**Chapter**** – 16, THERMODYNAMICS**

**Chapter**

**– 16, THERMODYNAMICS**

**Introduction :**

When we rub our palms, it gets warmer due to conversion of mechanical work done to heat energy. Why is the palms not to get warmer ? When the door of the refrigerator is opened for few minutes, why the room becomes warmer instead of getting colder ?

In this chapter, we will discuss all these type of phenomena occurring due to interconversion of heat, work and energy.

**Let us discuss a few important terms related to our topic.**

**1 ) Thermodynamics :** It is the branch of physics which deals with the study of inter relationship of work, heat and energy.

**2 ) Thermodynamical System :** The assembly of extremely large number of particles to have a certain value of pressure, volume and temperature, is called as the thermodynamical system.

**3 ) Surroundings :** Each and every thing outside the system which has direct contact on the system, is called as the Surrounding.

Let us consider that a gas is enclosed in a vessel. The gas collection of an extremely large number of particles ( or molecules or atom ) having some definite pressure, volume and temperature. The gas and the vessel contain the thermodynamical system and the outside of the vessel like air etc are regarded as its surrounding.

**4 ) Thermodynamical Parameters : ** Pressure, volume and the temperature of a thermodynamical system are regarded as the thermodynamical parameters. These are also known as the thermodynamical variables. Other than these physical quantities, internal energy, enthalpy and entropy etc are also thermodynamical variables but these can be expressed as the pressure, volume and temperature.

**5 ) Thermodynamic Process :**

**a ) Isothermal Process :** If the pressure and the volume of the thermodynamical system change but its temperature remains constant with time, then this type of change is termed as isothermal process.

Examples : Let a gas is enclosed in a vessel and it allows to expand very slowly so that its pressure and volume change but its temperature not. This type of change is considered as the isothermal process or isothermal change.

**b ) Adiabatic Process :** If the pressure, volume and the temperature of a thermodynamical system change but there is no heat exchange between the system and surrounding. This type of change is termed as the Adiabatic process.

Example : The water or milk or tea kept in a thermos flask, does not exchange heat with its surrounding with a particular time period. This can be considered as the adiabatic process.

**c ) Isobaric process :** If the temperature and the volume of a thermodynamical system change but its pressure remains constant. This type of change is called as the isobaric process.

**d ) Isochoric or Isovolumic process :** If the volume of a thermodynamical system kept constant while its pressure and temperature. Then the change is called as the isochoric process.

**e ) Cyclic Process :** The thermodynamic process in which the system returns in its initial position after undergoing a series of changes, is called as the Cyclic process.

f ) Non cyclic process : The thermodynamic process in which the system does not return in its initial position, is called as the non cyclic process.

**g ) Quasi static Process :** The thermodynamic process which is performed very slowly so that in each and every steps, the temperature and pressure remain constant, is called as the Quasi static process.

**h ) Isolated system :** The system which has no effect of surrounding completely, is called as the Isolated system.

**Work and Heat :**

The heat and work are the way of transformation of energy from one form to another form. Let us consider a cheque. The cheque is a form of money which is used to transfer the money from one place to another or from one man to another. But it is not money. It is just a form of money. Similarly, diamond or gold are very precious substances that can be used to conserved money for future use or to transfer of money form one place to another or from one man to anther. In the same manner, the heat is a transformation of thermal energy from one body to anotherbody due to temperature difference of two different bodies and the work is the transformation of mechanical energy.

Let H is the amount of heat and W is the amount of mechanical work. The greater amount of heat indicates the larger amount of mechanical work and vice versa. Hence, we can say that the work is directly proportional to the amount of heat.

W ∝ H

W = JH

where J is the proportionality constant and known as the Joule’s mechanical equivalent of heat.

J = W / H

If H = 1 unit, then J = W

Joule’s mechanical equivalent of heat is defined as the amount of mechanical work done to produce unit amount of heat.

Its SI unit is J / cal and CGS unit is erg / cal. Its value is 4.2 joule.

**Work done in a non cyclic process :**

Let an ideal gas is enclosed in a perfectly insulated cylinder fitted with a non conducting and frictionless piston. Let ‘ p ‘ pressure is applied on the gas so that the volume of the gas is changed by dv through change in temperature dT.

