2

If x and y are the distances moved by E and L respectively,  A₁x = A₂y

xy=A2A1=R2r2=V.R

Example 2:

An inclined plane of angle 15⁰ is used to raise a load of 4500N through a height of 2m. If the plane is 75% efficient, calculate (i) Velocity ratio of the plane (ii) Work done on the load

Solution:

M.A=LE=A2A1=R2r2LE=A2A1L=A2×EA1=200×4010=800N

EVALUATION

A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform      2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the

(i) Minimum effort applied in pulling up the box

(ii) Velocity ratio of the plane

(iii) Mechanical advantage of the plane

(iv) Efficiency of the plane

(v) Energy lost in the system

(vi) Work output of the man

(vii) power developed by the man given that the time taken to raise the box onto the platform is 50s. ( g = 10m/s²)

 

Screw

Definition of Screw

A screw is mechanical fastening device consisting essentially of an inclined plane wound spirally around a cylinder or a cone. The ridges formed by the winding planes are called threads, and depending on the intended use, the threads may be square, triangular, or rounded in cross section.

 

Parts of a Screw Thread

A screw thread is made of the following parts:

1. Thread or Lead: Is the distance along the screw’s axis that is covered by one complete rotation of the screw (360°).
2. Pitch: The distance between two corresponding points on adjacent threads
3. Plane or Start : It can be Single-start or double-start. Single-start- means that there is only one “ridge” wrapped around the cylinder of the screw’s body.Double-start” -means that there are two “ridges” wrapped around the cylinder of the screw’s body Each time that the screw’s body rotates one turn (360°), it has advanced axially by the width of two ridges.

Another application of the screw is Screw jack, Ratchet brace.
When a screw is turned through one complete revolution by the application of an effort ( on the screw head, using the handle of the ratchet brace, or the Tommy bar of the screw jack), the load moves a distance equal to the pitch(P), which is the distance between consecutive threads.

V.R=2πRP

R = Radius of the screw or length of the handle of the ratchet brace of length of the tommy bar.

 

Wheel and Axle

Definition of Wheel and Axle

A wheel and axle is really two machines in one because it consists of a wheel mounted rigidly upon an axle or drum of smaller diameter, the wheel and the axle having the same axis.

The wheel & axle device is similar to the lever simple machine. The wheel acts as the lever and the axle acts as a fulcrum. The lever applies force to the fulcrum, causing something to move.

If a rope is fixed onto the wheel and wound round it, leaving a free end where an effort E is to be applied, the rope attached to the axle is wound round in the opposite direction and the load attached to the free end,  For each complete rotation of the wheel, there is one complete rotation of the axle.

Let a and b be the radii of the wheel and axle respectively. Then

Velocity Ratio =Distance moved by EffortDistance moved by Load=2πa2πb=ab

Applications of the Wheel and Axle

Applications of the wheel and axle are:

1. Steering wheel of an automobile
2. Doorknob
3. Windlass.
4. Treadmill
5. Windmill
6. Waterwheel

 

Gear

Definition of Gear

A gear is a toothed wheel or cylinder used to transmit rotary or reciprocating motion from one part of a machine to another. Two or more gears, transmitting motion from one shaft to another, constitute a gear train.

Gears work on the wheel and axle principle. If  toothed wheel A drives wheel B resulting in turning forces,

Velocity Ratio =Number of teeth on the driven gearNumber of teeth on the driving gear

 

GENERAL EVALUATION

A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calculate the:

1. velocity ratio of the jack.[mediator_tech]
2. Mechanical advantage of the jack
3. Effort required in raising  the body.
4. Work done by the effort in raising the body (g = 10m/s², π = 22/7)

TEMPERATURE AND ITS MEASUREMENT

CONTENT

1. Definition of Temperature
2. The Thermometer
3. Types of Thermometer
4. Temperature Scales

 

Definition of Temperature

Temperature is a measure of the degree of hotness or coldness of a body.  It is measured by means of a thermometer.  The S. I. Unit of temperature is the Kelvin.  However, it is also measured in degree Celsius.

 

The Thermometer

The thermometer is the instrument used for measuring temperature.  There are various types of thermometer, e.g. liquid-in-glass thermometer and the thermocouple.  Each one makes use of the change in the physical properties of materials they are made of, to indicate temperature change.

 

Types of Thermometer

1. The liquid-in-glass thermometer

The liquid-in-glass thermometer uses liquid(mercury) as thermometric substance. In which the change in volume of the liquid measures the change in temperature.  Any liquid that will be used as a thermometric liquid must be a good conductor of heat, be easily seen in glass, have a high boiling point, have a low freezing point, have a low specific heat capacity, must not wet glass and must expand uniformly. The liquid-in-glass consists of the school thermometer, clinical thermometer, maximum and minimum thermometer.

(i) The clinical thermometer has a constriction and it has a short range (35⁰C – 43⁰C).  The narrow constriction prevents the mercury from flowing back into the bulb immediately after the thermometer has been removed from the patient’s body.

