Category: Physics Notes PPT

The transformer converts electrical energy from one circuit to another. Electromagnetic induction is used to do this. It’s known as a voltage converter because it can convert high voltage to low voltage and vice versa. A good-condition transformer is made up of two windings, the main and secondary windings. Step up and step down transformers are the two types of transformers provided.

Transformer Formula

A transformer is an electrical device that allows us to maintain power when increasing or decreasing the voltage in an alternating current electrical circuit. In the case of a perfect transformer, the power entering the equipment is equal to the power received at the output. There is a small percentage of losses in real machinery. Based on the phenomena of electromagnetic induction, it is a device that transforms alternating electrical energy of one voltage level into alternating electrical energy of another voltage level.

The power of an electric circuit is calculated by multiplying the voltage by the current intensity. The value of the power in the primary is the same as the power in the secondary, as in the case of a transformer.

(Input voltage on the primary coil) x (Input current on the primary coil)

(Output voltage on the secondary coil) x (Output current on the secondary coil)

Transformer Equation can be Written As,

[ V_{p} times I_{p} = V_{s} times I_{s} ]

If we know the input voltage and the number of turns on the primary and secondary coils, we can calculate the transformer output voltage.

[frac{Input, Voltage, on, the, Primary, Coil}{Output, Voltage, on, the, Secondary, Coil}] = [frac{Number, of, turns, of, Wire, on, the, Primary, Coil}{Number, of, turns, of, Wire, on, the, Secondary, Coil}]

Transformer Equation Can be Written as Follows:

[ frac{V_{p}}{V_{s}} = frac{N_{p}}{N_{s}} ]

Where

[V_{p}] = Primary voltage

[V_{s}] = Secondary voltage

[N_{p}] = number of turns in the primary

[N_{s}] = number of turns in the secondary

[I_{s}] = Input current on the secondary coil

[I_{p}] = Input current on the primary coil

Definition of Transformer Formula

The formula calculates the efficiency of a transformer. The transformer is an electrical device that transfers electricity from one circuit to another circuit using magnetic induction. The transformer has two coils, a primary coil and a secondary coil instead of wires with voltage differences in it. Transformers are used daily by people for different purposes as they use them as inductors or motor protection devices. The transformer has two types of transformer, step up and step down.

Types of Transformer Formulas

1. Step-up Transformer: A step-up transformer is used to increase the voltage of an electrical current. It does this by taking a low voltage input and increasing it to a higher voltage output. The step-up transformer accomplishes this by using a larger number of turns in the primary coil.

A step-up transformer takes a low voltage and steps it up to a higher voltage by increasing the number of turns on the primary coil. The image above shows how this is done with an input of 12 volts and an output of 120 volts. The increase in voltage is due to the increased number of turns on the primary coil which then causes a decrease in current. This is important as it allows for smaller wires to be used when transferring power over long distances.

The step-up transformer can also be used as a boost converter, which takes low DC voltage and converts it into high-voltage AC voltage. This is used to power devices that require a high voltage, such as an electric motor.

2. Step-down Transformer: A step-down transformer is used to decrease the voltage of an electrical current. It does this by using a larger number of turns in the secondary coil.

A step-down transformer takes high voltage and steps it down to a lower voltage by increasing the number of turns on the secondary coil. The image above shows how this is done with an input of 120 volts and an output of 12 volts. The increase in Voltage is due to the increased number of turns on the secondary coil which then causes less resistance or loss of energy, resulting in higher efficiency compared to its counterpart (step-up Transformer).

Turn Ratio

A measure for describing how many more or fewer windings there are in the Transformer’s secondary coil when compared to its primary. The ratio of turns is expressed as Ns/Np, where “Ns” represents the number of windings in the Secondary Coil and “Np” is equal to the number of windings on a Primary Coils

The Transformer Formula: Transformer Efficiency = Output Voltage / Input Voltage * Turn Ratio (Ns/Np)

An efficient transformer has a high turn ratio which means that it contains more coils or wires wrapped around each other inside with less resistance making them more power-efficient than low turn-ratio transformers. In addition, they can also be used for voltage step-up applications if their input is compared to the output. Transformers can be found in many devices such as microwaves, washers, and TVs.

