Physics Class 11 Notes Thermodynamics Chapter 10 for kpk

Phsycis New Notes 2021 for kpk boards shrot question, long question, and numerical problems.

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Short Questions Thermodynamics Physics 1st Year Notes 2021

  1. Why is the earth not in thermal equilibrium with the sun?

    Physics Class 11 Notes Thermodynamics

    Earth is not a perfect absorber. Sun produces a tremendous amount of heat due to several fusion reactions occurring in the sun. Earth does not absorb all of this heat but just a small portion of it. This is due to its low absorbing capacity as well as different atmosphere layers that surround the earth.

    Earth also radiates some of the heat in the night so loses some of the absorbed heat in this way. And as we know that two bodies can only be in thermal equilibrium when all of the heat emitted by one body will be absorbed by the other. This is the reason that earth and sun are not at thermal equilibrium with each other.

  2. When a block with a hole in it is heated, Why does not the material around the hole expand into the hole and make it small?

    Physics Class 11 Notes Thermodynamics

    When we heat up a block then every dimension of it will change equally and will have an increase.
    Therefore temperature will expand the hole of the block as well it increases every dimension of the block. So that is the reason that the hole of the block does not become small with increase in temperature.

  3. A thermometer is placed in direct sunlight. Will, it read the temperature of the air, or of the sun, or of something else?

    Physics Class 11 Notes Thermodynamics

    A thermometer is a device which we use to measure the temperature of a body. We place the thermometer in contact with the body and then the temperature will transfer between them until the temperature of both of them become the same and they get into thermal equilibrium with each other. In that way, it will tell us the temperature of the body before the contact thermometer has its own given temperature.
    When we place the thermometer in open air then it will be more in contact with the air than the sun or anything else, therefore it will measure the temperature of the air.

  4. The pressure in a gas cylinder containing hydrogen will leak more quickly than if it is containing oxygen. Why?

    Physics Class 11 Notes Thermodynamics

    Graham’s law of diffusion tells us that the rate of diffusion of a gas is inversely proportional to the molecular mass and density of the gas. This means that smaller the molecular mass and density more will be the diffusion.

    We know that hydrogen gas is lighter than that of oxygen gas. Therefore, the diffusion rate of hydrogen gas will be greater than the oxygen gas. That is the reason that gas cylinders containing hydrogen gas will leak more quickly than if they contain oxygen gas.

  5. What happens to the temperature of a room in which an air conditioner is left running on a table in the middle of the room?

    Physics Class 11 Notes Thermodynamics

    There will be no change in the temperature of the room. As the air conditioner is placed in the middle of the room, therefore absorption and rejection will occur at the same place. So overall change in temperature will be zero.

Q.6)  When a sealed thermos flask bottle full of hot coffee is shaken, what are the changes if any in

(a) The temperature of the coffee, and
(b) The internal energy of the coffee.
Answer :

(a) When we shake the sealed flask, full of coffee then the kinetic energy of the molecules will increase. Due to this increase in K.E of molecules, they will collide with each other and will raise the temperature of the coffee. Therefore the temperature of the coffee will increase.
(b) When we shake the sealed flask, full of coffee then the kinetic energy of the molecules and also the temperature of the coffee will increase. As the whole process is an adiabatic process, the internal energy of the coffee will increase.

Q.7) When an object is heated, not all the energy it absorbs goes into increasing the velocity of the molecules. Explain where does the remaining energy go?

Answer :

As in every object, some frictional force between molecules of that object exists. So when an object is heated then a part of this energy is used to overcome this frictional force and some part is used in order to increase the kinetic energy of the molecules of an object.

This is the reason that when an object is heated, not all the energy it absorbs goes into the increasing velocity of the molecules of the object.

Q.8) Why does the pressure of the air in automobile tires increase if the automobile is driven for a while?

 Answer :

When an automobile runs on a road then a frictional force exists between the road and tyres of the vehicle. So when an automobile is driven for a while then some work is done against that frictional force. Therefore due to this work, some heat energy is produced which results in an increase of temperature of the tyre and it will increase the air inside the tyre.

Due to this increase of air in tyre, molecules of air will collide with each other and also with the wall of the tyre and hence the pressure of the air will increase inside the tyre. 

Q.9) On removing the valve, the air escaping from a cycle tube cools. Why?