(ii) The school thermometer is used in school laboratory, it ranges from 0⁰C – 100⁰C. It has no constriction.

(iii) The Maximum and minimum thermometer is used to measure the maximum and minimum temperatures of the day. It contains alcohol at two upper part of the bent tube and mercury below.  A steel index is seen in the two tubes. When the temperature rises, since alcohol expands more than mercury, The alcohol expands and flows in clockwise direction and the mercury pushes steel index x upwards. The maximum temperature is read from the lower end of x. When the temperature falls the alcohol contracts and moves in anticlockwise direction, the mercury pushes steel index Y up. The lower end of Y indicates the minimum temperature.

EVALUATION

1. State three desirable properties of a thermometric liquid.
2. List four advantages of mercury over alcohol as a thermometric thermometer.

2. Constant – Volume Gas Thermometer

The constant – volume gas thermometer depends on the variation in the pressure of a gas at constant volume with changes in the temperature of the gas. On heating the bulb to a specific temperature, the gas expands and pushes the mercury down to tube A and consequently mercury level at C rises. The right side of AC of the manometer is moved up and down in order to bring the mercury level on the other side to its original position D to ensure that the volume of gas is constant. The pressure of gas is read from the manometer.

 

When the level of mercury in A is higher than that of D, then

P=h+H, where H is the atmospheric pressure.

But when the level of mercury is lower, then

P=H−h

To use the thermometer it must be calibrated by getting the gas pressure at 0⁰C , when the bulb containing is placed in pure melting ice. The pressure at 100⁰C is also gotten by placing it over stem.

θ−0100−0=Pθ−P0P100−P0

P_θ – resistance at temp θ⁰c

P₀ – resistance at O⁰C

P₁₀₀ – resistance at 100⁰C

Example 1:

A constant volume gas thermometer records pressure of a body as 250mm of Hg at 0⁰C and 350mmHg at 100⁰C. Calculate the temperature of the body when the gas pressure reads 300mmHg.

Solution:

p_θ = 300mmHg,   p₀ = 250mmHg,  P₁₀₀ = 350mmHg, θ = ?

θ=Pθ−P0P100−P0×100θ=300−250350−250×100θ=50100×100=50o

EVALUATION

1. A constant volume gas thermometer records pressure of a body as 300mm of Hg at 0⁰C and 400mmHg at 100⁰C. Calculate the temperature of the body when the gas pressure reads 350mmHg.

3. Platinum resistance thermometer

This thermometer depends on the variation in the electrical resistance of a conductor with temperature. The higher the temperature the greater the resistance. The ends of the wire are connected to a wheatstone bridge which measures the resistance at 0⁰C, at 100⁰C and the resistance at the desired temperature. The temperature can be calculated using the equation below.

θ−0100−0=Rθ−R0R100−R0θ=Rθ−R0R100−R0×100

R_θ – resistance at temp θ⁰c

R₀ – resistance at O⁰C

R₁₀₀ – resistance at 100⁰C

Example 2:

The electrical resistance of the element in a platinum resistance thermometer at 100⁰C, 0⁰C and room temperature are 75.000Ω, 63.000Ω and 64.992Ω respectively. Determine the room temperature.

R_θ = 64.992Ω,      R₀ = 63.000Ω,        R₁₀₀ = 75.000Ω,        θ = ?

θ=64.992−63.00075.000−63.000×100θ=1.9921.200×100=166.0oC

[mediator_tech]

4. The Thermocouple

A thermocouple consists of two different metals joined together by a circuit containing a galvanometer.  The working principle of a thermocouple depends on the variation of the electromotive force (e.m.f) between junctions.  The equation of the relationship is

E=a+bt+ct2, (a, b, c are constants)

GENERAL EVALUATION

1. Describe the action of resistance thermometer.
2. (a) Draw a labeled diagram of a clinical thermometer and explain how it works.
3. (b) Give the reasons for the following features of the thermometer

(i) a bulb of thin glass.

(ii) the constriction on the bulb.

(iii) a tube of fine bore.

1. State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.
2. (a) State three physical properties of  substances which may be used to measure temperature.

(b) State three desirable properties of a thermometric liquid.

(c) List four advantages and four disadvantages of mercury as thermometric liquid.

(d) Why is water considered as an unsuitable liquid for a thermometer?

(e) The resistance in the element in a platinum resistance thermometer is 6.750Ω at 0⁰c, 7.750Ω at 100⁰c and 6.900Ω at room temperature.

 

Temperature Scales

1. Celsius scale: The lower and upper fixed points are 0⁰C and 100⁰C.  The fundamental interval is divided into 100 equal parts.
2. Fahrenheit scale: The lower  and upper fixed points are 32⁰F and 212⁰F.  The fundamental interval is divided into 180 equal parts.
3. Absolute scale/Thermodynamic scale: The lower and upper fixed points are 273k and 373k.  The fundamental interval is divided into 100 equal parts.

The absolute scale of temperature is thermodynamic scale because it gives us the idea of the lowest possible temperature or absolute zero with the value of –273⁰C.

