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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Chapter 12.9 - COP of a Carnot Refrigerator

In the previous section, we saw the efficiency of the Carnot engine. In this section, we will see details about the Carnot refrigerator

• A Carnot engine can be operated in the reverse direction to work as a Carnot refrigerator
• The four segments in the PV diagram will reverse their directions. This is shown in fig.6.24 below:

Fig.12.24

The basic working can be explained in 4 steps:
1. The cycle begins at point D. The first segment is DC
    ♦ The engine is placed in contact with the cold reservoir of temperature T₂
    ♦ The engine extracts heat from the cold reservoir and expands isothermally
2. The second segment is CB
    ♦ The engine is placed inside an insulator
    ♦ The gas is compressed adiabatically
    ♦ The temperature increases to T₁
3. The third segment is BA
    ♦ The engine is placed in contact with the hot reservoir of temperature T₁
    ♦ The gas is compressed isothermally
    ♦ During this compression, the gas rejects some heat into the hot reservoir
4. The fourth segment is AD
    ♦ The engine is placed inside an insulator
    ♦ The gas is allowed to expand adiabatically
    ♦ The temperature falls to T₂
• Thus the initial point D is reached, completing one cycle

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Next we will calculate the work done in each segment:

First segment from D to C
• In this segment, the gas place in contact with T₂ and subjected to an isothermal expansion
    ♦ Volume increases from V_D to V_C
    ♦ Pressure decreases from P_D to P_C
    ♦ Temperature remains constant at T₂
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from C to B
• In this segment, the gas is subjected to an adiabatic compression
    ♦ Volume decreases from V_C to V_B
    ♦ Pressure increases from P_C to P_B
• We know that, the work done by the ideal gas during the adiabatic expansion will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$ (see Eq.12.6 at end of section 12.3)

Third segment from B to A
• In this segment, the gas is subjected to an isothermal compression
    ♦ Volume decreases from V_B to V_A
    ♦ Pressure increases from P_B to P_A
    ♦ Temperature remains constant at T₁
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$

Fourth segment from A to D
• In this segment, the gas is subjected to an adiabatic expansion
    ♦ Volume increases from V_A to V_D
    ♦ Pressure decreases from P_A to P_D
• We know that, the work done by the ideal gas during the adiabatic compression will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$

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Now we can find the coefficient of performance (COP) of the Carnot refrigerator. It can be written in steps:
1. Net work, W = Work done by the gas - Work done on the gas
   ♦ Work done by the gas = Work done during segments one and four
   ♦ Work done on the gas = Work done during segments two and three
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( \frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}+n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}-n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$

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• An interesting comparison:
    ♦ In the previous section for the engine, we obtained the net work as:
          $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
    ♦ In this section for the refrigerator, we obtain the net work as:
          $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}-n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$
    ♦ Both are equal in magnitude but opposite in sign
◼ That means:
    ♦ In the engine, work is done by the gas
    ♦ In the refrigerator (reversed engine), the same work is done on the gas

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2. We know that, COP of a refrigerator is given by:
$\mathbf\small{\rm{\alpha = \frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Work}}}$ (see Eq.12.12 in section 12.6)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\alpha = \frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Heat}}}$ (see Eq.12.13 in section 12.6)
• It is easier to find COP using this equation
4. So our next task is to find the 'Heat absorbed' and the 'Heat rejected'
• It can be done in steps:
(i) We know that, heat absorption takes place only during the first segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_C}{V_D}}}$
(ii) We know that, heat rejection takes place only during the third segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat rejected = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_B}{V_A}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}-n R T_2\,\,ln \frac{V_C}{V_D}}}}$
6. In the previous section, we compared the two adiabatic segments, and proved that: $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
• So the result in (5) becomes:
$\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_C}{V_D}-n R T_2\,\,ln \frac{V_C}{V_D}}}}$
⇒ $\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R \,\,ln \frac{V_C}{V_D}(T_1 - T_2)}}}$
• Thus we get:
Eq.12.16: $\mathbf\small{\rm{\alpha = \frac{T_2}{T_1 - T_2}}}$
7. Two expressions for COP:
   ♦ The above Eq.12.16 gives us an expression for COP
   ♦ Step (3) also gives us an expression for COP
• So we can equate the two. We get Eq.12.17:
$\mathbf\small{\rm{\frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Heat}=\frac{T_2}{T_1 - T_2}}}$

