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 12.3 - The Adiabatic process

In the previous section we saw the details about isothermal process. In this section, we will see the remaining three processes.

Isobaric process
This can be explained in 8 steps:
1. The word 'iso' in Greek means equal. 'Baros'  means pressure.
• So isobaric process is a process in which the pressure remains the same from start to finish.
2. Consider the gas inside the cylinder in fig.12.10(a) below.
• The piston is free to move. But in fig.(a), it is in equilibrium at a certain position.

Fig.12.10

3. Let us assume that the mass of the piston is negligible.
• Then we can write:
    ♦ The pressure exerted by the gas, P
    ♦ is equal to
    ♦ The sum of the 'pressure due to the 5 kg mass' and the 'atmospheric pressure P_a'
4. Let A be the area of cross section of the piston.
• Then the pressure due to the 5 kg mass = ⁵⁄_A
• So we get: P = ⁵⁄_A + P_a
    ♦ This is how we get the equilibrium mentioned in (2)
5. Now, if we heat the gas slowly (by supplying Q joules of heat), it's pressure will increase and become a new value P₁
• But we want the pressure to remain the same at (⁵⁄_A + P_a)
• This can be achieved if the piston is 'free to move'.
• How does the pressure remain constant when the piston is 'free to move' ?
The answer can be written in 3 steps:
(i) When the pressure increases to P₁, the gas pushes the piston upwards.
• So volume increases. This is shown in fig.b
(ii) Since more volume is available, the gas molecules have more space to move around.
• They will not hit the walls of the cylinder with the same force.
(iii) That means, the pressure has decreased.
• Thus, even when heat is applied, the pressure remains equal to (⁵⁄_A+P_a)
6. Since the pressure remains the same, the P-V Diagram will be a horizontal line.
This is shown in fig.c
• We know that, work done by the gas will be equal to the area under the P-V diagram.
• In our present case, it is easy to find the area because, it is a rectangular area.
7. In addition to the above graphical method, we can use the ideal gas equation also to calculate the work. It can be explained in 5 steps:
(i) We have: PV = nRT
(ii) Here P, n and R are constants. So we get:
PV_A = nRT_A and PV_B = nRT_B
(iii) Subtracting PV_A from both sides of 'PV_B = nRT_B', we get:
PV_B - PV_A = nRT_B - PV_A
(iv) But PV_A = nRT_A
• Using this on the right side of (iii), we get:
PV_B-PV_A = nRT_B - nRT_A
⇒ P(V_B-V_A) = nR(T_B-T_A)
(v) But P(V_B-V_A) = P ΔV, is the work done.
• So we get Eq.12.4: Work done in an isobaric process = nR(T_B-T_A)
8. Let us apply the first law of thermodynamics to an isobaric process. It can be written in 3 steps:
(i) When the gas expands, work is done by the gas.
• Eq.12.1 becomes: ΔU = Q - W
• So we get: ΔU = Q - P(V_B-V_A)
• The heat Q can be calculated using Eq.11.9 that we saw in the previous chapter (Details here).
Q = nC_PΔT
• Thus it is possible to calculate ΔU
(ii) When the gas contracts, work is done on the gas.
• Eq.12.1 becomes: ΔU = Q + W
• So we get: ΔU = Q + P(V_A-V_B)
• The heat Q can be calculated using Eq.11.9 as before.
(iii) Once we obtain W and Q, we can easily calculate ΔU.