Answer :

The pressure inside the tube is greater than outside. This high-pressure results in high temperature inside the tube. So when the valve is removed then the air will rush out of the tube to the outside atmosphere. As the pressure outside the tube is low and so as the temperature of the air coming out of the tube will become low and hence becomes cool.

Q.10) Can a room be cooled by leaving the door of an electric refrigerator open?

Answer :

No, a room cannot be cooled by leaving the door of an electric refrigerator open.

Reason:

        The function of a refrigerator is just opposite to that of a heat engine. It extracts heat from a cold body and transfers it to a hot body. But a refrigerator does not work with 100% efficiency i.e. it emits more heat to the room but extracts comparatively less heat from the cold body. Therefore the overall temperature of the room will increase and it cannot be cooled.

Q.11)  What are the conditions for a process to be reversible?

Answer :

Conditions for a process to e reversible are as follows:

  • The whole system should be in equilibrium or close to equilibrium.
  • The speed of the process to occur should be very slow.
  • There will be no loss of energy due to frictional forces during the process.

Q.12) Write the limitations of the first law of thermodynamics.

Answer :

Some limitations of first law of thermodynamics are as follows:

  • There is no information on the direction of heat in it.
  • It does not explain the condition under which heat is converted into work.
  • It cannot solve the problem that all of the heat energy cannot be converted into mechanical work in a continuous way.
  • It does not give any kind of information about the frictional effects.
  • We cannot get any knowledge about the thermal equilibrium and entropy of the system.

Q.13) Is it possible, according to the second law of thermodynamics, to construct a heat engine that is free from thermal pollution?

Answer :

No, it is not possible to construct a heat engine that is free from thermal pollution.

Reason:

      According to the second law of thermodynamics, there is always some loss of heat during the process. Part of it is due to some work done against the frictional effects. So it is impossible to construct a machine with a 100% efficiency i.e. which will take heat from a hot body and transfer all of it to work done without losing it.

Q.14) When two systems are in thermal equilibrium, do they have the same amount of kinetic energy?

Answer :

When two systems of different temperatures are placed together in contact with each other then there will be a transfer of heat. The heat will flow from the body of high temperature to the body of low temperature until both systems are at the same temperature. At this point, both bodies are said to be in thermal equilibrium with each other.

It is not necessary for two systems which are at thermal equilibrium with each other to have the same amount of kinetic energy.

Q.15) Work can be converted completely into heat, so can heat be converted completely into work?

Answer :

No, it is not possible to convert heat into work done completely.

This is due to the reason that during the conversion of heat into work some part of heat is lost against the frictional forces.

Q.16) Entropy has often been called as “time arrow”. Explain?

Answer :

For any process to happen, time must be consumed. As it is said that this process takes that much time to occur or happen.

Similarly, entropy is also defined in terms of time i.e. entropy of the universe either remains constant or increases with time. That is the reason that we usually call entropy as a time arrow.

Q.17) Can specific heat of a gas be zero or infinity? Can specific heat be negative?

Answer :

Yes, the specific heat of a gas can be zero as well as infinity.

As specific heat of a gas is given by,

                                ΔQ = nCΔT                ⇒               C = ΔQ/ΔT

So, specific heat will be zero if the term ‘ΔQ’ becomes zero and it will be infinity if the term ‘ΔT’ is zero i.e. if the temperature of the system remains constant throughout.

However, specific heat can never be negative. It always remains positive.

Q.18) An inventor claims to have developed a heat engine, working between 27°C and 227°C having an efficiency of 45 %. Is the claim valid? Why?

Answer :

Given Data:

                                  Initial temperature = T2 = 27oC = 27 + 273 = 300K

                                  Final temperature = T1 = 227oC = 227 + 273 = 500K

                                  Estimated efficiency = η = 45%

To Find:

                                 Is efficiency attainable = ?

Solution:

As the efficiency of a heat engine is given by,

1 5

Putting values, we get

2 4

And the percentage efficiency is,
                                                        η = 0.4 × 100% = 40%
Therefore the claim to attain the efficiency of 45% is wrong as maximum possible efficiency is 40%.


Comprehensive Questions Physics Class 11 Notes 2021

Q.1) Explain, briefly, the following terms used in thermodynamics: System, Surroundings, Boundary and State variables.

Answer :

System:

     The amount of material or area of space which are under scientific observation to understand their behaviour is called a system.

Surroundings:

      The atmosphere which surrounds the system is known as the surrounding.