Let us see a solved example
Solved example 12.11
A refrigerator is to maintain eatables kept inside at 9 ^०C. If room temperature is 36 ^०C, Calculate the coefficient of performance
Solution:
1. Given that:
   ♦ Temperature of the hot reservoir T₁ = 36 ^०C = (273 + 36) = 309 K
   ♦ Temperature of the cold reservoir T₂ = 9 ^०C = (273 + 9 ) = 282 K
2. We have Eq.12.16: $\mathbf\small{\rm{\alpha = \frac{T_2}{T_1 - T_2}}}$
• Substituting the values, we get:
$\mathbf\small{\rm{\alpha = \frac{282}{309 - 282}}}$ = 10.44

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In the next section, we will see the second law of thermodynamics and it's application to heat engines and refrigerators


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Chapter 12.8 - Efficiency of a Carnot Engine

In the previous section, we saw the basic processes in a Carnot engine. In this section, we will see the work done during each of those processes. For easy reference, the final PV diagram is shown below:

Fig.12.23

First segment from A to B
• In this segment, the gas is subjected to an isothermal expansion
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from B to C
• In this segment, the gas is subjected to an adiabatic expansion
• We know that, the work done by the ideal gas during the adiabatic expansion will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$ (see Eq.12.6 at end of section 12.3)

Third segment from C to D
• In this segment, the gas is subjected to an isothermal compression
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$

Fourth segment from D to A
• In this segment, the gas is subjected to an adiabatic compression
• We know that, the work done by the ideal gas during the adiabatic compression will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$

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Now we can find the efficiency of the Carnot engine. It can be written in 9 steps:
1. Net work, W = Work done by the gas - Work done on the gas
   ♦ Work done by the gas = Work done during segments one and two
   ♦ Work done on the gas = Work done during segments three and four
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
2. We know that, efficiency of a heat engine is given by:
$\mathbf\small{\rm{\eta = \frac{Net \; Work}{Heat \; Absorbed}}}$ (see Eq.12.9 in section 12.5)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\eta = 1-\frac{Heat \;Rejected}{Heat \; Absorbed}}}$ (see Eq.12.10 in section 12.5)
• It is easier to find efficiency using this equation
4. So our next task is to find the 'Heat absorbed' and the 'Heat rejected'
• It can be done in steps:
(i) We know that, heat absorption takes place only during the first segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(ii) We know that, heat rejection takes place only during the third segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat rejected = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\eta = 1-\frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}}}}$
⇒ $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right ) \left (\frac{ln \frac{V_C}{V_D}}{ln \frac{V_B}{V_A}}\right )}}$
6. We see that, the right side of the above equation involves temperatures and volumes. So our next task is to find the relations between them
• For that, we consider the two adiabatic processes in the cycle. It can be written in two steps:
(i) For the  adiabatic process BC, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(see Eq.12.7 of section 12.3)
(ii) Similarly, for the adiabatic process DA, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_D}{V_A} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(iii) The right sides in (i) and (ii) are the same. So we get:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{V_D}{V_A} \right )^{\gamma - 1}}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )=\left (\frac{V_D}{V_A} \right )}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
7. We see the above result in (5)
So the result in (5) becomes:
Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
8. Two expressions for efficiency:
   ♦ The above Eq.12.14 gives us an expression for efficiency
   ♦ Step (3) also gives us an expression for efficiency
• So we can equate the two:
$\mathbf\small{\rm{1-\frac{Heat \;Rejected}{Heat \; Absorbed} = 1- \left (\frac{T_2}{T_1} \right )}}$
• Thus we get Eq.12.15:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{T_2}{T_1} \right )}}$

⇒ $\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{Temperature \; of \; cold \; reservoir}{Temperature \; of \; hot \; reservoir} \right )}}$
9. Let us use try the results in (1) and (2) to find efficiency. it can be written in 3 steps:
(i) Using the result in (6), the result in (1) becomes:
W = $\mathbf\small{\rm{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}}$
(ii) Using the result in 4(i), we have:
Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(iii) Applying the above two results into (2), we get:
$\mathbf\small{\rm{\eta = \frac{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}{n R T_1\,\,ln \frac{V_B}{V_A}}=\frac{T_1 - T_2}{T_1}}}$
⇒ $\mathbf\small{\rm{\eta =1-\left (\frac{T_2}{T_1} \right )}}$
This is the same result as in Eq.12.14