−273oC=ok=273k

The absolute zero is the lowest possible temperature below which nothing can be cooled since temperature is the measure of the average or mean kinetic energy of the molecules of a substance.  It follows that as we subtract heat from a substance, its temperature drops and hence its kinetic energy until it eventually becomes zero under which the molecules remain stationary.

To convert from one scale to another, use interpolation technique.

Unknown temperature – Lower fixed pointUpper fixed point – Lower fixed point=Known temperature – Lower fixed pointUpper fixed point – Lower fixed point20−0100−0=θ−32212−32θ−32=20×180100=36θ−32=36θ=36+32=68oF

MEASUREMENT OF HEAT

CONTENT

1. Molecular Explanation of Temperature
2. Measurement of Heat Energy
3. Heat Capacity
4. Specific Heat Capacity
5. Determination of the Specific Heat Capacity of a Solid by Method of Mixtures
6. Calculations Using Method of Mixtures
7. Determination of Specific Heat Capacity of a Solid by Electrical Method
8. Latent Heat
9. Determination of Specific Latent Heat of Fusion of Ice
10. Determination of Specific Latent Heat of Vaporization of Steam

 

Molecular Explanation of Temperature

According to the molecular theory solids vibrate about their mean position, when heat is applied to solid, the amplitude of vibration of these molecules become larger and their kinetic energy increases. From kinetic theory, the total translational kinetic energy of the molecules is directly proportional to the absolute temperature, thus increase in kinetic energy implies increase in temperature.

Measurement of Heat Energy

Heat is a form of energy called thermal energy that flows due to temperature difference. It is measured in joules.

The quantity of heat Q received by a body is proportional to its mass (m), and temperature change (0₂ –  0₁) and on the nature of the material the body is made of.

Thus Q∝(θ2−θ1)Q=MC(θ2−θ1)

C is a constant of proportionality called the specific heat capacity of the body, which depends on the nature of the body.

The quantity of heat energy possessed by a body depends on these three quantities:

(i) the change in temperature (θ₂ – θ₁)

(ii) the specific heat capacity of the body (C)

(iii) mass of the body (m)

Heat Capacity

This is the quantity of heat required to raise the temperature of the entire body by one degree rise in temperature(1k).  It is measured in Joules/K.

H = MC

 

Specific Heat Capacity

Specific heat capacity of a substance is the heat required to raise the temperature of a unit mass(1kg)  of the substance through a degree rise in temperature(1⁰C or 1K)

Q=MC(θ2−θ1)

Q is quantity of heat,  C is specific heat capacity,  (θ2−θ1) change in temperature and m is the mass of the substance.

C=QM(θ2−θ1)

The unit is JKg^-1K^-1

Example 1:

What is meant by the statement: The specific heat capacity of copper is 400Jkg^-1k^-1?

Solution:

It means the quantity of heat required to raise the temperature of a unit mass of copper through one degree rise in temperature is 400J

Example 2:

How much heat is given out when a piece  of iron of mass 50g and specific heat capacity 460Jkg^-1K^-1 cools from to 25⁰C.

Solution:[mediator_tech]

M=50g=0.05kg,=460Jkg−1k−1,θ2−θ1=85−25=60oCQ=MC(θ2−θ1)Q=0.05×460×60=1380J

Determination of the Specific Heat Capacity of a Solid by Method of Mixtures

The solid lead block is weighed on a balance to be M₁.  A lagged calorimeter is dried and weighed to be M₂.  It is then reweighed to be M₃ when half filled with water.  The initial temperature of the water is taken to be 0₁.

The lead block is suspended in boiling water with a temperature 0₂ after which it is transferred to the calorimeter and the mixture stirred to maintain a uniform temperature 0₃.

The specific heat capacity of the lead can be calculated using the fact that heat loss by the lead = heat gained by calorimeter and water.  Given the specific heat capacity of calorimeter and water to be C_c and C_w respectively.

M1Cc(θ2−θ3)=M2Cc(θ3−θ1)+(M3−M2)Cw((θ3−θ1))C1=(M2Cc)(θ3−θ1)+(M3−M2)Cw(θ3−θ1)M1(θ2−θ3)

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The hot solid should be quickly transferred to prevent loss of heat.

Calculations Using Method of Mixtures

Example 3:

An iron rod of mass 2kg and at a temperature of 280⁰C is dropped into some quantity of water initially at a temperature of 30⁰C . If the temperature of the mixture is 70⁰C ,  calculate the mass of the water.   [Neglect heat losses to the surroundings.] [Specific heat capacity of iron = 460Jkg^-1K^-1_, Specific heat capacity of water = 4200Jkg^-1K^–]

Solution:

Mass of iron rod = M₁ = 2kg

Temperature of hot iron rod = θ₂ = 280⁰C

Initial temperature of water  = θ₁ = 30^oC

Final temperature of mixture = θ₃ = 70⁰

Specific heat capacity of iron = C_{I  =} 460Jkg^-1K^-1

Mass of water = M₂ = ?