The transformer formula is used to calculate the efficiency of a transformer. The transformer is an electrical device that transfers electricity from one circuit to another circuit using magnetic induction. The transformer has two coils, a primary coil and a secondary coil instead of wires with voltage differences in it. Transformers are used daily by people for different purposes as they use them as inductors or motor protection devices. The transformer has two types of the transformer, step up and step down

Some Common Applications for a Step-up Transformer Are as Follows

• Converting low voltages from solar panels or batteries to a higher voltage required by appliances or electrical equipment

• Step-up converter for 12 Volt DC systems to power 24 Volt or 48 Volt loads

• Boosting the voltage of an Alternating Current (AC) system to charge lead acid or lithium-ion batteries

Efficiency of Transformer Formula

The efficiency of a transformer is denoted by the letter ‘η’ and is defined as the ratio of output in watts (or kW) to input in watts (or kW) (and is also known as commercial efficiency).

The efficiency of transformer formula is simply as follows,

Efficiency = [frac{Output, Power}{Output , Power + Losses}] x 100%

Transformer Turns Ratio Formula

The number of turns on the primary winding divided by the number of turns on the secondary coil is the transformer turns ratio. The transformer turns ratio affects the transformer’s predicted functioning as well as the voltage required on the secondary winding. When a secondary voltage lower than the primary voltage is required step-down transformer – the number of turns on the secondary must be lower than in the primary, and vice versa for step-up transformers when the transformer turns ratio steps-down the voltage, it steps-up
the current, and vice versa, so that the voltage and a current ratio of an ideal transformer are directly related to the number of turns on the secondary.

The Transformer Ratio Formula for Voltage Is as Follows

[ K = frac{V_{1}}{V_{2}} ]

Where,

[V_{1}] = Primary voltage 

[V_{2}] = Secondary voltage 

The Transformer Ratio Formula for Current Is as Follows

[K = frac{I_{1}}{I_{2}} ]

Where,

[I_{1}] = Primary current

[I_{2}] = Secondary current

Step-up Transformer Formula

A step-up transformer is a type of transformer that transforms low voltage (LV) and high current from the primary side to high voltage (HV) and low current on the secondary side. 

The primary coil turns are smaller than the secondary coil turns in a step-up transformer, which converts a low primary voltage to a high secondary voltage.

The step-up transformer formula is as follows:

[V_{s} = frac{N_{s}}{N_{p}} times V_{p}]

Where,

[N_{p}] = number of turns in the primary

[N_{s}]  = number of turns in the secondN

[V_{p}] = Primary voltage,

[V_{s}] = Secondary voltage,

Step Down Transformer Formula

A step-down transformer converts a high primary voltage to a low secondary voltage. The primary winding of a coil in a Step Down Transformer has more turns than the secondary winding. 

The Step Down Transformer Formula Is as Follows

[V_{s} = frac{N_{s}}{N_{p}} times V_{p}]

Where 

[V_{p}] = Primary voltage

[V_{s}]= Secondary voltage

[N_{p}] = number of turns in the primary

[N_{s}]  = Number of turns in the secondary

Solved Examples

Ex.1. The number of primary and secondary windings is 90 and 120 respectively. The secondary voltage is given by 310V, which determines the primary voltage.

Solution:

Given:

Np = 90,

Ns= 120

Vs = 310V

By using the transformer calculation formula, we get:

Vp/Vs=Np/Ns

Vp=Ns/Np x VS

VP= 90/120  x 310

 Vp = 232.5 volt

Ex.2. The number of primary and secondary windings is 110 and 240, respectively. The primary voltage is given by 300V, which determines the secondary voltage.

Solution:

Given:

[N_{s}] = 110,

[N_{s}]= 240

[V_{p}] = 300V

The transformer formula is given by

[frac{V_{p}}{V_{s}} = frac{N_{p}}{N_{s}}]

[V_{s} = frac{N_{s}}{N_{p}} times V_{p} ]

[V_{s}] =240/ 110 x 300

[V_{s}] = 654.5 volts

The branch of physics that deals with electricity, electronics, and electromagnetic concepts, is known as electrical. Electrical formulae are very helpful in determining the value of a parameter in any electrical circuit. Voltage, current, power, resistance, and other electrical formulae are the most often used.