Workdone in changing the volume of the gas is given by

dw = force * displacement

dw = pressure * area * displacement

dw = p * ( A * dx )

**dw = pdv **

If we plot pressure versus volume graph, we get p-v diagram as shown above figure.

In the above figure, the two points i ( initial condition ) and f ( final condition ) can be joined by a number of different curves. The area of different curves will be different but the area under the p-v diagram represents the work done by the gas or on the gas. Hence, we can say that the work done by a gas or on the gas in a non cyclic process is function of path chosen.

**Work done in a cyclic process :**

Let an ideal gas is enclosed in a perfectly insulated cylinder fitted with a non conducting and frictionless piston. Let ‘ p ‘ pressure is applied on the gas so that the volume of the gas is changed by dv through change in temperature dT.

If we plot pressure versus volume graph, we get p-v diagram as shown above figure.

Work done by the gas during expansion from v_{1} to v_{2} following the path ijf is given by

W_{1} = area of ijfmli

work done by the gas during compression from volum v_{2} to v following the path fki is given by

W_{2} = area of fkilmf

During the whole cyclic process, the total work done by the gas is given by

W = W_{1} – W_{2}

W = area of ijfmli – area of fkilmf

W = area of ijfki

W = area of the loop

**Conclusion : **

* The total work done in a cyclic process is not zero. This is because, during expansion it absorbs heat which is converted into work.

* The work done per cycle is equal to the area of the loop .

* For clock wise direction, work is taken as positive ⇒ the work is done **by** the system

* For anticlock wise direction, the work is taken as negative ⇒ the work is done **on** the system.

**Conversion of Work in different cases :**

1 ) In case of a body having potential energy

w = JH

**mgh = JH**

where symbol has usual meaning.

2 ) In case of a body having kinetic energy

w = JH

**mv ^{2} / 2 = JH**

3 ) For both energy that is potential and kinetic energy

w = JH

**mgh + mv ^{2} / 2 = JH**

**Conversion of Heat :**

1 ) In case of change in temperature

H = mcdt

where m = mass of the substance

c = specific heat of the substance

dt = change in temperature

2 ) In case of change in temperature

H = mL

where L is latent heat of change of state

3 ) In both of the above cases

H = mcdt + mL

**Specific Heat Of Gas :**

Specific heat of a substance is defined as the amount of heat required to raise the temperature by unity of substance of unit mass.

There are two types of specific heat of gas.

1 ) Molar Specific Heat at constant pressure

2 ) Molar Specific Heat at constant volume

**1 ) Molar specific heat at constant pressure :**

It is defined as the amount of heat required to raise the temperature through unity of substance of one mole at constant pressure.

**2 ) Molar specific heat at constant volume :**

It is defined as the amount of heat required to raise the temperature through unity of a substance of one mole at constant volume.

**Laws Of Thermodynamics **

**Laws Of Thermodynamics**

**1 ) Zeroth Law Of Thermodynamics :**

**1 ) Zeroth Law Of Thermodynamics :**

Statement : ” If the two systems are in thermal equilibrium with a third a system separately, then the former two thermodynamic systems are in thermal equilibrium with each other. ”

The physical quantity which determines whether the systems will be in equilibrium or not, is the temperature.

Hence, Zeroth law of thermodynamics gives the concept of temperature which is a scalar quantity.

**2 ) First Law Of Thermodynamics : **

**2 ) First Law Of Thermodynamics :**

*** ****Statement :** ” If a amount of heat is given to a system, it is equal to the sum total of the increase in internal energy of the system and the external work done by the system. ”

*** ****Mathematical Expression :**

Let us consider that a gas is enclosed in a vessel fitted with a frictionless and non conducting piston. Let the vessel has thermally insulated walls except bottom. Let dQ amount of heat is given to the system containing the gas and the vessel. Due to supplied heat, the temperature of the system increases. Consequently, the internal energy of the system is also increased by dU and there is upward movement of the piston that is there is external work done by the system dW.

Hence, according to first law of thermodynamics,

dQ = dU + dW

*** ****Discussion on first law of thermodynamics :**

* The first law of thermodynamics is restatement of the law of conservation of energy that is the energy can neither be created nor be destroyed, it can be only transferred from one form to another form.

* By applying this law we can express all the three physical quantities pressure, volume and the temperature in the same unit.

* The internal energy dU may be any type of energy — mechanical energy, translational kinetic energy, rotationla kinetic energy and binding energy etc.