Heat lost by hot iron = heat gained by water

M1CI(θ2−θ3)=M2CW(θ3−θ1)M2=M1CI(θ2−θ3)CW(θ3−θ1)M2=2×460×(280−70)4200(70−30)M2=2×460×2104200×40M2=193200168000=1.15kg

Example 4:

A piece of copper of mass 120g is heated in an enclosure to a temperature of 125⁰C. it is then taken out of the enclosure and held in air for half a minute and dropped carefully into a copper calorimeter of mass 105g containing 200g of water at 20⁰C. The temperature of the water rises to 25⁰C. Calculate the rate at which heat is being lost from the piece of copper when it is held in air. (specific heat capacity of water is 4200Jkg ^{– 1 0}C ^{– 1} ,  specific heat capacity of copper is 400J kg ^{– 10}C ^{– 1}       WAEC)

Solution:

θ₁ = 125⁰C,  θ₂ = 25⁰C,  mass of copper (M_c) = 120g = (120/1000)kg = 0.12kg

Heat lost by copper =McCc(θ2−θ1)=0.12×400×(125−25)=480J

Mass of calorimeter (m_c) =105g=(1051000)kg=0.105kg

Specific heat capacity of calorimeter (c_c) =400Jkg−10C−1

Mass of water =(2001000)kg=0.2kg

Change in temperature Δθ=(25−20)oC

Heat gained by calorimeter and water

=McCcΔθ+MwCwΔθ=0.105×400×(25−20)+0.2×4200×(25−20)=4410J

Heat lost to air =4800−4410=390J

Therefore, rate of lost of heat to air

=39030=13Js−1

Determination of Specific Heat Capacity of a Solid by Electrical Method

To calculate the specific heat capacity C_b of a solid brass block, we make two holes in a weighed brass block into which a thermometer and a heating element connected to a source of power supply are inserted.  Oil is poured in the holes to ensure thermal conductivity.  Assuming no heat is lost to the surrounding, the total amount of electrical heat energy supplied by the coil, Ivt = heat gained by the brass, MC_b0

Ivt=MCbθ ——(1)

From v = IR (Ohms law)

I2Rt=MCbθ ——(2)

v2tR=MCbθ ——(3)

Example 5:

A liquid of specific heat capacity 3Jg ^{– 1} k ^{– 1} rises from 15⁰C to 65⁰C in one minute when an electric heater is used. If the heater generates 63KJ per minute, calculate the mass of the liquid.

Solution:

Specific heat capacity of liquid

C1=3Jg−1k−1=3000Jkg−1k−1Δθ=65−15=50oC

Heat supplied by heater = heat gained by water

Ivt = M_l x C_l Δθ where M_l = mass of liquid

63000=M1×3000×50M1=630003000×50=0.42kg

Example 6: A certain metal of mass 1.5kg at initial temperature of 27⁰C, absorb heat from electric heater of 75W rating for 4 minutes. If the final temperature was 47⁰C, calculate the specific heat capacity of the metal and its heat capacity.

Solution:

Time ‘t’ = 4 minutes = 4 × 60 = 240s. power IV = 75W,  mass of metal ‘m’ = 1.5kg.

Heat supplied by electric heater = heat gained by the metal

IVt=mcΔθ75×240=1.5×c×(47–27)75×240=1.5×c×20c=75×2401.5×20=600Jkg−1K−1

Heat capacity =mc=1.5×600=900JK−1

GENERAL EVALUATION

1. 250g of lead at 170⁰C is dropped into 100g of water at 0⁰C. If the final steady temperature is 12⁰C, calculate the specific heat capacity of lead.  (C_w = 4.2 x 10³ J/kgk)
2. A 2000W electric heater is used to heat a metal object of mass 5kg initially at . If a temperature rise  of  is obtained after 10min, calculate the heat capacity of the material.

 

Determination of Specific Heat Capacity of Liquid by Electrical Method

Apparatus:

Thermometer, liquid, calorimeter, heater, stop clock, chemical/beam balance

Method:

A calorimeter of known heat capacity (M_cC_c) is used and a known mass of liquid( M₁)  is placed in the calorimeter, the temperature of the liquid is recorded (θ₁₎). The known quantity of heat (VIt) is  recorded by taking readings from the voltmeter, ammeter and stop watch. The final temperature is recorded (θ₂).

Calculations:

Electrical energy supplied by the heater = Heat energy absorbed by the calorimeter and water.[mediator_tech]

VIt=M1CL(θ2−θ1)+McCc(θ2−θ1)CL=VIt−McCc(θ2−θ1)C1(θ2−θ1)

Latent Heat

Latent heat or hidden heat is experienced when there is a change of state ( melting, vaporization, condensation, freezing, sublimation), it is not visible in the thermometer because there is no change in temperature. There are latent heat of fusion and latent heat of vaporization.

Latent heat of fusion is the heat energy required to convert a substance from its solid form to its liquid form without change in temperature.

Specific latent heat of fusion of a substance is the quantity of heat required to change unit mass of a substance from solid to liquid without change in temperature. The unit is Jkg^-1.