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Let us see a solved example
Solved example 12.10
Three heat engines operate between the following temperature differences
    ♦ Engine A: T₁ = 1000 K, T₂ = 700 K
    ♦ Engine A: T₁ = 800 K, T₂ = 500 K
    ♦ Engine A: T₁ = 600 K, T₂ = 300 K
• Which is the most efficient engine?
• Which is the least efficient engine?
Solution:
1. Efficiency of a heat engine is given by Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
• So we get:
    ♦ Efficiency of Engine A = $\mathbf\small{\rm{\eta = 1- \left (\frac{700}{1000} \right )}}$ = 0.3
    ♦ Efficiency of Engine B = $\mathbf\small{\rm{\eta = 1- \left (\frac{500}{800} \right )}}$ = 0.375
    ♦ Efficiency of Engine C = $\mathbf\small{\rm{\eta = 1- \left (\frac{300}{600} \right )}}$ = 0.5
2. We can write:
    ♦ Engine C is the most efficient
    ♦ Engine A is the least efficient
3. Dependence of efficiency on temperature:
    ♦ We see that, the temperature difference is the same for all three engines
    ♦ In such a situation, the engine with the smallest T₂ has the most efficiency 

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In the next section, we will see Carnot refrigerator


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Chapter 12.7 - The Carnot Engine

In the previous section, we saw the basics about refrigerator. In this section, we will see the Carnot engine

Carnot Engine

The basic processes involved in the working of a Carnot engine can be written in 14 steps:
1. A Carnot engine is an ideal engine
• To make it ideal, two restrictions are imposed:
(i) Every process that occur during the working of a Carnot engine must be reversible
(ii) The working gas inside the engine must be an ideal gas
2. We have seen the details about reversible process in the first section of this chapter (see fig.12.4 of the first section)
• Based on that, we can write:
    ♦ At any instant during the working of a Carnot engine, the gas inside the engine will have the same temperature as the surroundings
    ♦ We can change the direction of the reversible process any time we want
    ♦ A reversible process can proceed only very slowly
    ♦ So the Carnot engine will be very slow. But it gives us a base to study other real irreversible engines
3. Initially, the gas is at state A
• For the engine to start operating, we have to supply heat
   ♦ So the cylinder is placed in contact with a hot reservoir
   ♦ This is shown in fig. 12.19(a) below
   ♦ The temperature of this hot reservoir is maintained at T₁ K
• So at this state, the V, P, T values of the gas will be: (V_A, P_A, T_A)
    ♦ Where T_A = T₁

Fig.12.19

4. The gas absorbs heat and expands. Thus work is done on the piston
• This expansion process should be isothermal
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason can be seen here
• At the end of this isothermal process, the gas reaches state B
    ♦ Pressure P_A decreases to a lower value P_B
    ♦ Volume V_A increases to a higher value V_B
    ♦ Temperature remains the same because it is an isothermal process
• So at this state, the V, P, T values of the gas will be: (V_B, P_B, T_B)
    ♦ Where T_B = T₁
• This process is represented by the red curve AB in fig.12.19(b)
• The first segment of the cycle is complete
5. Next we will see not the second segment, but another important segment
• Some quantity of heat is to be dumped into a cold reservoir
• Only then, can the gas return to the initial state and become ready for the next cycle
   ♦ For that, we place the cylinder in contact with a cold reservoir
   ♦ This is shown in fig.12.20(a) below
   ♦ The temperature of the cold reservoir is maintained at T₂ K
• So at this state, the V, P, T values of the gas will be: (V_C, P_C, T_C)
    ♦ Where T_C = T₂

Fig.12.20

6. The gas is compressed isothermally. Thus work is done on the gas
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason is same (but in reverse order)as that we saw for the first segment
    ♦ The discussion can be seen here
• At the end of this isothermal process, the gas reaches state D
    ♦ Pressure P_C increases to a higher value P_D
    ♦ Volume V_C decreases to a lower value V_D
    ♦ Temperature remains the same because it is an isothermal process
• So at this stage, the V, P, T values of the gas will be: (P_D, V_D, T_D)
    ♦ Where T_D = T₂
• This process is represented by the red curve CD in fig.12.20(b)
• This segment of the cycle is complete
7. The two segments mentioned above are important and inevitable segments
• During AB, the gas absorbs heat and does useful work
• But the gas has to return to it's initial state to start the next cycle
   ♦ For that, it has to dump the heat
   ♦ This dumping is done during CD
8. But we see a problem. It can be explained in 2 steps:
(i) The curve AB is separated away from curve CD
• This is obvious because, they are isotherms
    ♦ Every point in AB will be at temperature T₁
    ♦ Every point in CD will be at temperature T₂
    ♦ Two isotherms will never meet
(ii) So our problem is this:
How to reduce the temperature of the gas from T₁ (at state B) to T₂ (at state C)
9. Just after state B, we must subject the gas to a suitable process so that, the temperature falls to T₂
◼ Which process is suitable?
• An adiabatic process is the only suitable process
• Processes like isobaric, isochoric or isothermal cannot be used
    ♦ The reason can be seen here
10. An adiabatic process is most suitable for achieving our goal
• Just after state B, the cylinder is placed in a thermal insulator
This is shown in fig.6.21 below:

Fig.6.21

• The gas is then allowed to expand adiabatically
   ♦ So no heat will be lost or gained
• The gas will use it's own internal energy to expand
   ♦ This results in the decrease in temperature of the gas from T_B to T_C (T₁ to T₂)
• Thus we successfully established the connection between the first and third segments
11. Next we want to establish the connection between the third and first segments
• That is., between points D and A
• During the process from D to A,
    ♦ The volume decreases from V_D to V_A
    ♦ The pressure increases from P_D to P_A
    ♦ Temperature increases from T₂ to T₁
• So our problem is this:
How to increase the temperature of the gas from T₂ (at state D) to T₁ (at state A)
12. Just after state D, we must subject the gas to a suitable process so that, the temperature rises to T₁
◼ Which process is suitable?
• An adiabatic process is the only suitable process
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason is same (but in reverse order)as that we saw for the second segment
    ♦ The discussion can be seen here
13. An adiabatic process is most suitable for achieving our goal
• Just after state C, the cylinder is placed in a thermal insulator
This is shown in fig.6.22 below:

Fig.6.22

• The gas is then compressed adiabatically
   ♦ This results in the increase in temperature of the gas from T_D to T_A (T₂ to T₁)
• Thus we successfully established the connection between the third and first segments
• The cycle is now complete
14. Let us write a summary. It can be written in 4 steps:
(i) We supply energy to the gas during the first segment
    ♦ During this segment, the gas expands and does useful work
    ♦ The supply of energy is stopped at the end of the first segment
(ii) During the second segment also, useful work is done by the gas
    ♦ But this time, the gas expands using it's internal energy
    ♦ The temperature falls to T₂ during this segment
(iii) The gas has to return to it's original state so that another cycle can begin
    ♦ So in the third and fourth segments, the gas undergoes compression
    ♦ In the third segment, the compression is carried out at a lower temperature T₂
        ✰ So it is easier to achieve compression
(iv) In the fourth segment, the temperature rises back to T₁
    ♦ This is because, no heat is allowed to escape during the compression

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• So now we know the four processes involved in the Carnot cycle
• Next we want to calculate the work done in each of the four segments in the cycle
• We will see it in the next section

Discussion on the notes can be started here

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Chapter 12.2 - Thermodynamic Work

In the previous section we saw the first law of thermodynamics. In this section, we will see work done during a thermodynamic process.

The method of calculating the work can be explained in 18 steps:
1. We have seen the gas inside the cylinder in fig.12.5 at the beginning of the previous section.
• Suppose that, the gas is expanding.
    ♦ The expanding gas will push the piston upwards.
    ♦ That means, the gas is doing work on the piston.
2. By a suitable mechanism, the piston can be connected to the wheel of an automobile.
• Then the upward motion of the piston will rotate the wheel.
• So in effect, we are using the ‘expansion of the gas’ to rotate the wheel.
• That means, the work done by the gas is converted into kinetic energy of the automobile.
3. We want to calculate the amount of work done by the gas.
• In fig.12.7(b) below, the green arrows indicate the pressure exerted by the gas on the piston.
    ♦ Let this pressure be P.