Understanding how the various units of electricity can work together can certainly help from a system of water pipes. Voltage represents water pressure, the current is represented flow rate, and resistance represents pipe size in this analogy. Ohm’s Law, volts, amps, ohms, and watts are all significant fundamental components of electricity. According to Ohm’s Law, the voltage is equal to the current flowing in a circuit multiplied by the resistance of that circuit. The base unit for measuring voltage is known as volts. The ampere, abbreviated as “amp” or “A,” is the fundamental unit of electric current in the International System of Units.

Some commonly used electrical formulae are included below, and they may be useful to you.

Electric Field Formula

An electric field is a region created by an electric charge around it, the influence of which can be observed when another charge is introduced into the field’s region.

The Electric Field formula is given by,

E = F/q

Where,

F = Force

q = Charge

[]

Potential Difference Formula 

The potential between two points (E) in an electrical circuit is defined as the amount of work (W) done by an external agent in moving a unit charge (Q) from one point to another. The potential difference formula is expressed as, 

E = W/Q

Where,

E =The electrical potential difference between two points.

W = Work done in moving a charge from one point to another.

Q = Quantity of charge in coulombs.

[]

Electric Power Formula

Electric power may be defined as the rate at which work is completed. The watt is the SI unit for power and is written as P. The time, voltage, and charge are all connected by the power formula. Ohm’s law can be used to change the formula. The formula of electric power is as follows:

P = VI

The formula of electric power in term of Ohm’s law is as follows,

P = I²R

Or

P = V²/R

Where

Q= Electric charge

V= Voltage

t= Time

R= Resistance

Electric Potential Formula

The charge possessed by an object and the relative position of an object with respect to other electrically charged objects is the two elements that give an object its electric potential energy.

When an item moves against an electric field, it gains energy that is known as electric potential energy. The electric potential is calculated by dividing the potential energy by the quantity of charge for any charge.

The electric potential energy formula at any point around a point charge is given by:

[V=ktimes[frac{q}{r}]]

Where,

V = Electric potential energy

q = Point charge

r = Distance between any point around the charge to the point charge

k = Coulomb constant; k = 9.0 × 10⁹ N

[]

Electric Flux Formula

The electric flux is the total number of electric field lines passing through a given area in a given period of time.

The electric flux formula is expressed as,

𝜑_p = EA

When the same plane is tilted at an angle ϴ, the projected area is Acosϴ, and the total flux through the surface is:

𝜑 = EA cosƟ

Where,

E = Magnitude of the electric field.

A = Area of the surface.

Ɵ = Angle made by the plane.

[]

Electric Current Formula

An electric current is the constant flow of electrons in an electric circuit. When a potential difference is applied across a wire or terminal, electrons move. The rate of change of electric charge via a circuit is known as electric current. This current is equal to the circuit’s voltage and resistance. The symbol for it is I, and the SI unit is Amperes. The electric charge and the time are related to the electric current.

The electric current formula, according to Ohm’s law, will be,

[I=frac{V}{R}]

Where,

V = Voltage

R = Resistance

I = Current

[]

Electric Charge Formula

Electric Charge is the property of subatomic particles that causes to experience a force when placed in an electromagnetic field. The S.I unit of electric charge is coulomb and the symbol is Q. The electric charge formula is given by,

Q = I x t

Q = Electric Charge

I = Electric current 

t = Time

[]

Solved Examples

Ex.1. A Wire Carrying a Voltage of 21 Volts is Having a Resistance of 7⍵. Calculate the Electric Current.

Solution:

Given: Voltage = 21 V,

Resistance R = 7 ω

The electric current formula is,

[I=frac{V}{R}]

[I=frac{21}{7}]

I = 3 Amperes

Hence electric current is 3 Amp.

Ex.2. A force of 13 N is Acting on the Charge at 9 μ C at any Point. Determine the Electric Field Intensity at that Point.

Solution

Given

Force F = 13 N

Charge q = 9 μ C

Electric field formula is given by

[E=frac{F}{q}]

[E=frac{13}{9times10^{6}}]

E = 1.45 10^-6 N/C

Ex.3.  If The Current And Voltage of An Electric Circuit Are Given As 3.5A And 16V Respectively. Calculate The Electrical Power?

Answer:

Given measures are,

I = 3.5A and V = 16V

The formula of electric power is,

P = VI

P = 16 × 3.5 = 25

P = 56 watts.

Specific heat is the heat energy required to change the temperature of one unit mass of a substance of a constant volume by 1 °C. 

The specific heat capacity of a substance is the amount of energy needed to change the temperature by 1 unit of material of 1 kg mass.