Q = quantity of heat(in joules)

m = mass of substance(kg)

Q = ml

l=Qm

l = specific latent heat of fusion

Specific latent heat of vaporization of a substance is the quantity of heat required to change unit mass of substance from liquid to vapour without change in temperature.

Q=mll=Qm

Example 1:

How much heat is required to convert 20g of ice at 0℃  to water at the same  temperature? (Specific latent heat of ice = 336Jg^-1)

M=20gl=336Jg−1l=QmQ=mlQ=336×20=6720J

Example 2:

Calculate the quantity of heat released  when 100g of steam at 100℃ condenses to water. [Take specific latent heat of vaporization of water as Jkg^-1]

M=100g=0.1kgl=2.3×106Jkg−1l=QmQ=mlQ=2.3×106×0.1×105J

EVALUATION

Calculate the energy required to vapourise 50g of water initially at 80⁰C. [Specific heat capacity of  water = 4.2Jg^-1K^-1; specific latent  heat of vapourisation of water =  2260Jg^-1]

 

Determination of Specific Latent Heat of Fusion of Ice

Dried ice is dropped in weighed calorimeter(M₁) containing water of known mass(M₂-M₁) and known temperature(θ₁). The mixture is stirred continuously and more ice is added until the temperature of the mixture falls to about 10⁰C below the room temperature(θ₂). The content is reweighed to find the mass of the ice.

Heat lost by calorimeter and water in cooling from θ₁ to θ₂ = Heat gained by ice in melting to water at 0⁰c  + Heat gained by melted ice when its temperature rises from 0⁰C to θ_2.

M1CI(θ1−θ2)+(M2−M1)CW(θ1−θ)=(M3−M2)L+(M3−M2)CWθ1L=M1CI(θ1−θ2)+(M2−M1)CW(θ1−θ)−(M3−M2)CWθ1(M3−M2)

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The ice must be dried before it is put in the calorimeter.

Example 3:

Calculate the heat energy required to   change 0.1kg of ice at  to water  boiling at . [Specific heat  capacity of water = 4200Jkg^-1K^-1;  specific latent heat of fusion of ice =336000Jkg^-1].

Solution:

Q = Heat required to melt ice at 0⁰C  +  Heat required to change temperature of ice from 0⁰C to 100⁰C.

Q=ml+mcw(θ2−θ1)Q=0.1×336000+0.1×4200×(100−0)Q=33600+42000=75600=7.56×104J

EVALUATION

1. Calculate the heat required to convert 20g of ice at to water at .  [Specific latent heat of fusion   of ice =  336Jg^-1; specific heat capacity of water  = 4.2Jg^-1K^-1]
2. Explain what is meant by the following statements: The specific latent heat of fusion of ice is 3.4 x 10⁵Jkg^-1.

 

Determination of Specific Latent Heat of Vaporization of Steam

The calorimeter is weighed empty and the mass(M₁) is recorded. Water is poured into the calorimeter and the mass(M₂) recorded. Dried steam is passed into the lagged calorimeter containing water until the temperature of water rises to 25⁰and the steam is removed and the content stirred. The mass(M₃ )is recorded and the final steady temperature taken(θ₂).

Mass of water = M₂ – M₁ ,   Mass  of steam = M₃ – M₂

Heat lost by stem in condensing  + Heat lost by condensed stem in cooling from 100⁰C to θ₂  = Heat gained by water and calorimeter during the experiment.

(M3−M2)L+(M3−M2)(100−θ2)CW=(M2−M1)CW(θ2−θ1)+M1C(θ2−θ1)L=(M2−M1)CW(θ2−θ1)+M1C(θ2−θ1)−(M3−M2)(100−θ2)CW(M3−M2)

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The Steam must be dried

Example 4:[mediator_tech]

Calculate the energy required to vaporize 50g of water initially at 80⁰C . [Specific heat capacity of water = 4.2Jg^-1K^-1; specific latent  heat of vaporization of water = 2260Jg^-1]

Solution:

Q = Heat required to raise the temperature of water from 80⁰C to 100⁰C  +  Heat required to vaporize water

Q=mcw(θ2−θ1)+mlQ=50×42×(100−80)+50×2260Q=42000+113000=155000=1.55×105J

Example 5:

Heat is supplied to a liquid of mass 500g contained in a can by passing a current of 4A through a heating coil of resistance 12.5Ω immersed in the liquid. The initial temperature of the liquid is 24⁰C. The liquid reaches its boiling point in 10 minutes after the current is switched on. It takes a further 2 minutes after the liquid starts to boil away. Calculate

(a). The specific heat capacity of the liquid

(b). The specific latent heat of vaporization of the liquid (boiling point of liquid = 84⁰C, thermal capacity of can = 400J/K)

Solution:

(a) Mass of liquid = 500g = 0.5kg

Heat required to raise temperature of liquid from 24⁰C to 84⁰C (boiling point of liquid) is given as

Q=mc(θ2−θ1)=0.5×c×(84−24)=30c

c is the specific heat capacity of the liquid.

