Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 113. In which case will the efficiency of a Carnot cycle be higher: when the hot body temperature is increased by ΔT, or when the cold body temperature is decreased by the same magnitude? 

Solution. 113. The efficiency is given by

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Now in the two cases the efficiencies are

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus      Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 114. Hydrogen is used iu a Carnot cycle as a working substance. Find the efficiency of the cycle, if as a result of an adiabatic expansion
 (a) the gas volume increases n = 2.0 times;
 (b) the pressure decreases n = 2.0 times.

Solution. 114.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Also    Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Finally   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

(b) Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

So we get the formulae here by  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev in the previous case. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 115. A heat engine employing a Carnot cycle with an efficiency of η = 10% is used as a refrigerating machine, the thermal reservoirs being the same. Find its refrigerating efficiency ε. 

Solution. 115.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Used as a refrigerator, the refrigerating efficiency of a heat engine is given by

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

where η is the efficiency of the heat engine.


Q. 116. An ideal gas goes through a cycle consisting of alternate isothermal and adiabatic curves (Fig. 2.2). The isothermal processes proceed at the temperatures T1, T2, and T3. Find the efficiency of such a cycle, if in each isothermal expansion the gas volume increases in the same proportion. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Solution. 116. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Q1 = Heat taken at the upper temperature

Now   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Similarly   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus Q2 = heat ejected at the lower temperature   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 117. Find the efficiency of a cycle consisting of two isochoric and two adiabatic lines, if the volume of the ideal gas changes n = 10 times within the cycle. The working substance is nitrogen. 

Solution. 117. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 118. Find the efficiency of a cycle consisting of two isobaric and two adiabatic lines, if the pressure changes n times within the cycle. The working substance is an ideal gas whose adiabatic exponent is equal to γ.

Solution. 118. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 119. An ideal gas whose adiabatic exponent equals γ goes through a cycle consisting of two isochoric and two isobaric lines. Find the efficiency of such a cycle, if the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion.

Solution. 119. Since the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Heat rejected is Q'2 = Q'21 + Q’22 where

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 120. An ideal gas goes through a cycle consisting of
 (a) isochoric, adiabatic, and isothermal lines;
 (b) isobaric, adiabatic, and isothermal lines, with the isothermal process proceeding at the minimum temperature of the whole cycle. Find the efficiency of each cycle if the abso- lute temperature varies n-fold within the cycle. 

Solution. 120.  

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Thus    Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 121. The conditions are the same as in the foregoing problem with the exception that the isothermal process proceeds at the maximum temperature of the whole cycle.

Solution. 121. Here the isothermal process proceeds at the maximum temperature instead of at the minimum temperature of the cycle as in Q.120.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
i.e.  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 122. An ideal gas goes through a cycle consisting of isothermal, polytropic, and adiabatic lines, with the isothermal process proceeding at the maximum temperature of the whole cycle. Find the efficiency of such a cycle if the absolute temperature varies n-fold within the cycle.

Solution. 122. The section from Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev is a polytropic process of index a. We shall assume that the corresponding specific heat C is + ve.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Here,  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 123.  An ideal gas with the adiabatic exponent γ goes through a direct (clockwise) cycle consisting of adiabatic, isobaric, and isochoric lines. Find the efficiency of the cycle if in the adiabatic process the volume of the ideal gas (a) increases n-fold; (b) decreases n-fold. 

Solution. 123.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Along the adiabatic line  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

so   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

(b)  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Along the adiabatic line   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 124. Calculate the efficiency of a cycle consisting of isothermal, isobaric, and isochoric lines, if in the isothermal process the volume of the ideal gas with the adiabatic exponent γ
 (a) increases n-fold;
 (b) decreases n-fold.

Solution. 124.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

So   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

So  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 125. Find the efficiency of a cycle consisting of two isochoric and two isothermal lines if the volume varies v-fold and the absolute temperature τ-fold within the cycle. The working substance is an ideal gas with the adiabatic exponent γ. 

Solution. 125. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

We have

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 126. Find the efficiency of a cycle consisting of two isobaric and two isothermal lines if the pressure varies n-fold and the absolute temperature τ-fold within the cycle. The working substance is an ideal gas with the adiabatic exponent γ.

Solution. 126. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 127. An ideal gas with the adiabatic exponent γ goes through a cycle (Fig. 2.3) within which the absolute temperature varies τ-fold. Find the efficiency of this cycle. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Solution. 127. 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Because of the linearity of the section
BC whose equation is

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Along BC, the specific heat C is given by

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Thus  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Finally   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 128. Making use of the Clausius inequality, demonstrate that all cycles having the same maximum temperature Tmax  and the same minimum temperature Tmin  are less efficient compared to the Carnot cycle with the same Tmax  and Tmin• 

Solution. 128. We write Claussius inequality in the form 

  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

where  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev is the heat transeferred to the system but Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev is heat rejected by the system, both are +ve and this explains the minus sign before Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

In this inequality Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev and we can write 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Thus  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

or  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 129. Making use of the Carnot theorem, show that in the case of a physically uniform substance whose state is defined by the parameters T and V 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

where U (T, V) is the internal energy of the substance.

Instruction. Consider the infinitesimal Carnot cycle in the variables p, V. 

Solution. 129. We consider an infinitesimal carnot cycle with isothermal process at temperatures T + dT and T.

Let δA be the work done in the cycle and δQ, be the heat received at the higher temperature. Then by Carnot's theorem

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

On the other hand  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

while  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
Hence    Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 130. Find the entropy increment of one mole of carbon dioxide when its absolute temperature increases n = 2.0 times if the process of heating is (a) isochoric; (b) isobaric. The gas is to be regarded as ideal. 

Solution. 130. (a) In an isochoric process the entropy change will be

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

For carbon dioxide  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

so,  Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

(b) For an isobaric process,

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 131. The entropy of v = 4.0 moles of an ideal gas increases by ΔS = 23 J/K due to the isothermal expansion. How many times should the volume v = 4.0 moles of the gas be increased? 

Solution. 131. In an isothermal expansion 

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev


Q. 132. Two moles of an ideal gas are cooled isochorically and then expanded isobarically to lower the gas temperature back to the initial value. Find the entropy increment of the gas if in this process the gas pressure changed n = 3.3 times. 

Solution. 132.

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

The entropy change depends on the final & initial states only, so we can calculate it directly along the isotherm, it is Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev
(assuming that the final volume is n times the inital volume)


Q. 133. Helium of mass m =1.7 g is expanded adiabatically n = 3.0 times and then compressed isobarically down to the initial volume. Find the entropy increment of the gas in this process.

Solution. 133. If the initial temperature is T0 and volume is V0 then in adiabatic expansion,

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

so,   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

V1 being the volume at the end of the adiabatic process. There is no entropy change in this process. Next the gas is compressed isobarically and the net entropy change is

Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

But   Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

So    Irodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRevIrodov Solutions: The Second Law of Thermodynamics Entropy- 1 Notes | EduRev

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