                                                    OR

     Anything other than the system is called surrounding.

Boundary:

     The partition line or area which separates the system from the surrounding is called the boundary of the system.

State Variables:

   The variable quantities or functions which determine the physical state of the system are called state variables or state functions.

Q.2) Distinguish among the three forms of energy: work; heat and internal energy.

Answer :

Heat, work and internal energy are so distinct as well as closely related to each other as they all are related to the temperature of a body.

Heat:

        “Something which flows from the hotter body to colder body till the temperature of two bodies become equal is called heat”

To raise the temperature of water then we will place the water on some heat source, sThe needs on a burner or a flame. We can also do that by placing the water reservoir in contact with a hotter body. This happens due to the flow of heat from the hotter body or heat source to the colder body. Need keeps on flowing from the hotter body to the colder body until the thermal equilibrium state is attained.

So we can say that heat is the energy which is transferred between two substances. It is energy in transit. Heat provides energy to the particles of material which store in them as kinetic energy. So, we can say that heat represents the total K.E of molecules of a body.

Units of heat are joule and calorie.

Work:

             There is another way to increase the temperature of a body instead of providing heat to that body. That method or way is to perform work done on the body. Some examples of work done are as follows:

1. If we want to raise the temperature of our hand then all we have to do is to rub the hands with each other. In that way, we are doing work which results in the rising temperature of the hands and in that way we can warm up our hands.

2. When we have to fix a nail to something then we hit it in the head. By hitting the nail, its temperature goes up and the nail becomes a little hot.

3. Similarly, when we shake a curd in a blender or any vessel then the curd gets hot i.e. its temperature will increase due to the work done of shaking or churning.

4. Another example of raising the temperature with the help of work done is when we put the air into the tire with a hand pump then the temperature of the tire increases which warms up the tire.

So overall we can conclude that heat and work both are the ways to increase the temperature of a body. The main difference between both methods is that work can increase the temperature of the body just by changing the displacement of the body while for the heat we need a temperature difference between the two bodies or between the body and the surrounding.

Internal Energy:

        “The sum of kinetic and potential energies associated with the random motion of the atoms of the substance is the internal energy of the substance”

As matter consists of atoms and molecules, which are in constant motion. Atoms have a back and forth motion about their mean position. Due to their motion, the matter particles have both kinetic energy and potential energy.

Kinetic energy can be in the form of translational, rotational and vibrational kinetic energy. When we heat a substance then it will provide energy to its molecules so their random motion will increase. In this way, the provided heat energy will convert into internal energy of the substance.

Similarly, some kind of work can also be done on the substance to increase its internal energy. Once heat and work or both are transferred to a substance, they are no longer distinguishable as heat energy and work energy in the substance.

Q.3) State and explain the first law of thermodynamics.

Answer :

First Law of Thermodynamics:

Statement:

        “This law states that every thermodynamic system possesses a state variable (U) called the internal energy”

Explanation:

The first law of thermodynamics explains the fact that energy can neither be created nor be destroyed in any thermodynamic system. So we can say that the first law of thermodynamics is another form of the law of conservation of energy, which deals only with heat energy.

In any thermodynamic system, when we add a certain amount of energy ΔQ in it then this addition of energy will result in an increase in internal energy ΔU of the system plus work is done ΔW by the system. We can express all these phenomena as,

                                                    ΔQ = ΔU + ΔW

Above equation is the mathematical statement of the first law of thermodynamics.

From this equation we can also define the internal energy as follows:

                                                    ΔU = ΔQ – ΔW

The internal energy of the system will be positive if the temperature of the system rises and will be negative if the temperature of the system falls.


Q.4) In the light of the first law of thermodynamics describe the processes:

a. Isochoric process
b. Isobaric process
c. Isothermal process and
d. Adiabatic process.

Answer :
Isochoric Process:

“The thermodynamic process during which the volume of the system remains constant is called isochoric process”
Explanation:
  Consider a gas in a cylinder of conducting base where the walls of the cylinder are non-conducting. Let its piston is not movable as shown in figure (a).

Thermodynamics Physics 1st Year Notes 2021

Now add some amount of heat ΔQ to the gas such that the gas will heat at constant volume due to the fixed piston. Suppose the pressure of the gas changes from P1 to pressure P2 and its temperature changes from T1 to T2.
As first law of thermodynamics gives us,
                                                                        ΔQ = ΔU + ΔW
But as the volume of the gas remains constant so there will be no work done by the system. Therefore,
                                                                        ΔQ = ΔU
So we can say that in an isochoric process, the amount of heat supplied to the system will convert totally into internal energy of the system.
The graph shown in figure (b) of isochoric process is called “isochor”, which is always a straight line parallel to pressure axis.