Heat required to raise temperature of can from 24⁰C to 84⁰C =400×60=24000J  (thermal capacity × change in temperature).

Heat supplied by heating coil is

IVt=I2Rt=4×4×12.5×10×60=120000J

Since this heat is used to raise the temperature of the can and the liquid to boiling point, we have

30c+24000=12000030c=120000–24000c=3200Jkg−1

(b) Let L be the specific latent heat of vaporization of the liquid.

Heat required to vaporize liquid =mL=0.5L

Heat supplied by current =I2Rt=4×4×12.5×2×60=24000J

Since this heat is required to boil away the liquid at 84⁰C, we have

0.5L=24000 (neglecting heat loss to the surrounding)

L=240000.5=4.8×104Jk−1

 

GENERAL EVALUATION

1. A cup containing 100g of pure water at 20^oC is placed in a refrigerator. If the refrigerator extracts heat at the rate of  840J per minute, calculate the time taken for the water to freeze. [Neglect  the heat capacity of the material of  the cup.] [Specific heat capacity of water = 4.2Jg^-1K^-1; specific latent heat of fusion of ice = 336Jg^-1].

EVAPORATION

CONTENT

1. Definition of Evaporation
2. Factors Affecting the Rate of Evaporation
3. Application of the Cooling Effect of Evaporation and Latent Heat in Refrigerators
4. Boiling Point
5. Effect of Temperature on S.V.P
6. Determination of the Boiling Point of a Liquid
7. Effect of Pressure and Dissolved Substances on Boiling and Freezing Point
8. Humidity
9. Relative Humidity
10. Measurement of Relative Humidity (Hygrometer)
11. Dew Point, Mist and Cloud

 

Definition of Evaporation

Evaporation is a change of state from the liquid to the gaseous state. It takes place at all temperatures and from the surface of the liquid. Volatile liquids such as Freon, ether, alcohol, vaporize easily. Evaporation causes cooling because latent heat is taken from the body to change the liquid to the gaseous state. Example the human utilizes the process of evaporation in cooling after exercise.

 

Factors Affecting the Rate of Evaporation

1. Temperature: The rate of evaporation increases as temperature increase.
2. Pressure: The rate of evaporation decreases with increase in pressure.
3. Area of liquid surface exposed: The greater the surface area of liquid exposed, the more rapid will be the evaporation.
4. The nature of the liquid: The lower the boiling point of the liquid, the greater will be the rate of evaporation.
5. Wind and dryness of the air: Dryness of the air around the body causes rapid evaporation from the body. Wind blows away liquid vapour and causes rapid evaporation.

 

Application of the Cooling Effect of Evaporation and Latent Heat

1. Refrigerator
2. Air conditioner

Application in Refrigerators

Refrigerators make use of the cooling effect of evaporation. The volatile liquid such as liquid ammonia or freon evaporates inside copper coil surrounding the freezing compartment, supported by electric pump which reduced the pressure.  As the volatile liquid evaporates in those coils, it absorbs heat from the surround air, consequently cooling the inside of the refrigerator and its content.

The vapour produced is pumped off into the condenser, where it is compressed by the pump and condenses back to liquid.  The latent heat given out during this condensation is quickly dissipated by an arrangement of cooling fins at the back of refrigerator.

Heat is eliminated by convection and radiation to the surroundings and by conduction into fins.  The liquid is again passed into the evaporator coil and thus the level of cooling is regulated by a thermostat connected to the switch.

EVALUATION

1. What is evaporation?
2. Mention and explain that affects the rate of evaluation.
3. Explain the mode of operation of the refrigerator.

 

Boiling Point

Meaning of Boiling Point

Boiling happens at a particular temperature and pressure, the temperature of the liquid remains steady during boiling. Boiling occurs throughout the entire volume of the liquid and wind has no effect on boiling.

Boiling point of a liquid is that temperature at which its saturated vapour pressure is equal to the atmospheric pressure.

Saturated vapour is a vapour that is in contact with its own liquid within a confined space. The pressure exerted by the saturated vapour is known as saturated vapour pressure.

Saturated vapour pressure can be defined as the pressure exerted by the vapour of the liquid which is in dynamic equilibrium with its own liquid.

[mediator_tech]

Effect of Temperature on S.V.P

Increase in the temperature will increase the S.V.P of the liquid.

 

Determination of the Boiling Point of a Liquid

Determination of the Boiling Point of a Liquid

The J-shaped tube is filled with mercury and a drop of liquid is introduced into the space on top of the mercury in the shorter arm. The liquid evaporates and the pressure of the vapour depresses the mercury level slightly. The tube is heated in a beaker of water as the temperature increases, the liquid evaporates and more drops introduced. The s.v.p of the liquid increases. At a stage the mercury levels in both arms of the tubes become equal and the temperature is read at this point. This temperature is the boiling point of the liquid because at this temperature s.v.p of the liquid is equal to the atmospheric pressure.