* dW may be any type of work. It depends on the manner how the system been changed from one state to another state.

significance : It explains that without giving an an equivalent amount of energy to a machine, it is impossible to get work from it. It is the restatement of the law of conservation of energy. **This law gives the concept of the INTERNAL ENERGY**

*** ****Internal Energy ( U )** = The sum total of all types of energy possess by a body is called as its internal energy.

* Internal energy of an ideal gas is wholly kinetic in nature. It depends only on the temperature of the gas.

* Internal energy of a real gas is the sum total of kinetic and potential energy of the gas molecules. It depends on both temperature and volume.

*** ****Application of First Law of Thermodynamics :**

**1 ) Work done in Iso thermal process :**

In isothermal process, the temperature remains constant while its pressure and volume can change according to the given condition. As there is no change in temperature, there is no change in internal energy.

∴ dU = 0

So, the whole amount of heat is used to perform external work done.

According to first law of third law of thermodynamics, we have

##### dQ = dW + dU

##### dQ = dW + 0

##### dQ = dW

Now, we know that the work done is given by

##### dW = pdV

where p represents the pressure and dV is the change in volume.

From real gas equation, we have

pV = nRT

p = nRT / V

where n = no of moles

R = real gas constant

T = temperature of the gas

Total work done is given by

##### W = ∫_{V}_{1}_{}^{V2}^{} dW = ∫^{V}pdv

##### W = ∫_{V}^{V2}( nRT / V ) dV

##### W = nRT ∫_{V1}^{V2} dV / V

##### W = nRT [ lnV ]_{V1}^{V2}

##### W = nRT [ lnV_{2} – lnV_{1} ]

##### W = nRT ln ( V_{2} / V_{1} )

##### W = 2.303 nRT log ( V_{2} / V_{1} )

In terms of pressure,

##### W = 2.303 nRT log ( p_{2} / p_{1} )

**2 ) Work done in Adiabatic Process :**

For adiabatic process, there is no exchange of heat between the surrounding and system.

Therefore, the change in heat i.e dQ = 0

Hence, from first law of thermodynamics, we have

##### dQ = dU + dW

##### 0 = dU + dW

##### dW = – dU

dW = – C_{v}dT

where C_{v} is the specific heat at constant volume and dT is the change in temperature.

Total workdone in a adiabatic process is given by

##### W = ∫^{T2}_{T1} – C_{v}dT

##### W = – C_{v} ∫^{T2}_{T1} dT

##### W = -C_{v} ( T_{2} -T_{1} )

##### W = C_{v} ( T_{1} – T_{2} )

##### we know that C_{p} – C_{v} = R

##### C_{v} = R / ( Y – 1 )

##### where Y = C_{p} / C_{v}

##### Therefore**, W = R( T**_{1} – T_{2} ) / ( Y – 1 )

_{1}– T

_{2}) / ( Y – 1 )

**3 ) Relation Between C**_{p} and C_{v} :

_{p}and C

_{v}:

If dQ amount of heat is given to a system of an ideal gas, it uses this amount of heat in two ways. Some part of the heat is used to increase internal energy and remaining part of heat is used to perform external work. This is in accordance with the first law of thermodynamics.

At constant volume, the external work is zero that is all the amount of supplied heat is used to increase the internal energy.

##### dw = pdV = p * 0 = 0

##### and dQ = dW + dU

##### dQ = dU

##### dU = C_{v} dT

Let the gas be heated at constant pressure, then the external work done is not zero and given by

dw = pdV

From first law of thermodynamics

##### dQ = dU + dW

##### C_{p} dT = C_{v}dT + pdV

##### ( C_{p} – C_{v} ) dT = pdV

##### ( C_{p} – C_{v} ) = p( dV / dT )———————————- ( 1 )

##### form ideal gas equation

##### pV = nRT

##### differentiating both sides w. r .t T

##### p ( dV / dT ) = nR

##### putting this value in equation ( 1 ) we get

##### ( C_{p} – C_{v} ) = nR

##### where n = no of moles

R = real gas constant

for one mole of the gas

**( C**_{p} – C_{v} ) = R

_{p}– C

_{v}) = R

**Physical Significance of First Law of Thermodynamics :**

1 ) From this law, we get the idea of heat which is form of energy in transit.

2 ) Energy is conserved in thermodynamics.