Isobaric Process:
        “The thermodynamic process during which the pressure is kept constant is called an isobaric process”
Explanation:
         Consider a gas in a cylinder of conducting base whose walls are non-conducting. Let its piston is movable as shown in figure (c).

 Physics 1st Year Notes 2021

Let T1, V1 and P be the initial temperature, volume and pressure of the gas. Now heat up the gas by adding a certain amount of heat ΔQ to it. Due to addition of heat, the gas will expand and piston of cylinder will move in upward direction. If the displacement of the piston is very small then the pressure of the gas will almost remain the same. Now consider V2 and Tbe the volume and temperature of the gas after expansion.
For very small displacement of the piston, work done is given by
                                    ΔW = Force × Distance = F × ΔY
And as force is a product of pressure and are so we can write,
                                    ΔW = PA × ΔY = P (AΔY)
Or we can write it as,
                                    ΔW = PΔV = P (V2 – V1)
So putting this value in the equation of first law of thermodynamics, we get
                                    ΔQ = ΔU + ΔW = ΔU + PΔV
So we can say that isobaric expansion of a system is usually used to convert heat into work done i.e. work done performed by the gas comes from one or both: heat supplied or internal energy of the gas.
The graph of isobaric process shown in figure (d) is called “isobar”, which is also a straight line, parallel to the volume axis.
Isothermal Process:
“The thermodynamic process which is carried out in such a way that a system undergoes changes but its temperature remains constant is called an isothermal process”
Explanation:
                         Consider a cylinder, filled with an ideal gas covered by a moveable piston as shown in figure (e).

 Physics FA fsc notes for class 11

Let the internal energy of the gas depends only on the temperature of the gas. Now place the cylinder on a heat reservoir at temperature T1. Then due to heat, the gas will expand which leads to a decrease in pressure, as a result, the gas will cool down.
    As heat is continuously added from the reservoir, therefore the temperature will remain constant during the expansion and will be the same as that of the reservoir. Take P1, V1 and T be the initial pressure, volume and temperature and P2, V2 be the final pressure and volume of the system.
    Now as the internal energy remain constant throughout the process because there is no change in temperature, therefore, there will be no change in internal energy. So the equation of the first law of thermodynamics will become,
                                    ΔQ = 0 + ΔW                     ⇒                    ΔQ = ΔW
This equation shows that we have to supply an equal amount of energy to the gas in order to get the same work done
Adiabatic Process:
    “The thermodynamic process during which no heat enters or leave a system is called an adiabatic process”
Explanation:
        For a process to be adiabatic, ΔQ should be zero i.e. there should be no flow of heat to or out of the system.
Consider a gas-filled in a completely isolated cylinder with a movable piston as shown in figure (g).

Notes for Phsycis

Let allow the gas to expand by removing some weight from the piston then due to expansion, temperature of the gas will fall and so as the internal energy of the gas. Similarly by compressing the gas both the temperature and internal energy of the system will increase.
    As, ΔQ = 0, so the equation of first law of thermodynamics will become,
                    0 = ΔU + ΔW                 ⇒                   ΔU = -ΔW
This tells us that in an adiabatic process, work is done at the cost of internal energy. The graph as shown in figure (f) which explains the adiabatic process is known as an “adiabate”.


Q.5) Define the molar heat capacities Cp and Cv for a gas. Show that, for a mole of an ideal gas, Cp – Cv = R

Answer :
Specific Heat of a Gas at Constant Volume (Cv):
        “The amount of heat required to raise the temperature of one mole of gas by 1 K while keeping its volume constant is called the constant volume molar specific heat Cv of the gas”
We can write it as,
                                ΔQv = nCvΔT
Specific Heat of a Gas at Constant Pressure (CP):
    “The amount of heat required to raise the temperature of one mole of gas by 1 K while keeping in pressure constant is called the constant pressure molar specific heat CP of that gas”
This can be expressed as,
                                ΔQP = nCPΔT
Proof (CP – Cv = R):
 