The SI unit of specific heat and specific heat capacity is J/Kg. The specific heat formula and the formula of specific heat capacity will be discussed here.

Thermal Capacity Formula

The thermal capacity or heat capacity is a physical property of matter or substance. We define this property as the amount of heat supplied to a given mass of a material to generate a change of unit temperature.

The thermal capacity SI unit is Joule per Kelvin or J/Kg. Besides this, the heat capacity formula or the thermal capacity formula is:

[C = lim frac{Delta Q}{Delta T}]

[Delta T rightarrow 0]

Here,

Delta Q = It is the amount of heat that must be added to the object of mass  ‘M” to raise its temperature by delta T.

Also, the value of the above parameter varies considerably depending on the initial temperature {ΔT} T of the object and the pressure {ΔP}P applied to it.

Molar Heat Capacity Formula

Molar heat capacity is the amount of heat needed for the temperature rise of a given substance by 1 ⁰C. The molar heat capacity formula is given by:

C_m = C/n

Here,

C_m = molar heat capacity

C = heat capacity

And,

n = number of moles

We see that “n” is the number of moles of the sample. The number of moles can be determined by the following formula:

[= frac{Quantity , of , a , sample , or , mass , of , the , substance}{molar , mass} = frac{m}{M}]

Now, let’s understand the specific heat capacity equation:

Specific Heat Formula 

The specific heat capacity formula is:

[Q = mc Delta t]

Or,

[c = frac{Q}{m Delta t}]……(1)

Here,

Q  is the heat energy

m = mass in Kg

c = specific heat capacity, and

[Delta ]

[Delta ]

t is the temperature change in Kelvin

Also, the change in temperature is given by:

Δ T = (T_f – T_i)

Where Tf is the final temperature and Ti is the initial temperature in K.

Unit of Specific Heat Capacity

The unit of specific heat capacity is J/Kg. K, or J/Kg °C.  

Dimensional Formula of Specific Heat

The dimensional formula of specific heat is calculated as;

The dimensional formula of heat is [[M^{1} L^{2} T^{-2}]]

The dimensional formula of m is [M^{1} L^{0} K^{0}]

The dimensional formula of [Delta T] is [M^{0} L^{0} K^{1}]

Step 1: Now, putting the dimensional formulas of Q, m, and [Delta T]:

[= frac{[M^{1} L^{2} T^{-2}]}{[M^{1} L^{0} K^{0}][M^{0} L^{0} K^{1}]}]

Step 2: Canceling out the common terms:

[= M^{0} L^{2} T^{-2} K^{-1}]

So, the dimensional formula of specific heat capacity is [= M^{0} L^{2} T^{-2} K^{-1}]

Specific Latent Heat Formula

Specific latent heat is the measure of the heat energy (Q) released/absorbed per mass (m) during a phase/state change.

The specific latent heat formula is:

[Q = mL]

Or,

[L = frac{Q}{m}]

Here,    

Q = the heat absorbed or released depending on the direction of the transition. It is    measured in  KJ

L = specific latent heat of the material. It is measured in KJ/kg.

m = mass of the substance. It is measured in Kg.

Points to Remember

• Transferred heat depends on three factors which are – The change in temperature, the mass of the system, and the phase of a substance.

• The temperature change is directly proportional to the amount of heat transferred.

• Heat can be added twice to double the temperature and mass.

• Mass is directly proportional to the amount of heat transferred.

• Heat has to be added in order to cause an equivalent temperature change in a doubled mass.

• The amount of heat transferred depends on the materials and the phase.

• The specific heat power of water is 4.2 joules per gram per Celsius degree 

• The specific heat capacity of water is higher than than the metal

• The application of specific heat is used in the swimming pools.

• The constant volume of the specific heat capacity of steam is 1.4108 kJ/kg K.

• The constant pressure of the specific heat capacity of steam is 1.8723 kJ/kg K.

• The heat needed to raise a substance’s temperature by 1 degree Celcius is called the specific heat capacity.

• The specific heat power of water is 4.2 joules per gram per Celsius degree.

• Q = C m ∆t is the formula for the specific heat capacity

Tips for Understanding the Topic

• Specific Heat capacity is an important chapter for competitive exams such as JEE Mains and BITSAT so this chapter is important in Physics.