3 ) Every thermodynamics system possess internal energy which is the function of state only.

4 ) This law establishes an exact relation between heat and other form of energy.

**Draw Back of First Law of Thermodynamics :**

1 ) This can not say the condition under which the system can transfer the heat into work.

2 ) It does not specify that how much heat energy is converted into work.

**3 ) Second Law of Thermodynamics :**

There are two statements regarding the second law of thermodynamics which are given below

*** ****Kelvin Plank Statement :**

” It is impossible to construct an heat engine which can convert heat into work without any change in anywhere.”

*** ****Clausius Statement :**

” Heat can not be transformed from higher temperature region without any change in anywhere.”

Let us discuss the different terms that may be used to understand the second law of thermodynamics.

*** ****Reversible Process :**

A process which is performed in such a way that on reversing the system and surrounding may be restored to initial condition without producing any change in anywhere, is called as reversible process.

**Condition for Reversible Process :**

1 ) The process should be Quasistatic.

2 ) Dissipative force ( frictional force and viscosity force etc ) must be absent.

3 ) No loss of energy due to conduction, convection or radiation during the process.

*** ****Irreversible Process :**

Any process which can not be brought to initial condition without any change in anywhere is called as irreversible process.

All natural process are irreversible process.

*** ****Heat Engine :**

In thermodynamics, a heat engine is a system that converts heat or thermal energy and chemical energy to mechanical energy which can be used to do mechanical work.

**Question – Answer Zone**

**Question – Answer Zone**

**1 ) Out of the parameters – temperature, pressure, work and volume which parameter does not characterize the thermodynamic state of matter ? **

Ans : Out of the parameters – temperature, pressure, work and volume, work does not characterize the thermodynamic state of matter.

**2 ) What is the total internal energy possessed by molecules of an ideal gas ? **

Ans : The total internal energy possessed by molecule of an ideal gas is 3kT / 2, where k is Boltzmann constant and T is the absolute temperature of the gas.

**3 ) What is the change in the internal energy of a system over complete cycle of a cyclic process ? **

Ans : The change in the internal energy of a system over complete cycle of a cyclic process is zero.

**4 ) What is the amount of work done in a cyclic process ? **

Ans : The amount of work done in a cyclic process is the area of the loop in the PV – diagram.

**5 ) A piece of metal is hammered. Does its internal energy increase ? **

Ans : When a piece of metal is hammered, the work is converted in to heat energy. So, its internal energy increases.

**6 ) What is the relation between heat energy, work done and change in internal energy ? **

Ans : According to First Law of Thermodynamics, the relation between heat energy, work done and change in internal energy is given by

dQ = dW + dU

where dQ = Change in amount of heat

dW = work done

dU = change in internal energy

** 7 ) Heat is supplied to a system. but its internal energy does not increase, What is the process involved ? **

Ans : Internal energy depends only on the temperature of the gas. Heat is supplied to a system, but its internal energy does not increase. It means that its temperature remains constant. This implies the isothermal process.

**8 ) What is the change in the internal energy of a gas, which is compressed isothermally ? **

Ans : As we know that the internal energy of a gas depends only on its temperature, there is no change in the internal energy of the gas, which is compressed isothermally.

**9 ) Heat is supplied to a system, but the system does not perform any external work. What is the process involved ? **

Ans : Heat is supplied to a system, but the system does not perform any external work. That is

dw = 0

PdV = 0

dV = 0

On integrating both sides, we get

V = constant

Hence, the isochoric process is involved in the given case.

**10 ) Is it possible to increase the temperature of a gas without giving it heat ? **

Ans : Yes, it is possible to increase the temperature of a gas without giving it heat ( Adiabatic change ).

**11 ) Represent equation of an adiabatic process in terms of T and V.**

Ans : P V ^{γ} = constant —————————————- ( 1 )

We know that

P V = n R T

P = n R T / V

putting the values of P in equation ( 1 ), we get

P V = Constant

( n R T / V ) * ( V^{γ} ) = constant

V^{γ – 1}T = constant

**12 ) Represent equation of an adiabatic process in terms of T and P.**

Ans :

P V ^{γ} = constant —————————————- ( 1 )

We know that

P V = n R T

V = n R T / P

putting the values of P in equation ( 1 ), we get

P V = Constant

P * ( nRT / P )^{γ} = constant

P^{1 – γ}T^{γ} = constant

**13 ) Why the gas has two specific heat, one is at constant pressure and other is at constant volume practically ? **

Ans : The change in temperature negligible affects the volume and pressure of a solid material. In case of liquid substance, the change in pressure and volume may be neglected due to change in temperature. But in case of gaseous substance, there is considerable change in pressure and volume due to slight change in temperature. So, the gas should be heated in two ways, one at constant pressure and other at constant volume. Therefore, the gas has two specific heat, one is at constant pressure and other is at constant volume practically.