    Consider “n” moles of an ideal gas filled in a cylinder. Now supply some heat to the gas at constant volume and pressure. As we know that at constant volume, equation of first law of thermodynamics is “ΔQv = ΔU”. Putting the value of ΔQv, we get
                             nCPΔT = ΔU   …………………………….. (1)
Similarly, at constant pressure, the first law of thermodynamics is
                                ΔQP = ΔU + ΔW
Putting the value of ΔU from equation (1), we get
                              ΔQP = nCvΔT + ΔW                            ………………………………… (2)
As, ΔQP = nCPΔT, so putting this value in equation (2), we get
                        nCPΔT = nCvΔT + ΔW
Also, we know that ΔW = PΔV, so putting this in the above equation, we get
                        nCPΔT = nCvΔT + PΔV                           ………………………………….. (3)
As ideal gas equation is,
                            PV = nRT                   OR                    PΔV = nRΔT
So putting this in equation (3), we get
                    nCPΔT = nCvΔT + nRΔT
                ⇒      CP = Cv + R                                                                                                                     ⇒   CP – Cv = R           


Q.6) Explain with examples reversible and irreversible processes.

Answer :
Reversible Process:

        “A process is said to be reversible if it can be retraced exactly in reverse order without producing any change in the surroundings”
Explanation:
           In a reversible process, the system undergoes through the same changes as it would go in a direct process. The main difference is that the thermal and mechanical effects would be completely reversed at each and every stage of the process.
        We can understand it as if during the process, some amount of energy is been absorbed then it will be delivered back during its reverse stage. Similarly, in direct process duration, some work is been done on the system then in its reverse duration, the same amount of work will be done by the system.
        I real life, no change in exactly reversible so we can consider only slowly occurring processes to be reversible processes.
Examples:
         One example of a reversible process is liquefaction of a substance. Vaporization of any material is also an example of reversible process.
Similarly, if we compress a gas slowly, then it will also expand in reverse in exactly same manner so is an example of a reversible process.
Irreversible Process:
        “A process which cannot be traced in the backward direction by reversing the controlling factors is called an irreversible process”
Explanation:
         In an irreversible process, the system does not undergo through the same changes as it does in a direct process.
Usually, the changes which happen suddenly in any process or which have some frictional factors in them or some kind of dissipation energy process are irreversible in nature.
In real life, almost all the processes are irreversible.
Example:
          Dissipation of energy through conduction, convection and radiation are all examples of irreversible processes. A very obvious example of an irreversible process is an explosion.
Production of heat in an electric circuit is another example of the irreversible process.



Q.7) What is meant by a heat engine? What is its main purpose? How is its efficiency defined?

Answer :
Heat Engine:

    “A heat engine is a device that converts heat energy into mechanical work”
Examples:
      The steam engine, petrol engine and diesel engine are all examples of heat engines.
A heat engine consists of the following parts:
Heat Source or Heat Reservoir:
       It is a heat source or we can say that a very large reservoir at a very high temperature says T1, whose temperature remains constant during the process. We also call it a High-Temperature Reservoir (HTR).
Heat Sink or Cold Reservoir:
         t is a reservoir at very low temperature say T2. Its temperature also remains constant throughout the process and we also call it a Low-Temperature Reservoir (LTR).

Physics Notes 1st year

Working Substance:
         For working substance in a heat engine, we usually use a gas. This working substance undergoes a cyclic process. This process is performed in such a way that,
Some heat says Q1 is absorbed from a high-temperature reservoir (HTR) whose temperature is T1 and maximum work will be done to the surrounding due to this absorption. Then heat Q2 will be rejected to the low-temperature reservoir (LTR) or to the sink whose temperature is T2 as shown in figure.
As we say that heat engine work in a cycle then change in internal energy should be zero for that as system will return to its original state after completing the cycle. So we have ΔU = 0. Therefore equation of first law of thermodynamics will become,
                                                ΔQ = ΔU + ΔW = 0 + ΔW
i.e.                      
                                               ΔQ = ΔW
Or we can write it as,
                                        Q1 – Q2 = ΔW         …………………………………. (1)
The efficiency of a heat engine is defined as the ratio of net work done by heat engine to heat Q1 absorbed in cycle i.e.

11 1

Putting value of work done from equation (1), we get

12 2

And if Q2 = 0, then the efficiency of the engine will be equal to 1, which means that percentage efficiency will be 100%. But in actual it is impossible to attain 100% efficiency as some amount of energy will always lost and efficiency of engine will always drop from 1.