Solved example 12.4
Find the change in internal energy of 1 gram of boiling water when it is changed from liquid state to gaseous state.
Solution:
1. Take 1 gram of 'water at 100 °C' in an open vessel.
• This water is at 100 °C. So it has an initial internal energy, U_i
• If we supply heat to this water, it will vaporize into steam.
2. Let us see the pressure acting on the system.
• Our system is the 1 g of water at 100 °C
• It is in an open vessel. So the atmospheric pressure P_a is compressing it.
• We can assume that, a 'mass less imaginary piston' is applying a pressure of P_a on the water.
3. Since heat is being supplied, the water gradually turns to steam.
• This steam pushes the imaginary piston upwards.
    ♦ So pressure will not build up below the piston.
    ♦ The system will be under the constant pressure of P_a all the while.
• So it is an isobaric process.
4. For an isobaric process, we can calculate the work done easily.
• In our present case, the equation will be: P_a(V_f - V_i)
    ♦ Vi is the initial volume occupied by 1 gram of water.
          ✰ When the pressure is P_a, this volume is equal to 1 cm^₃
    ♦ Vf is the volume occupied by the steam produced from 1 gram water.
          ✰ When the pressure is P_a, this volume is equal to 1671 cm^₃
• Thus W = P_a(V_f - V_i) = 1.013 × 10^₅ × (1671 - 1) × 10^_-6 = 169.2 J
5. Next we want the quantity of heat supplied to the 1 gram water.
• We know that, water at boiling point will require the latent heat to vaporize (Details here).
• This latent heat will be equal to: (Mass of water × Latent heat of water).
= (1 gram × 2256 J/gram) = 2256 J
6. Now we can apply the first law of thermodynamics.
• We have: ΔU = Q - W
• Substituting the values, we get: ΔU = 2256 - 169.2 = 2086.8 J
7. Remember that, we supplied 2256 J heat.
• But a major portion (2086.8 J) of that heat is used to increase the internal energy of the system.
• The system did a work of only 169.2 J
• If instead of water, the system was a gas, we could obtain more work.

Isochoric process
This can be explained in 4 steps:
1. In an isochoric process, volume remains the same.
• This can be achieved by locking the piston at a suitable position.
2. Since there is no change in volume, work, W will be zero.
• This zero work can be confirmed graphically also:
    ♦ Since the volume is constant, the P-V Diagram will be a vertical line.
    ♦ A vertical line has zero area below it.
3. When W is zero, we can apply the first law of thermodynamics easily.
• We have: ΔU = Q + W
⇒ ΔU = Q + 0
⇒ ΔU = Q
4. We can find Q easily:
• Since the gas is at a constant volume, the heat absorbed by it will be equal to mC_V(T_B-T_A)
    ♦ Where m is the mass of the gas.
    ♦ C_V is the specific heat of the gas at constant volume.
    ♦ T_A is the initial temperature and T_B is the final temperature.
• If this much heat is supplied to the gas, all that heat will be utilized to increase the internal energy of the gas.
• If this much heat is released from the gas, there will be a corresponding decrease in the internal energy of the gas.

Adiabatic process
This can be explained in 7 steps:
1. In an adiabatic process, there is no flow of heat into or out of the system.
• So we assume that, there is an imaginary insulating layer on these two areas:
    ♦ the inside surface of the cylinder (green layer in fig.12.11 below)
    ♦ the underside of the piston (pink layer in fig.12.11 below)