• This chapter is also covered in the school so students are suggested to clear the concepts during the school which will help them for the further exams too.

• As this topic is important from an exams point of view the students are recommended to cover this topic.

• The students should understand the concept of specific heat capacity and can make notes regarding this topic.

• This chapter consists of the problems in which mathematical solutions are to be solved so students need to practice the problems so that they won’t face problems during the exams.

• 
  Make notes while the teacher teaches this will help them to make the running notes. This will make a brief understanding of this chapter.

• They can make their own personal notes by studying the chapters and referring to the external notes which are available on the website. This will make the students remember the topic and can refer for quick study.

• Practice the problems given in the textbooks, don’t just stay with a single book for solving the questions. They should solve multiple questions from different books this will make the student more acquainted with a topic. 

• They can refer to the solution books for the solutions of the problems after they try to solve the question. By double-checking the steps used to answer the problem is correct or not and comparing the answer to that question will help the students to be sure if they can solve a problem or not. 

• Some students may face problems they should practice more that will make their concepts more clear.

• Clear the basic concepts of the topic so that it will be easy to understand and answer the questions.

• Students are recommended to have external guidance to prepare for the exam and they can guide them to study the right topic with better knowledge.

• If they get stuck in a problem or have any doubts they can ask the teachers or tutors so that they can help the students regarding the topic.

• Students should always revise the chapter after they finish studying. This will help the students to come up with doubts which they might have while revising the topic. Doubts help the student to know and gain more knowledge about the subject that might help them.

• Strengthen the math skills because this will help to solve the problems in Physics. 

• Students also have to remember the formula and have the knowledge of the formula to use it in the solution so that they can get the answers right.

• Students should learn the SI units for using the equations in the correct way. By this, it will be easy for them to simplify the equations and can get the answers they want.

has everything prepared for the students so that students can have the materials that they need. Free notes and study materials of this are available on the website of and can be downloaded with the help of a mobile or laptop. It can be downloaded in PDF form. Students can refer to the study material for their exams.

Conclusion

The specific heat is the measure of heat per unit mass needed for the temperature rise by one degree Celsius. The connection between heat and temperature change is typically communicated in the structure that appeared underneath where c is specific heat.

In this section, you will be able to learn more about the Momentum formula and how to use this particular formula to solve some of the most important questions in the exam. Momentum is a really important topic for the students and they definitely need to be aware of the formula. This section deals with the exact formula for momentum and it also tells you about the change in momentum and how one can calculate it.

How to Calculate Momentum?

One of the most important topics that students seem to focus on is the momentum formula physics. So, we are going to provide some information on that. According to the Laws of Motion as denoted by Newton, all the bodies that are in a state of motion or rest, continue to be in that position until and unless they are being interfered with by some sort of an external force. Now, this is the same principle that we are going to use in order to get the formula to find momentum. For example, if the velocity and the mass of a particular object remain unchanged, then there will be no change in the momentum of the object as well. This is a very simple formula that students can use in order to find out the momentum of any object.
 

How to Calculate Momentum?

Newton’s Laws of Motion says that all the bodies that are in a state of motion or at rest, continue to be in that position till such times when they are being interfered with, or by some external force.  In order to get the formula to find momentum, the same principle is used.
 

If the velocity and the mass of any particular object remain unchanged, then we will see no change in the momentum of the object.
 

Equation for Momentum

As we know the formula for momentum is given as: p=mv
 

Where, p can be denoted as the momentum that a body has,
 

m can be denoted as the mass that the body has and v can be denoted as the velocity that the body has.
 

How Do You Write The Formula of Momentum? 

Before you know all about the formula for momentum, it is important to define it. As per its definition, it can be said that momentum is very closely associated with the entire mass that a moving body has. This can actually be defined as the motion quantity. It is measured by creating the product of the velocity and the mass of the moving body. Momentum can also be called the measurement of a vector. The direction of Velocity and momentum remains the same. So, the formula can be written only when we know about the velocity and the mass of the body.
 

What is the Equation for Momentum 

If you want to know what is the formula of change in momentum then we are going to provide you with a very simple solution right here. As we know the formula for momentum is given as: 
 

p=mv
 

Where,
 

p can be denoted as the momentum that a body has,
 

m can be denoted as the mass that the body has
 

and v can be denoted as the velocity that the body has.
 