Q.8) State the second law of thermodynamics in its alternative forms. Discuss the assertions of the first and second laws about heat and work energies.

Answer :
Second Law of Thermodynamics:

        The first law of thermodynamics is basically a generalized form of the law of conservation of energy. As it explains the process of interconversion of the heat and mechanical work is done.
    But the second law of thermodynamics tells us about the way and process that how heat energy can be converted into some useful work. There are several ways to state the second law of thermodynamics. Two of them are as follows:
1.     Lord Kelvin Statement:
    “It is impossible to construct a heat engine, operating continuously in a cycle, which takes heat from a heat source at a higher temperature and performs an equivalent amount of work without rejecting any heat to a heat sink at low temperature”
Explanation:
       According to Lord Kelvin’s statement, it is essential to have a heat source at a higher temperature and a heat sink at a lower temperature to attain conversion of heat into mechanical work. It is a fact that it’s impossible to convert 100% amount of heat into mechanical work as there is always some loss of heat due to absorption by the sink. So we can say that the output of that heat engine will always be less than that of input or in other words, the efficiency of the heat engine is always less than 100%.
2.     Rudolf Clausius Statement:
    “It is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work”
The assertion about Heat and Work energies:
       The first law of thermodynamics tells us about the equivalence of heat and work did. It asserts that in a cyclic process total heat that flows into the system is equal to the work done by that system.
i.e.                                                           Q = W
However, the second law of thermodynamics is more concerned with that conversion process of heat into work done. The second law asserts that work done by the system will always be less than the absorbed energy by the system.
i.e.                                
                     Q > W                    or             W < Q


Q.9) What were the basic questions that led Carnot to invent Carnot engine?

Answer :
    As heat engine is basically a closed system that exchanges only heat and work with its surrounding environment in cycles. In all cyclic heat engines, the process of conversion that it takes heat from a heat source and use some part of it into useful work and then rejects its remaining part to some heat sink. In this way, the efficiency of the heat engine will be reduced considerably.
    The fact is that all engines actually dissipate some amount of energy. So we can say that the efficiency of all engines is always less than 100%.
If we talk about the efficiency then some questions do arise about the efficiency of the engine, such as:
1.       For an engine, working at different temperatures between two reservoirs, how much maximum efficiency one can get?
2.       Can the efficiency of a heat engine be made to increase?
These were the basic questions which led a French military engineer Sadi Carnot to invent the Carnot engine in 1824.


Q. 10) What is meant by Carnot cycle and by Carnot engine?

Answer :
Carnot Engine:

        “Carnot engine is a hypothetical, idealized heat engine which is free from all sorts of heat losses and friction and possess maximum efficiency than all other heat engines”
        It consists of a gas cylinder with perfectly insulating walls and has a perfectly conducting base. A perfectly insulated, weightless and frictionless piston is also fitted to the cylinder.
Carnot Cycle:
        When after going through a process, the system returns to its original state at the end then we say that the system has complete on cyclic and the process is called a cyclic process. Therefore the operating cycle of the most efficient engine i.e. the Carnot engine is known as the Carnot cycle.
        Carnot cycle consists of four processes. Two of these processes are isothermal and two of them are adiabatic. And after going through all four processes, the Carnot engine will return to its original state from where it took the start.

Carnot Cycle

The Carnot cycle will behave during these processes are as follows:
Isothermal Expansion:
              Consider the working substance of an engine is at pressure P1, volume V1 and temperature T1. Now place the cylinder on a heat source or on a high-temperature reservoir, such that Q1 amount of heat is added to the gas such that the temperature of the gas will remain constant. After placing to the source the states of the gas will change to P2 and V2. This is indicated by isotherm A-B in the figure.
Adiabatic Expansion:
              Now place the gas cylinder on an insulating stand such that no heat will flow into or out of the system. Let the gas is now allowed to expand from volume V2 to V3. Due to this expansion, temperature of the gas will drop from T1 to T2 and pressure will decrease from P2 to P3. This adiabatic process is shown in the figure by adiabat B-C.
Isothermal Compression:
              Now gas is placed on a low-temperature reservoir at Tand is suppressed by increasing the weight on the piston and heat Q2 will leave the system. This process is represented by the isotherm C-D in the figure.
Adiabatic Compression:
             Now the gas cylinder is again placed on an insulating stand and no heat s entering or leaving the system. Gas is compressed again to the initial state and also to the state variable P1, V1 and T1. In this way, the cycle will be completed.


