Chapter 12.10 - Second Law of Thermodynamics

In the previous section, we saw the COP of the Carnot refrigerator. In this section, we will see the second law of thermodynamics and it's applications in Carnot engine and refrigerator

The Second law of thermodynamics

◼  The second law of thermodynamics states that:
When a spontaneous process occurs, the entropy of the universe always increases
   ♦ So a process will take place only in that direction which causes an increase in entropy
   ♦ Entropy is a measure of randomness or disorder
• Links to detailed notes on entropy and some related topics are given below:

   ♦ Spontaneity

   ♦ Entropy

   ♦ Second law of thermodynamics

   ♦ Gibbs free energy

• Based on the above discussion, we can write the reason for phenomena such as:
   ♦ Heat never flows from a cold object to a hot object
         ✰ Heat always flow from hot object to cold object
         ✰ A hot object has greater entropy than a cold object
   ♦ Water never spontaneously become ice
         ✰ Water has greater entropy than ice

• Application of second law to heat engines gives the explanation for why no heat engine can have 100 % efficiency • Application of second law to refrigerator gives the explanation for why no refrigerator can have a COP of infinity

 

Application to Carnot Engine

• Let us find the entropy changes in a Carnot engine
    ♦ For that, we consider one complete cycle of the engine
    ♦ There are two isothermal processes and two adiabatic processes in one complete cycle
• The PV diagram is shown again below:

• There is no exchange of heat during the adiabatic processes
    ♦ So there is no entropy changes during the adiabatic processes
• We need to consider the isothermal processes only. We can write it in 9 steps:
1. First isothermal process AB:
(i) In the first isothermal process, the hot reservoir gives away some heat
   ♦ Let this heat given away be Q_h
   ♦ Let Q_i(HR) be the initial heat content of the hot reservoir
   ♦ Let Q_f(HR) be the final heat content of the hot reservoir
• Then Q_h = Q_f(HR) – Q_i(HR)
• This Q_h will be a negative quantity because, Q_f(HR) will be less than Q_i(HR)
◼  So the entropy change suffered by the hot reservoir = $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$
(ii) The same heat content Q_h is gained by the gas
   ♦ Let Q_i(HG) be the initial heat content of the hot gas
   ♦ Let Q_f(HG) be the final heat content of the hot gas
   ♦ Then Q_h = Q_f(HG) – Q_i(HG)
• This Q_h will be a positive quantity because, Q_f(HG) will be greater than Q_i(HG)
◼  So the entropy change suffered by the hot gas = $\mathbf\small{\rm{+\frac{Q_h}{T_1}}}$
2. Second isothermal process AB:
(i) In the second isothermal process, the cold reservoir receives some heat
   ♦ Let this received heat be Q_c
   ♦ Let Q_i(CR) be the initial heat content of the cold reservoir
   ♦ Let Q_f(CR) be the final heat content of the cold reservoir
• Then Q_c = Q_f(CR) – Q_i(CR)
• This Qc will be a positive quantity because, Q_f(CR) will be greater than Q_i(CR)
◼  So the entropy change suffered by the cold reservoir = $\mathbf\small{\rm{+\frac{Q_c}{T_2}}}$
(ii) The same heat content Q_c is lost by the gas
   ♦ Let Q_i(CG) be the initial heat content of the cold gas
   ♦ Let Q_f(CG) be the final heat content of the cold gas
   ♦ Then Q_c = Q_f(CG) – Q_i(CG)
• This Q_c will be a negative quantity because, Q_f(CG) will be less than Q_i(CG)
◼  So the entropy change suffered by the cold gas = $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$
3. So now we have four entropy changes:
(i) Entropy change suffered by the hot reservoir, $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$. This we calculated in 1(i)
(ii) Entropy change suffered by the hot gas, $\mathbf\small{\rm{\frac{Q_h}{T_1}}}$. This we calculated in 1(ii)
(iii) Entropy change suffered by the cold reservoir, $\mathbf\small{\rm{+\frac{Q_c}{T_2}}}$. This we calculated in 2(i)
(iv) Entropy change suffered by the cold gas, $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$. This we calculated in 2(ii)
4. We have the basic equation:
Entropy change suffered by the universe
= Entropy change suffered by the system + Entropy change suffered by the surroundings
5. Let us calculate each item on the right side of the above equation:
(i) Entropy change suffered by the system
= Entropy change suffered by the gas = Item 3(ii) + Item 3(iv)
= $\mathbf\small{\rm{\frac{Q_h}{T_1}-\frac{Q_c}{T_2}}}$
(ii) Entropy change suffered by the surroundings
= Entropy change suffered by the reservoirs = Item 3(i) + Item 3(iii)
= $\mathbf\small{\rm{-\frac{Q_h}{T_1}+\frac{Q_c}{T_2}}}$
6. Consider the Eq.12.15 that we derived in section 12.8:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \frac{T_2}{T_1}}}$
This is same as: $\mathbf\small{\rm{\frac{Q_c}{Q_h} =  \frac{T_2}{T_1} }}$
⇒ $\mathbf\small{\rm{\frac{Q_c}{T_2} =  \frac{Q_h}{T_1} }}$
• Based on this result,
   ♦ 5(i) will become zero
   ♦ 5(ii) will also become zero
7. So we can write:
◼  Entropy change suffered by the system = 0
◼  Entropy change suffered by the surroundings = 0
• So the result in (4) becomes:
◼  Entropy change suffered by the universe = 0
• That means, when one cycle of the Carnot engine is complete, the universe suffers zero entropy change
8. This is an ideal situation
• It shows that, the Carnot engine is most efficient
• All real engines will cause the universe to suffer a positive entropy change
• This is because, all real engines dump some net heat (caused due to friction) into the surroundings
• The Carnot engine is a theoretical engine. We assume that, there is no friction
9. Now let us see if we can attain 100 % efficiency. It can be written in 6 steps:
(i) If there is to be 100% efficiency, Qc must be zero
• That is., there should be no heat exchange with a cold reservoir
• This condition of 'no heat exchange with cold reservoir' can be easily seen from the schematic diagram of a heat engine that we saw in a previous section. It is shown again below:

• It is clear that, for 100% efficiency, Qc must be zero
(ii) If Q_c is not to be given, the cold reservoir will be absent in the engine
• So there will not be any need for the three segments BC, CD and DA
• All the heat received from the hot reservoir will be converted into work
(iii) In such a situation,
    ♦ Entropy change suffered by the surroundings (hot reservoir) = $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$
    ♦ Entropy change suffered by the gas (system) = $\mathbf\small{\rm{\frac{Q_h}{T_1}}}$
(iv) Here also, the net entropy change of system and surroundings is equal to zero
• That means, the entropy change suffered by the universe is zero
• This does not violate the second law
(v) So we are inclined to adopt such an engine, where all heat is converted into work
• But such an engine can perform work only in one segment AB
• It will not return to it's original state
(If we try to return along the same path BA, the same work will have to be done on the gas, resulting in zero net work)
◼ So it cannot perform the next cycle. We can obtain continuous work only if the engine operates in cycles
(vi) Thus it is clear that, to obtain a cyclic process, some heat has to be definitely given to the cold reservoir
• When some heat is given to the cold reservoir, efficiency will become less than 100 %
• That is why, we cannot obtain 100 % efficiency

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Application to Carnot Refrigerator

• Let us find the entropy changes in a Carnot refrigerator
    ♦ For that, we consider one complete cycle of the refrigerator
    ♦ There are two isothermal processes and two adiabatic processes in one complete cycle
• The PV diagram is shown again below:

• There is no exchange of heat during the adiabatic processes
    ♦ So there is no entropy changes during the adiabatic processes
• We need to consider the isothermal processes only
• Those isothermal processes are just the reverse of what we saw in the engine
• So it is easy to prove that, in the case of refrigerator also, the entropy change suffered by the universe is zero
   ♦ The steps are left to the reader
• Our next aim is to check whether it is possible to make a refrigerator with COP infinity. It can be written in 6 steps:
(i) If there is to be infinite COP, W must be zero
• That is., there should be zero work requirement
    ♦ The cooling must be accomplished with out external work
• This condition can be easily seen from the schematic diagram of a refrigerator that we saw in a previous section. It is shown again below:

• It is clear that, for infinite COP, W must be zero
(ii) If W is not to be given, the segments CB and BA will be absent in the PV diagram below
   ♦ Because, W is applied during those segments
   ♦ The gas gets compressed during those segments
(iii) If those two segments are absent, entropy changes occur only during DC
    ♦ During DC, the entropy change suffered by the surroundings (hot reservoir) = $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$
    ♦ During DC, the entropy change suffered by the gas (system) = $\mathbf\small{\rm{\frac{Q_c}{T_2}}}$
(iv) Here also, the net entropy change of system and surroundings is equal to zero
• That means, the entropy change suffered by the universe is zero
• This does not violate the second law
(v) So we are inclined to adopt such a refrigerator, where no external work is required
(vi) But, if we avoid paths CB and BA, the gas cannot reach the hot temperature T₁
• It is essential to reach the hot reservoir temperature T₁ to dump the heat
• If we decide to retrace the path CD, the absorbed heat will be given back to the  cold reservoir
   ♦ This will not effect cooling
◼ We can obtain continuous extraction of heat only if the refrigerator operates in cycles
(vi) Thus it is clear that, to obtain a cyclic process, some work has to be definitely done
• When some work is done, COP will become less than infinity
• That is why, we cannot obtain infinite COP

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We have completed our present discussion on thermodynamics. In the next section, we will see some solved examples related to the various topics that we saw in this chapter



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