This is a very simple equation and it will definitely answer your question on how to find momentum. It will also help you to know that the S.I unit for momentum can be termed as Kg ms^-1

Science is an interesting subject that is based on application through experimentation and observation. Students are required to read and learn the Concepts of science to understand the basic things happening in and around our surroundings. There are mainly three divisions of science namely, Physics, Chemistry and Biology.

Physics is considered to be a stream of Science that studies matter and its entities of energy and force. Physics is an important part of science that reasons and solves  the basic and most important aspects of real life like Electricity, car seat belts, electrical appliances and much more. In this article, students will find all the necessary details about Aberration of lens.

Suppose I have a lens and an object in front of it. Generally, we see the object with white light (having a spectrum of colours), so we call it a white light object. 

When I see this object through the lens, the rays of different focal lengths form different images, and we get a blurred image of the object. 

Ideally, these rays should focus at a point on the lens and form a single image, but we get different images at different points. This image error is caused because of the defect called the ‘Aberration of a lens’.

In this article, we will study the types of aberration in lenses and the ways to reduce them.

Types of Aberration of Lenses

The types of aberrations are:

1. Spherical Aberration

2. Chromatic Aberration

3. Astigmatism

4. Distortion

5. Field Curvature

6. Coma

7. Zernike Polynomials

1. Spherical Aberration

An optical defect in a lens is called the spherical aberration because we cannot see the objects with clarity. The two causes of spherical aberration are:

• Low-quality Lens

• Large aperture Lens

In this type of aberration, the light rays passing through the lens don’t converge at the common point. There are two causes of spherical aberration in lenses. Let’s discuss these one-by-one:

The rays coming from the margin (or far away) are called the marginal lines. These lines meet at a point close to the centre of the axis at a focal length fm. 

The rays nearer to the centre are called paraxial rays. These paraxial rays of different focal lengths form images at different focal points on the axis. 

Because of the formation of multiple images of the same object, we get a low-resolution/blur image on the lens.

2. Chromatic Aberration                         

We know that white light is composed of seven colours and each colour has its focal length & wavelength. When this light passes through the lens, the colours of different focal lengths form images at different spots on the axis. Because of this, we get a blurred image of the object. This type of defect is called chromatic aberration.

There are two types of chromatic aberrations, these are:

Longitudinal

Longitudinal chromatic aberration is also called the axial chromatic aberration. A lens cannot focus rays of light with different focal lengths on the same focal plane. 

Lateral Aberration

This aberration is also called the transverse chromatic aberration. 

Lateral aberration is a type of aberration that causes colour fringing as a result of image magnification which varies with colour wavelength. 

A secondary chromatic aberration is also related to lateral chromatic aberration. This secondary chromatic aberration causes difficulty in the simultaneous correction of blue, green, and red light rays.

3. Astigmatism

This aberration is similar to chromatic aberration because, in both, the object lies off the principal axis. This type of aberration causes a sharp image to appear as an ellipse away from the focal plane, with the long axis of the ellipse shifting by 90° on the opposite sides of the focal plane.

Combination of 2 Lenses to Reduce Chromatic Aberration

Let’s consider a white light after passing through the lens dispersed into three colours, i.e., violet, red, and yellow.

We know that the focal length of red is greater than the violet colour, i.e. Fr > Fv, so chromatic aberration = Fr – Fv

The condition for a minimum chromatic aberration is: Fr – Fv should be equal to zero. This is what we are going to prove further.

Here, we will use the lens maker’s formula for red and violet light forming images at different points on the lens:

 [frac {1}{Fr} = (mu_{r} – 1) (frac {1}{R1} – frac {1}{R2})] ..(a)  

[frac {1}{Fv} = (mu_{v} – 1) (frac {1}{R1} – frac {1}{R2})] …(b)

F_r – F_v =  F_r F_v[(mu_{v} – mu_{r}) (frac {1}{R1} – frac {1}{R2})]…(c)

For yellow light, the formula is:

[frac {1}{Fy} = (mu_{y} – 1 ) (frac {1}{R1} – frac {1}{R2})] ….(d)

[frac {Fr – Fv}{FrFv} = (mu_{v} – mu_{r} ) (frac {1}{R1} – frac {1}{R2}) * frac {(mu{y} -1)} {(mu{y} -1)} = frac {(mu{v} – mu{r})} {(mu{y} -1)} * (mu{y} -1) * (frac {1}{R1} – frac {1}{R2}) ]