**14 ) Can the value of specific heat be infinity ? **

Ans : Yes, the value of specific heat can be infinity. As for example, in case of isothermal change, the temperature remains constant that is change in temperature is zero.

specific heat = amount of heat / ( mass * change in temperature ) = amount of heat / ( mass * 0 ) = amount of heat / 0 = infinity

**15 ) What are the conditions required for the two bodies to be in thermal equilibrium ? **

Ans : The required conditions for thermal equilibrium are given below

1 ) The temperature remains constant of the two bodies.

2 ) There is no net transfer of internal energy from one body to another body.

**16 ) What are the main conditions required for isothermal process ? **

Ans : The main conditions required for isothermal process are in the following.

1 ) The vessel containing the gas should be good conductor of heat.

2 ) The wall of the vessel should be thin.

3 ) The process should be performed very slowly so that there would be sufficient chance to exchange the heat between the surrounding and the system and vise versa, so that the temperature of the system remains constant.

**17 ) What are the important conditions for the adiabatic change ? **

Ans : The important conditions for the adiabatic change are given below.

1 ) The vessel containing the gas should be bad conductor of the heat or insulator.

2 ) The wall of the vessel should be thick.

3 ) The process should be performed so rapidly so that there would be no chance to exchange of heat between surrounding and the system.

**18 ) Why molar specific heat of a gas at constant volume is less than the molar specific heat at constant pressure ? **

Ans : When a gas is heated at constant pressure, it uses this amount of heat to raise its internal energy and to perform external work. But the same gas is heated through same range of temperature, at constant volume, there is only increase in internal energy and no performance of external work. Therefore, molar specific heat of a gas at constant volume is less than the molar specific heat at constant pressure.

**19 ) What is the relation between two molar specific heat at constant volume and at constant pressure ? **

Ans : C_{p} – C_{v} = R

**20 ) State Zeroth law of thermodynamics.**

Ans : It states, ” If the two systems are in thermal equilibrium with a third a system separately, then the former two thermodynamic systems are in thermal equilibrium with each other. ”

**21 ) State first law of thermodynamics. **

Ans : It sates, ” If a amount of heat is given to a system, it is equal to the sum total of the increase in internal energy of the system and the external work done by the system. ”

**22 ) Which law of thermodynamics gives the concept of temperature ? **

Ans : Zeroth law of thermodynamic give the concept of temperature.

**23 ) Define temperature according to zeroth law of thermodynamics. **

Ans : According to thermodynamics, the temperature is defined as the physical quantity which determines whether or not a system is in thermal equilibrium with another system.

**24 ) Which law of thermodynamics gives the concept of internal energy ? **

Ans : First law of thermodynamics gives the concept of internal energy.

**25 ) Derive the expression for the isothermal work done. **

Ans :

In isothermal process, the temperature remains constant while its pressure and volume can change according to the given condition. As there is no change in temperature, there is no change in internal energy.

∴ dU = 0

So, the whole amount of heat is used to perform external work done.

According to first law of third law of thermodynamics, we have

##### dQ = dW + dU

##### dQ = dW + 0

##### dQ = dW

Now, we know that the work done is given by

##### dW = pdV

where p represents the pressure and dV is the change in volume.

From real gas equation, we have

pV = nRT

p = nRT / V

where n = no of moles

R = real gas constant

T = temperature of the gas

Total work done is given by

##### W = ∫_{V}_{1}_{}^{V2}^{} dW = ∫^{V}pdv

##### W = ∫_{V}^{V2}( nRT / V ) dV

##### W = nRT ∫_{V1}^{V2} dV / V

##### W = nRT [ lnV ]_{V1}^{V2}

##### W = nRT [ lnV_{2} – lnV_{1} ]

##### W = nRT ln ( V_{2} / V_{1} )

##### W = 2.303 nRT log ( V_{2} / V_{1} )

**26 ) Derive the expression for the adiabatic work done. **

Ans :

For adiabatic process, there is no exchange of heat between the surrounding and system.