Fig.12.7

• The piston will be having a certain ‘cross sectional area’ A.
    ♦ So the pressure P is acting on an area A.
• We know that: Pressure = ^Force⁄_Area
    ♦ So Force = Pressure × Area
    ♦ So in our present case, force exerted by the gas = P × A
4. As a result of this force, let the piston move up a distance of Δd.
    ♦ This is shown in fig.12.7(c)
• We know that: Work done = Force × distance
• So the work done by the gas in moving the piston by a distance of Δd.
= (P × A)  × Δd
• We can rearrange this as:
Work done = P × (A × Δd)
5. But (A  × Δd) will be the volume of a prism (in our present case, the prism is a cylinder) whose base area is A and height is Δd.
    ♦ This is shown in fig.12.7(d)
• In our present case, it is the volume between the initial and final positions of the piston.
• So (A × Δd) is the change in volume ΔV
6. Substituting for '(A × Δd)' in (4), we get:
Eq.12.2: Work done by the gas in moving the piston by a distance of Δd = P × ΔV
• So we can find the work done by multiplying pressure and change in volume.
7. But there is a problem. It can be explained in 3 steps:
(i) Consider fig.a. It shows the final position of the piston.
    ♦ For better comparison, it is drawn to the left of the initial state (fig.b)
• The piston has moved through a large distance d.
(ii) In such a case, volume has increased by a large amount.
• So pressure will have changed.
    ♦ The pressure in fig.a will be different from the pressure in fig.b.
• If the piston is at some intermediate point, then also the pressure will be different from the pressure in fig.b.
• It is clear that, P varies continuously as the thermodynamic process proceeds.
(iii) We want to use the formula: Work done = P × ΔV
• But if P is continuously changing, we cannot input a value for P.
■ So we adopt a new procedure. The following steps from (8) to (14) explains this new procedure:
8. Fig.12.8(a) below shows a P-V Diagram.
    ♦ A (V_A,P_A) is the initial point.
    ♦ B (V_B, P_B) is the final point.
• It is clear that, pressure varies with volume.
(If pressure was a constant, the graph would have been a straight horizontal line)
• A number of points (yellow dots) are marked on the graph.

Fig.12.8

9. Next, we draw some rectangles based on those dots.
• The rectangles must satisfy an important condition:
    ♦ The dots must be situated at the top-left corner of each rectangle.
• When the rectangles are drawn in this way, we get:
    ♦ The heights of the rectangles are equal to the pressure values at the dots.
10. The bases of the rectangles are also important.
• The bases of the rectangles are on the volume axis. The widths of those bases will give the changes in volume.
• For example:
    ♦ At the first yellow dot, pressure is P₁ and volume is V₁
    ♦ At the second yellow dot, pressure is P₂ and volume is V₂
    ♦ So the width of the base of the second rectangle = V₂ – V₁
    ♦ This is the change in volume ΔV
          ✰ This is the change when,
          ✰ the gas moves from the first yellow dot to the second yellow dot.
11. Thus we can obtain the heights and widths of all rectangles.
• Now, when the gas moves from one dot to the next, we assume that the pressure remains constant.
• For example:
    ♦ When the gas moves from the first dot to the second, the pressure surely changes from P₁ to P₂
    ♦ But since the width of the rectangle is small, we assume that: Pressure remains constant at P₁
    ♦ So for the volume change (ΔV = V₂ – V₁), the pressure is P₁
12. Since the pressure is constant at P₁, we can use the equation 12.2
• We get: work done while moving the piston from V₁ to V₂ = P₁ ΔV = P₁ (V₂ – V₁)
• Clearly,  'P₁ (V₂ – V₁)' is the area of the second rectangle.
13. So it is clear that:
• Work done when the volume changes from V₁ to V₂ is equal to the area of the second rectangle.
[Let us do a dimensional check:
• We want to check whether
    ♦ The dimensions of (P ΔV)
    ♦ is same as
    ♦ The dimensions of work
• The dimensions of (P ΔV) are:
$\mathbf\small{\rm{\frac{MLT^{-2}}{L^2} \times L^3\;=\;ML^2T^{-2}}}$
• The dimensions of work are:
$\mathbf\small{\rm{MLT^{-2}\times L\;=\;ML^2 T^{-2}}}$
• Both dimensions are the same]
14. The total work:
• Since we know the heights and widths of all the rectangles, we can calculate the areas of all the rectangles.
• The sum of all those areas will give the total work done when the piston moves from A to B.
• Thus we succeed in calculating the work done.
15. But now another problem arises. It can be explained in 2 steps:
(i) In fig.12.8 (a), the top right corners of the rectangles are above the curve.
    ♦ So a small triangular portion of each rectangle is outside the curve.
(ii) Thus some 'unwanted areas' are included while calculating the total work.
    ♦ This will lead to error.
16. To solve this problem, we reduce the widths of the rectangles.
• Then the triangular portions will become smaller.
• As the widths decrease, sizes of the unwanted triangles also decrease.
■ If we make the widths infinitesimal, there will not be any triangular portions projecting out.
• The total area calculated by such a method will give the accurate amount of work.
17. But when the widths are infinitesimal, the number of rectangles will be very large.
    ♦ A few such rectangles are shown in fig.b
• We will have to calculate the areas of all rectangles from point A to point B.
• Here, calculus comes to our help. Using calculus, we can easily find the area below the curve.
(At this stage, we do not have to learn the mathematical details of calculus. All we need is the final expression for determining the work. We will see such an expression later in this section)
18. Thus the above 17 steps help us to arrive at the fact that:
The area below the P-V Diagram will give the work done by the gas or the work done on the gas.