From equation (d), we have:

[frac {Fr – Fv}{FrFv} = frac {(mu{v} – mu{r} )} {(mu{y} -1)} * frac {1}{Fy} ] …..(e)

We know that prism is a special type of lens whose dispersive power, ω = [frac {(mu {v} – mu{r})} {({mu{y} -1})}]

So, from equation (e), we get,

 [frac {Fr – Fv}{FrFv} = omega * frac {1}{Fy} ]

Now, if we do F_r * F_v,  we can see that the geometric progression for F_y, i.e. F_r * F_v = Fy²

So, [frac {Fr – Fv}{Fy^2} = omega * frac {1}{Fy} Rightarrow F_r -F_v = omega * F_y ]

We also know that more is the dispersive power (ω), more will be an aberration.

On putting the value of ω as zero, we get  F_r – F_v = zero. 

Now, we can see that there is no difference between the focal lengths of light and the rays that meet at the common point.

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A thermodynamic system in which there is a change in the state of matter due to the change in the Pressure, Volume, Temperature (P, V, T) without transferring heat or mass with the thermodynamic system or its surroundings.

 

The process in which the work done is in the form of a change in internal energy (U) and the amount of heat transferred, Q is zero (or there is no gain or loss of heat)

 

On this page, we shall learn about the various topics:

• Adiabatic process 

• Related examples 

• Derivation of the Adiabatic Process formula

• Work done

• Adiabatic expansion and compression

• Adiabatic- Reversible and Irreversible Process.

 

There are four types of processes in a thermodynamic system, which are shown via an image below:

 

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Adiabatic Process: It can be considered as an ideal process in which there is no heat transfer and the change in internal energy would be equal to the work done.

 

Here, ∆U = U₂ – U₁ ,and W =- ∆U = U₁ – U₂

 

W = – ∆U

 

Since there is no heat transfer that’s why the Entropy (∆S) = 0.

 

Used in –

 

Turbines, Hot water flask, compression, and expansion of charge in IC.

 

Adiabatic Compression and Expansion

Adiabatic Compression

Pressure ( P) > Volume (V)

 

T, S, and U increase during compression.

 

In cylinders of a car, the compression of the gas-air mixture takes place in no time that there could be an exchange of heat between them.

 

Adiabatic Expansion 

The work is done by the gas in expanding the volume and there’s a decrease in the temperature.

 

Let’s take an example here,

 

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In picture 1, the molecules of a gas are tightly bound together as soon as the membrane is punctured, the volume of the gas expands to Volume dV, and a  drop in temperature by dT.

 

In this case, the work done by a gas W =  PdV and dQ = 0.

 

For an Adiabatic Process for an ideal gas is given by:

 

PV r = Constant 

 

Work Done in an Adiabatic Expansion

Consider a mole of gas contained in a cylinder having insulating walls, provided with a frictionless and insulating piston

 

Let P be the pressure of the gas when the piston moves up through a small distance dx, the work done will be:

 

dW = PAdx = PdV

 

Where A is the cross-sectional area of the piston and dV = Adx is the increase in the volume.

 

As the gas expands adiabatically, there will be a change in its initial stage (P₁, V₁, T₁) to (P₂, V₂, T₂).

 

The total work done will be:

        v₂

W = ∫ PdV……(1)

       v₁

 For an adiabatic change,

 

PV r = G  or P = G* V – r Putting it in eq..1

$w = int_{v1}^{v2}G*V^{-r} dv$

$= G int_{v1}^{v2}V^{-r}dv$

            v₂

= G * [  V(1-r)1-r ]  

             v₁

= G/ (1-r) *  [ V₂ (1 – r)  –  v₁ (1 – r)]

= G/ (r -1) * [v₁ (1 – r) – v₂ ( 1 – r)]

= 1/ (r -1) *  [G* v₁ (1 – r) – G * v₂ (1 – r)] …….(2)

G = P₁ V₁ r = P₂ V₂ r…putting in eq (2)

W (adia) = 1/ (r – 1) * [(  P₁Vr₁ * V₁ (1-r ) – P₂Vr₂ * V₂ (1-r)]

W(adia) = 1 / (r – 1) *

(P₁V₁−P₂V₂)

P₁V₁−P₂V₂)……(3)

 

