Physics Class 11 Notes Thermodynamics Chapter 10 for kpk

Physics Class 11 Notes Thermodynamics Chapter 10 for kpk

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

Short Questions Thermodynamics Physics 1st Year Notes 2021

(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,

Putting values, we get

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

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).

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