Q.11) State Carnot Theorem about the characteristics of a Carnot engine.

Answer :
Carnot Theorem:

This law states that,
    “No real heat engine operating between two heat reservoirs can be more efficient than a Carnot engine, operating between the same two reservoirs”
Explanation:
The efficiency of a Carnot engine is given by,

uyh

Above equation tells us that the efficiency of a Carnot engine depends on the temperature difference of the two reservoirs. The efficiency of a Carnot engine can be improved in two ways,
1.       By increasing the temperature of hot reservoir or
2.       By decreasing the temperature of cold reservoir.
From above equation, we can also see that if the temperature of cold reservoir will be ‘0 K’ i.e. T = 0 K  then the efficiency will be 100%. But practically it is impossible to drop the temperature of a body to absolute zero, therefore It is not possible to attain 100% efficiency. From that fact, Carnot concluded that it is not possible to develop an engine with 100% efficiency.
However the Carnot efficiency is also not attainable as there always exists some frictional effects, that results into drop of efficiency.


Q.12) What do you mean by a refrigerator? How does it function? Derive an expression for the Coefficient of performance of a refrigerator.

Answer :
Refrigerator:

    “The device which either cool or maintain a body temperature below that of surrounding is called a refrigerating device”
                                                OR
    “Device in which the working substance performs the cycle in a direction opposite to that of a heat engine is called a refrigerator”
Explanation:
   In the functioning of a refrigerator, we can say that heat must always have to flow from a body at a low temperature of surrounding at a high temperature. We know that naturally heat always flows from a hot body to a cold body i.e. from high temperature to a low temperature. Therefore it is not possible to have a flow of heat from the cold body at low temperature to surroundings at high temperature naturally. In other words, we can say that in order to complete a thermodynamic cycle, the work done will not be a natural work done.
Diagram:

What do you mean by a refrigerator?

Working:
                 Working of a refrigerator performs in a cycle. Consider T2 be the temperature of the low-temperature source, then a heat Q2 will be removed from that source. This work is done by the compressor of the refrigerator by using a working substance known as a refrigerant. In result of this work done, heat Q1 will flow out of the source to hot surroundings. Take the temperature of the surrounding be T1. Then work done will be expressed as,
                                W = Q1 – Q2                    ………………………………… (1)
And extracted heat will be,
                                Q1 = W + Q2
Coefficient of Performance:
        “Ratio of the amount of heat removed from the heat sink to the required work is known as the coefficient of heat”
The coefficient of heat describes the performance of the refrigerator. For cooling process coefficient of performance is given as,

Coefficient of Performance
Coefficient of Performance

Q.13) Explain the concept of entropy. Mention its major properties. How is the second law of thermodynamics expressed in terms of entropy?

Answer :
Entropy:

The amount of all kinds of disorders in a system is called the entropy of the system.
If a system undergoes a reversible process, such that it ΔQ amount of heat at temperature T, then the state variable which will increase as a result of this is known as entropy.
Mathematical Expression:

op

ΔS will be positive if we add heat to the system and its value will be negative if heat is removed by the system.
Explanation:
                   In study of thermodynamics, the concept of entropy was first used by the scientist Rudolf Clasusis in 1856. He proposed this concept in order to provide the quantitative basis for second law of thermodynamics. It also provides another variable in order to further describe the state of a system.
Consider a system is placed at temperature T1 and then extract a certain amount of heat Q from the system through some conducting rod so that the temperature of the system became T2 such that    T> T2. During al this process, the net change in entropy will be given by

lpo

We know that here T1 > T2, therefore from above equation we can see that entropy will be positive.
Hence we can rephrase the 2nd law of thermodynamics in form of entropy such that,
“In all natural phenomena where heat flows from one system to another, there is always a net increase in entropy”
Properties:
1.       Whenever there is a change in a system such that heat is flowing out of the system or provided to the system, then the change in entropy will always be positive.
2.       In all natural phenomena, entropy has always a positive value, or we can say that all naturally occurring phenomena always happen in a direction of disorder.
3.       Only those processes are probable for which the entropy of the system increases or remain constant.
4.       For all irreversible processes, entropy of the system will always increase.
5.       We can also define entropy as, measure of disorder of the system.


Numerical Problems Physics Class 11 Notes 2021 Chapter 10