Precautions:

1. A pure sample of the liquid must be used.
2. The mixture should be well stirred to ensure even distribution of heat.

 

Effect of Pressure and Dissolved Substances on Boiling and Freezing Point

Effect of Pressure on Boiling Point

1. Reduced pressure lowers the boiling point

If a flask of boiling water is allowed to cool until the water stops boiling and cold water is now poured over the bottom of the flask, boiling resumes and cease when the pouring of water stops. The poured cold water causes condensation of the vapour inside the flask, which consequently reduced the pressure on the surface of the water.  This reduced pressure lowers the boiling point of the liquid and make it boils again.  Hence reduced pressure always lowers the boiling point of a liquid.  Dissolved impurities raises the boiling point of liquid.

2. Increased pressure raises the boiling point

The fact that increased pressure raises the boiling point is put into a useful application in the pressure cooker.  The increased pressure of the trapped gas above the liquid raises the boiling of the liquid inside the cooker.  This provides a high cooking temperature needed to conserve fuel and save time. Pressure cookers are advantageous especially in mountainous area where atmospheric pressure is low.

Effect of Pressure on Freezing Point

Increase pressure lowers the melting point of all substances that expand on solidifying(ice block), but for substances that contracts on solidifying(paraffin), their melting point is raised by increased pressure.

 

Effect of Dissolved Impurities on Boiling Point

Dissolved impurities raises the boiling point of pure liquid.

 

Effect of Dissolved Impurities on Melting/Freezing Point

Dissolved impurities lowers the melting point of pure solid or the freezing point of pure liquid.

EVALUATION

1. Define the boiling point of a liquid.
2. Describe an experiment to determine the boiling point of a small quantity of a liquid.

 

Humidity

It is the water vapour content in the atmosphere. If there is high percentage of water in the air, the air is humid or moist.

Relative Humidity

It describes how moist the air is. It is defined as the ratio of the mass of water vapour present in a certain volume of air to the mass of water vapour required to saturate the same volume of air at the same temperature. It is expressed in percentage. Very high humidity or very low humidity makes one uncomfortable.

Measurement of Relative Humidity (Hygrometer)

Hygrometer is used to measure relative humidity. It consists of the dry bulb thermometer which reads temperature of dry air, and the wet bulb thermometer which records a lower temperature because evaporation of water from the  moist cloth cools the bulb. The lower the humidity, the lower the temperature of the wet bulb thermometer relative to the dry bulb thermometer. The difference between the two temperatures is high. If the humidity is high water evaporates slowly from the wet bulb. The difference between the two temperatures is small. The difference in the readings of the thermometers is a measure of the relative humidity.

Dew Point

It is the temperature at which the water vapour present in the air is just sufficient to saturate it.

Mist

They are condensed  water vapour( droplets)  suspended  in air near the earth surface when air have been  cooled below its dew point. Mist reduces visibility. In severe mist situation, we have fog.

Cloud

Cloud is a mass of small water droplets that float in the air. Cloud is high up in the atmosphere.

 

GENERAL EVALUATION

1. Explain the following terms: (i) Humidity (ii) Mist (iii) Dew point (iv) Cloud

GAS LAWS

CONTENT

1. Measurement of Gas Pressure
2. Gas Laws
3. Pressure Law or Gay-Lussac’s Law
4. Experimental Verification of Pressure Law
5. Boyle’s Law and its Application
6. Experimental Verification of Boyle’s Law
7. Charles’ Law and its Application
8. Experimental Verification of Charles’ Law
9. Cubic or Volume Expansivity (γ) of Gas
10. General Gas Law
11. The Ideal Gas Equation

 

Measurement of Gas Pressure

A manometer is used to measure Gas pressure gas pressure. One end of the tube is open to the atmosphere while the other end is connected the gas supply. When the gas is released, the gas exerts pressure on the water causing it to rise up to a height  h, When the water levels are steady;

Pressure of gas = pressure of liquid at B    (pressure at the level in a liquid are equal)

Pressure of gas = H + h      (H = Atmospheric pressure)

 

Gas Laws

In an attempt to study the behaviour of gases in relation to volume, temperature and pressure, the following conditions were investigated

1. Variation of volume with pressure at constant temperature, Boyle’s law (Pv=constant)
2. Variation of pressure with temperature at constant volume, pressure law (PT=K)
3. Variation of volume with temperature at constant pressure, Charles law (VT=constant)

 

Pressure Law or Gay-Lussac’s Law

Gay-Lussac’s law states that the pressure of a fixed mass of gas at constant volume is proportional to the absolute temperature of the gas.

PT=Constant,P1T1=P2T2

 

Experimental Verification of Pressure Law

It is performed using a constant volume Gas thermometer. The pressure of the gas at 0⁰C is first obtained by addition of ice shavings to water originally cooled to 0⁰C and the manometer is adjusted until the mercury level  coincides with the fixed mark C. The difference between the mercury levels (h)  is the pressure of the gas at that temperature. The water is heated increasing the temperature at intervals of 20⁰C up to 100⁰C and corresponding heights (h) taken. A graph of pressure against temperature plotted gives a straight line proving that pressure increases linearly with temperature.

Precautions:

1. The bulb must be dry and contain only dry air.
2. Position of the fixed mark must not shift throughout the experiment.