Therefore, the change in heat i.e dQ = 0

Hence, from first law of thermodynamics, we have

##### dQ = dU + dW

##### 0 = dU + dW

##### dW = – dU

dW = – C_{v}dT

where C_{v} is the specific heat at constant volume and dT is the change in temperature.

Total workdone in a adiabatic process is given by

##### W = ∫^{T2}_{T1} – C_{v}dT

##### W = – C_{v} ∫^{T2}_{T1} dT

##### W = -C_{v} ( T_{2} -T_{1} )

##### W = C_{v} ( T_{1} – T_{2} )

##### we know that C_{p} – C_{v} = R

##### C_{v} = R / ( Y – 1 )

##### where Y = C_{p} / C_{v}

##### Therefore**, W = R( T**_{1} – T_{2} ) / ( Y – 1 )

_{1}– T

_{2}) / ( Y – 1 )

**27 ) Prove that the work done in a cyclic process is path function.**

Ans :

Let an ideal gas is enclosed in a perfectly insulated cylinder fitted with a non conducting and frictionless piston. Let ‘ p ‘ pressure is applied on the gas so that the volume of the gas is changed by dv through change in temperature dT.

If we plot pressure versus volume graph, we get p-v diagram as shown above figure.

Work done by the gas during expansion from v_{1} to v_{2} following the path ijf is given by

W_{1} = area of ijfmli

work done by the gas during compression from volum v_{2} to v following the path fki is given by

W_{2} = area of fkilmf

During the whole cyclic process, the total work done by the gas is given by

W = W_{1} – W_{2}

W = area of ijfmli – area of fkilmf

W = area of ijfki

W = area of the loop

The work done per cycle is equal to the area of the loop.

**28 ) Show that the work done in a cyclic process is equal to the area enclosed between the P – V diagram and volume axis. **

Ans :

If we plot pressure versus volume graph, we get p-v diagram as shown above figure.

Work done by the gas during expansion from v_{1} to v_{2} following the path ijf is given by

W_{1} = area of ijfmli

work done by the gas during compression from volum v_{2} to v following the path fki is given by

W_{2} = area of fkilmf

During the whole cyclic process, the total work done by the gas is given by

W = W_{1} – W_{2}

W = area of ijfmli – area of fkilmf

W = area of ijfki

W = area of the loop

**28 ) Is work done in a cyclic process zero ? Explain your answer. **

Ans : No, work done in a cyclic process is not zero.This is because, during expansion it absorbs heat which is converted into work.

**29 ) What will be the expression for the change in internal energy in melting process ? **

Ans : During melting process, the change in volume is too small to neglect it and the temperature remains constant. So, from first law of thermodynamics, we have

dQ = dW + dU

mL = PdV + dU

mL = P * 0 + dU

dU = mL

where m is the mass of the substance and L is the latent heat of fusion.

**30 ) Compare the slopes of the isothermal and adiabatic change. **

Ans :

For isothermal process

PV = constant

On differentiating both sides, we get

PdV + VdP = 0

dP / dV = – ( P / V )

Therefore the slope of isothermal process is given by

S_{i} = – ( P / V )

For adiabatic change, the equation of motion is given by

PV^{γ} = constant

On differentiating both sides, we get

VdP + YV^{γ-1}PdV = 0

dP / dV = -Y ( P / V )

Therefore, the slope of the adiabatic change is given by

S_{a} = -Y ( P/ V )

S_{a} = Y S_{i}

Hence, the slope of the adiabatic change is steeper than the isothermal process.

**31 ) State Kelvin’s’ statement of second law of thermodynamics.**

**32 ) State Clausius’s statement of second law of thermodynamics. **

**33 ) Is there any equivalence between the Kelvin’s’ statement and Clausius’s statement of second law of thermodynamics ? Explain the answer. **

**34 ) What is meant by the reversible process ? **

**35 ) What do you mean by the irreversible process ? **

**36 ) Derive the expression for the efficiency of heat engine. **

**37 ) On which factors the efficiency of a Carnot engine depends ? **

**38 ) Give the Carnot’s theorem. **

**39 ) What are the limitation of first law of thermodynamics ? **

**40 ) Give the expression for the coefficient of performance of a refrigerator ? **