Fig.12.11

2. Since there is no heat flow in either direction, we can apply the first law of thermodynamics easily:
• We have: ΔU = Q + W
⇒ ΔU = 0 + W
⇒ ΔU = W
3. So we can write:
• If we do any work on the gas, all those work will be utilized to increase the internal energy of the gas.
• If the gas does any work, that work will be at the expense of the internal energy.
4. Note that, in an adiabatic process, neither pressure nor volume is a constant.
• So the P-V Diagram will be a curve.
(Recall that, if P is constant, the diagram will be a horizontal line. If V is constant, the diagram will be a vertical line. In our present case, both P and V vary)
• We can use calculus to find the area below the curve
    ♦ We will need the equation of the curve.
(Recall the steps that we did in the case of isothermal process. See fig.12.9 of the previous section)
5. For an adiabatic process, the equation is: $\mathbf\small{\rm{PV^{\gamma}=A \, Constant}}$
    ♦ Where $\mathbf\small{\rm{\gamma=\frac{C_P}{C_V}}}$
6. Now we can apply calculus. We get the following result:
The area below the curve = $\mathbf\small{\rm{\frac{1}{1\, - \,\gamma}\left[P_B V_B \, - \, P_A V_A \right]}}$
7. Thus we get Eq.12.5: $\mathbf\small{\rm{W=\frac{1}{1\, - \,\gamma}\left[P_B V_B \, - \, P_A V_A \right]}}$
8. We saw that, an ideal gas subjected to an adiabatic process will satisfy the equation in (5).
• But any ideal gas will satisfy the equation PV = nRT also.
• Based on this, we can derive another equation for W. It can be written in 5 steps:
(i) P_AV_A = nRT_A
(ii) P_BV_B = nRT_B
(iii) Substracting P_BV_B from both sides of (i), we get:
P_AV_A – P_BV_B = nRT_A – P_BV_B
(iv) But from (ii), we have: P_BV_B = nRT_B
• So the right side of (iii) can be changed:
P_AV_A – P_BV_B = nRT_A – nRT_B
⇒ P_AV_A – P_BV_B = nR(T_A – T_B)
(v) So Eq.12.5 can be modified as: $\mathbf\small{\rm{W=\frac{-1}{1\, - \,\gamma}\left[nR(T_A \, - \, T_B) \right]}}$
■ Thus we get Eq.12.6: $\mathbf\small{\rm{W=\frac{nR(T_A \, - \, T_B)}{\gamma\, - \,1}}}$  
• Where:
    ♦ W is the work done in an adiabatic process.
    ♦ T_A and T_B are the initial and final temperatures.
9. We can derive another interesting equation related to adiabatic processes. It can be written in 4 steps:
(i) In the previous step (8), we have used subscripts A and B.
• ‘A’ denotes the initial state of the gas and B denotes the final state of the gas.
    ♦ We successfully calculated the work done from A to B.
• We can use subscripts 1, 2, 3 etc., also.
    ♦ They will denote the different states of the gas.
(ii) Any ideal gas will satisfy the equation PV = nRT
• n and R are constants. So we get:
P₁V₁ = nRT₁ and P₂V₂ = nRT₂
• Dividing the first equation by P₂V₂, we get: $\mathbf\small{\rm{\frac{P_1 V_1}{P_2 V_2}=\frac{nRT1}{P_2 V_2}}}$
• But P₂V₂ = nRT₂
So we get: $\mathbf\small{\rm{\frac{P_1 V_1}{P_2 V_2}=\frac{nRT_1}{nRT_2}=\frac{T_1}{T_2}}}$
(iii) When this ideal gas is subjected to an adiabatic process, it will satisfy the relation $\mathbf\small{\rm{PV^{\gamma}=A \, Constant}}$ also
So we get: $\mathbf\small{\rm{P_1 V_1^{\gamma}=P_2 V_2^{\gamma}}}$
$\mathbf\small{\rm{\frac{P_1}{P_2}=\left ( \frac{V_2}{V_1} \right )^{\gamma}}}$
(iv) Substituting this in (ii), we get:
$\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma} \times \frac{V_1}{V_2}=\frac{T_1}{T_2}}}$
■ Thus we get Eq.12.7:
$\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$

Let us see two solved examples
Solved example 12.5
During an adiabatic process, a gas expands from state 1 to state 2. The initial temperature T₁ is 292 K. Calculate the final temperature T₂, if the volume expansion ratio is 1.28. Given that 𝛄 = 1.40
Solution:
1. We have Eq.12.7: $\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$
2. Given that $\mathbf\small{\rm{\frac{V_2}{V_1}}}$ = 1.28
3. Substituting all the known values in Eq.12.7, we get:
$\mathbf\small{\rm{(1.28)^{1.40 - 1}=\frac{292}{T_2}}}$
• Thus we get: T₂ = 265 K

Solved example 12.6
In an automobile engine, the initial values related to the gas are as follows:
T₁ = 293 K, P₁ = 1.00 × 10^₅ N/m2 and V₁ = 240 cm^₃. This gas is compressed adiabatically to a final volume V₂, which is 40 cm^₃. Find the values of P₂ and T₂. Also find the work done on the gas. Given that, 𝛄 = 1.40
Solution:
Part (a). To find P₂:
• For an adiabatic process, we have: $\mathbf\small{\rm{P_1 V_1^{\gamma}=P_2 V_2^{\gamma}}}$
⇒ $\mathbf\small{\rm{P_2=P_1\left ( \frac{V_1}{V_2} \right )^{\gamma}}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{P_2=1.00 \times 10^5 \times \left ( \frac{240}{40} \right )^{1.4}}}$ = 12.3 × 10^₅ N/m2
Part (b). To find T₂
• We have Eq.12.7: $\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{\left(\frac{40}{240} \right )^{1.40 - 1}=\frac{293}{T_2}}}$
• Thus we get: T₂ = 600 K
Part (c). To find W
• We have Eq.12.5: $\mathbf\small{\rm{W=\frac{1}{1\, - \,\gamma}\left[P_2 V_2 \, - \, P_1 V_1 \right]}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{W=\frac{1}{1\, - \,1.4}\left[(12.3) (40) \, - \, (1)(240) \right]\times 10^5 \times 10^{-6}}}$
• Thus we get: W = -63 J

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In the next section, we will see cyclic processes


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