Question 1:

A given mass of gas has a pressure of  80Nm^-2 at a temperature of. If the temperature is reduced to  with the volume remaining constant, the new pressure is

Precautions:

1. The bulb must be dry and contain only dry air.
2. Position of the fixed mark must not shift throughout the experiment.

Question 1:

A given mass of gas has a pressure of  80Nm^-2 at a temperature of . If the temperature isreduced to  with the volume remaining constant, the new pressure is?

Solution:

PT=Constant,P1T1=P2T2

T₁ = 47 + 273 = 320K,     T₂ = 27 + 273 = 300K,   P₁ = 80Nm^-2,   p₂ = ?

P1T1=P2T280320=P2300P2=80×300320=75Nm−2

EVALUATION

1. The pressure of air in a tyre is 22.5Nm^-2 At 27⁰c. If the air in the tyre heats up to 47⁰c, calculate the new pressure of the air, assuming that no air leaks out and that the change in volume of the air can be neglected.

[mediator_tech]

Boyle’s Law and its Application

Boyle’s Law

It states that the volume of a fixed mass of gas varies inversely as its pressure, provided the temperature remains constant.

P1V1=P2V2

Experimental Verification of Boyle’s Law

Dry air is introduced into the tube B and the atmospheric pressure noted(H). The tube is kept steady while tube A is raised or lowered to obtain at least five different heights(h) of mercury and the corresponding lengths of the dry air in tube B. Since the tube is of uniform cross-sectional area, The length L is proportional to the volume of air.

A graph of pressure against inverse of volume gives a straight line.

Example 1:

The set-up illustrated above shows a capillary tube of uniform cross-sectional area in two different arrangements.Using the data in the diagrams,  calculate the pressure of the atmosphere.

Solution:

P₁  = ( A + 15)cmHg, V₁ = 30cm,  P₂ = (A – 15)cmHg,  V₂ = 45cm

P1V1=P2V2(A+15)(30)=(A−15)(45)30A+450=45A−675450+675=45A−30A1125=15AA=75cmHg

 

Charles’ Law and its Application

Charles’ Law

It states that the volume of a fixed mass of gas is directly proportional to its absolute temperature (T), provided the pressure remains constant.

V1T1=V2T2

Experimental Verification of Charles’ Law

Dry gas is trapped by pellet of mercury in the capillary tube of uniform diameter. The temperature of the gas at 0⁰C is obtained by adding ice shavings to the water in the beaker until  the temperature is 0⁰C. The water is heated and lengths of the gas column at different temperatures taken and recorded. The water must be stirred before taken the readings. The length of the gas column is taken as the volume of the gas because the capillary tube is of uniform cross sectional area.

A graph of length of gas column against corresponding temperature is plotted, a straight line is obtained with slope approximately 1/273.

 

Cubic or Volume Expansivity (γ) of Gas

It is defined as the increase in volume per unit volume at 0⁰C per degree rise in temperature

γ=Increase in volumeVolume 0^oC × Temperature rise=V1−V0V0(T1−T0)

V₁ – V₀ = Increase in volume or expansion

T₁ – T₀ = Temperature rise

V₀ = Volume at 0^oC

V1=V0γ(T1−T0)+1

Question 1:

Dry hydrogen is trapped by a pellet of mercury in a uniform capillary tube closed at one end. If the lenght of the  column of hydrogen at 270⁰C is 1.0m, at what temperature will the lenght be 1.20m.

T₁ = 270 + 273 = 543K,     T₂ = ?,   V₁ = 1.0m,   V₂ = 1.20m

V1T1=V2T21.0543=1.20T2T2=543×1.201.0=651.6k

Question 2:

The cubic expansivity of a certain gas at  constant pressure is 1/273K^-1. If a given mass of the gas is held at constant pressure and its volume at 0⁰c is 273m³. Determine the volume of the  gas at 273⁰c.

γ=Increase in volumeVolume 0^oC × Temperature rise=V1−V0V0(T1−T0)V1=V0[γ(T1−T0)+1]V1=273[1273(273−0)+1]V1=273[1273×273+1]V1=273[1+1]V1=273×2=546cm3

 

General Gas Law

The general gas law is the combination of the Boyle’s, Charles and Pressure Law.

It follows that PVT=Constant,P1V1T1=P2V2T2

[mediator_tech]

The Ideal Gas Equation

For an ideal gas of n moles, having a volume V and pressure P, the ideal gas equation is given as:

PV=nRT

V –  volume of gas in m³.

n –  number of mole of gas in mol

T –  temperature of gas in Kelvin

R –  molar constant for an ideal gas  ( R = 0.082 Jmol^-1T^-1)

Example 1: 500cm³ of a gas is collected at 0⁰c and  at a pressure of 72.0cm of mercury. What is the volume of the gas at the same temperature and a pressure of 76.0cm of mercury?

Solution:

P1V1T1=P2V2T272×500273=76×V2273V2×76×273=72×500×273V2=72×500×27376×273V2=473.7cm