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• The gas inside the cylinder can be subjected to four types of thermodynamic processes. They are:
(i) Isothermal process
(ii) Isobaric process
(iii) Isochoric process
(iv) Adiabatic process
We will now see the basics of each of the above four processes.

Isothermal process
This can be explained in 7 steps:
1. The word 'iso' means equal.
• So isothermal process is a process in which the temperature remains the same from start to finish.
2. Consider the gas inside the cylinder.
• Let us compress the gas using the piston.
    ♦ So work is done on the gas. It's internal energy will increase.
• Due to the increase in internal energy, it's temperature will also increase.
3. For the process to be isothermal, there must be no increase in temperature.
• So we put the cylinder in contact with a thermal reservoir.
• A thermal reservoir is a body of large mass.
    ♦ It also has a large specific heat capacity.
• So it can remain at the same temperature for a long time.
    ♦ A lake of water is a good example for a thermal reservoir.
■ When the cylinder is kept in contact with a cold reservoir, the heat will flow out to the reservoir, and so the temperature will remain constant.
4. Similarly, when work is done by the gas, it’s internal energy will decrease and so the temperature will decrease.
• To keep the temperature constant, we put the cylinder in contact with a hot reservoir.
• So heat will flow into the cylinder and temperature will remain the same.
5. We know that, if there is no temperature change, there will not be any change in internal energy. That means, ΔU = 0
• So applying Eq.12.1, (when work is done on the gas) we get:
0 = Q + W
⇒ W = -Q
• That means, all the work done on the gas will be released as 'heat from the gas'.
6. Applying Eq.12.1 (when work is done by the gas) we get:
0 = Q – W
⇒ W = Q
• That means, all the heat supplied to the gas is converted into 'work done by the gas'.
7. Let us calculate the work done on or by the gas. It can be written in 9 steps:
(i) Fig.12.9(a) below shows the P-V Diagram of an isothermal process.
• The red curve indicates that, the gas experiences various pressures and volumes.
• But all the while, the temperature will remain constant.
• So in this case, the red curve is called an isotherm.

Fig.12.9

(ii) In fig.(b), vertical dashed lines are drawn through V_A and V_B
(iii) So we have a region with well defined boundaries.
• This region is shaded with magenta hatch lines.
    ♦ The vertical dashed lines through V_A and V_B form the left and right boundaries.
    ♦ The x-axis forms the bottom boundary.
    ♦ The red curve forms the top boundary.
(iv) So three boundaries are horizontal/vertical straight lines. Only the top boundary is curve shaped.
• If the top boundary was also a straight line, we could have calculated the area very easily.
(v) However, we need not worry much because, the curve is well defined.
• The curve is a well defined function.
• In simple terms, we can say: Our curve has a unique equation.
• This is just like a line having a unique equation: y = mx + c.
(vi) Let us find the equation of our curve:
    ♦ We know the ideal gas equation: PV = nRT.
    ♦ If there is no leakage of gas, n will be a constant.
    ♦ R is the universal gas constant.
    ♦ Since our present process is isothermal, T is also a constant.
(vii) So in effect, the equation of the red curve in fig.12.9 is: PV = A constant.
    ♦ This is same as: $\mathbf\small{\rm{P=\frac{A \, Constant}{V}}}$
    ♦ It is of the form: $\mathbf\small{\rm{y=\frac{A \, Constant}{x}}}$
• In math classes, we see the graph of $\mathbf\small{\rm{y=\frac{A \, Constant}{x}}}$
    ♦ That graph has a shape similar to the red curve in fig.12.9
(viii) Also recall that, $\mathbf\small{\rm{P=\frac{A \, Constant}{V}}}$  is the relation given by Boyle’s law
• Robert Boyle did experiments on the gas while keeping n and T constant.
(ix) Now, since the top boundary is well defined, calculus will do the mathematical works.
• The result is:
Area of the magenta hatched region = $\mathbf\small{\rm{n R T\,\,ln \frac{V_B}{V_A}}}$
So we get Eq.12.3:
Work done in an isothermal process = $\mathbf\small{\rm{n R T\,\,ln \frac{V_B}{V_A}}}$

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In the next section, we will see the remaining three processes

 

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