Deriving an Adiabatic Equation

Adiabatic relationship between P and V

According to the first law of thermodynamics:

dQ = dU + dW….(1)

For one mole of gas, the equation is:

dW = PdV

dU = nCVdT, and CV = dU/dT => dU =  nCVdT = CVdT (as n=1)

Also, for an adiabatic process, dQ = 0;

According to the ideal gas equation:

PV= RT…..(2) (n =1)

Now putting the values of dU and dW in equation(1):

0 = CVdT + PdV…..(3)

Differentiating both the sides in equation (2),  we get:

PdV + VdP = RdT

dT= (PdV + VdP)/ R

Putting the value of dT in q..(3)

CV (PdV + VdP)/ R + PdV = 0

CV (VdP) + CV + R (PdV)…(4)

As we know that CP = CV + R putting it in eq..(4)

CV (VdP) + CP (PdV) = 0

Dividing both the sides by CVPV, we get:

dP/P + (Cp/Cv)* dV/V = 0

We know that Cp / Cv = r

dP/P + r* dV/V = 0…….(5)

Integrating both the sides in eq…(5)

= ∫ dP/P + r  ∫ dV/V = C (∵ C, an integral constant)

= Log_e P + r Logₑ V = C

= Log_e PV r = C 

= PV r = ec

=  PV r = G 

Adiabatic Relation Between P and T

We already know that for an ideal gas, the equation is given by,

PV = RT ( for one mol gas)

V = RT/P…(1)

Putting the value of equation (1)….in the equation  PVr= K

P(RT/P) r = G

= P (1 – r) Tr = G (Constant)

Adiabatic Relation Between V and T

For one mole of gas,  PV= RT

P =RT/V

Putting PV r =G, we get

RT/V * Vr = G or T*V(r – 1) = G/ R

= TV(r – 1)  = G  (Constant)

This equation describes the adiabatic relation between V and T for an ideal gas.

 

Adiabatic – Reversible and Irreversible process

Reversible Adiabatic Process

A reversible adiabatic process is basically an isentropic process.

 

What is an Isentropic Process?

Since in an adiabatic process, the heat transfer is zero which means the change in entropy is also zero. That’s why the process is said to be isentropic by nature.

As, S (entropy) =dQ/ dT

dQ will be zero because there is no heat transfer, dQ/ dT = 0

d(S) = 0

Let’s Consider Some Real-Life Examples:

PVr is constant along a reversible adiabatic process.

 

Irreversible Adiabatic Process

As the name suggests, the process can’t be traced back to its original state.         

During an irreversible adiabatic process of expansion. There will be a change in entropy because of frictional dissipation.

Irreversible expansion cannot be performed at equilibrium.       

 

()

 

Summary

An Irreversible process can’t be in a quasi-static state (the process that happens slowly to keep it in an equilibrium)

 

A reversible process is a quasi-static process.

 

Application-Based Questions      

1. Let’s take an example of the daily based activity, of pumping air inside the tire of a bicycle using a hand pump. Here, we would consider air inside the pump is a thermodynamic system having Volume V at an atmospheric pressure P, and the room temperature, 30°C. Suppose the nozzle of the tire is blocked and you push the pump to ⅕ of the volume V. Calculate the final temperature of the air in the pump.

Solution:  Since the nozzle of the tire is blocked air will not flow, we will consider this as an Adiabatic Process

Ti Vi (r -1) = Tf Vf (r – 1)

Ti = 30°C = (273 + 30) K = 303 K

Vi = V , so Vf = V/5

Tf =  Ti *(Vi / Vf)(r -1)

= 303 K* 5(1.4 -1) 

= 303 K * 50.4

= 303 * 1.90365

Tf = 576.81 K 

                         

Since the temperature T is very high it is dangerous to touch the nozzle.

 

Q2: Two cylinders A and B of equal size (both fitted with piston) are filled with an equal amount of ideal gas at room temperature. In-cylinder A the piston is free to
move, while in B, it is fixed. When some amount of heat is added to cylinder A raises by 30 K. What will be the rising temperature of the gas in cylinder B?

Solution: For gas in cylinder A,

Q (heat) = n x CpdT₁, and that for cylinder B = n x CvdT₂

So, dT₂ = Cp/ Cv x dT₁

= 7/5 x 30. 

(Here, Cp/ Cv = 7/5)

dT₂  = 42 K (dT₂ = Rise in temperature in cylinder B).