We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative effects, as remarked in the preceding section, and lowers efficiency.
A process is reversible if it is quasi-static and non-dissipative. We have seen that a process is not quasi-static if it involves finite temperature difference between the system and the reservoir. This implies that in a reversible heat engine operating between two temperatures, heat should be absorbed (from the hot reservoir) isothermally and released (to the cold reservoir) isothermally.
We thus have identified two steps of the reversible heat engine : isothermal process at temperature `T_1` absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature `T_2` releasing heat `Q_2` to the cold reservoir.
To complete a cycle, we need to take the system from temperature `T_1` to `T_2` and then back from temperature `T_2` to `T_1`. Which processes should we employ for this purpose that are reversible? A little reflection shows that we can only adopt reversible adiabatic processes for these purposes, which involve no heat flow from any reservoir.
If we employ any other process that is not adiabatic, say an isochoric process, to take the system from one temperature to another, we shall need a series of reservoirs in the temperature range `T_2` to `T_1` to ensure that at each stage the process is quasi-static.
(Remember again that for a process to be quasi-static and reversible, there should be no finite temperature difference between the system and the reservoir.) But we are considering a reversible engine that operates between only two temperatures. Thus adiabatic processes must bring about the temperature change in the system from `T_1` to `T_2` and `T_2` to `T_1` in this engine.
A reversible heat engine operating between two temperatures is called a Carnot engine. We have just argued that such an engine must have the following sequence of steps constituting one cycle, called the Carnot cycle, shown in Fig. 12.11. We have taken the working substance of the Carnot engine to be an ideal gas.
(a) `color{purple} "Step 1 → 2"`
Isothermal expansion of the gas taking its state from `color{green} {(P_1, V_1, T_1)}` to `color{green} {(P_2, V_2, T_1)}`.
The heat absorbed by the gas `(Q_1)` from the reservoir at temperature `T_1` is given by Eq. (12.12). This is also the work done `(W_(1 →2))` by the gas on the environment.
`color{orange} { W_(1 → 2) = Q_1 = μ R T_1 ln (V_2/V_1)}` ............12.23
(b) `color{purple} "Step 2 → 3"`
Adiabatic expansion of the gas from `color{green} {(P_2, V_2, T_1)}` to `color{green} {(P_3, V_3, T_2)}` .
Work done by the gas, using Eq. (12.16), is
`color{orange} {W_(2 ->3) = (mu R(T_1 - T_2))/(gamma -1)}` .............12.24
(c) `color{purple} "Step 3 → 4"`
Isothermal compression of the gas from `color{green} {(P_3, V_3, T_2)}` to `color{green} {(P_4, V_4, T_2)}`.
Heat released `(Q_2)` by the gas to the reservoir at temperature `T_2` is given by Eq. (12.12). This is also the work done `(W_(3 → 4))` on the gas by the environment.
`color{orange} { W_(3->4) = Q_2 = muRT_2 In (V_3/V_4)}` ...............12.25
(d) `color{purple} "Step 4 → 1"`
Adiabatic compression of the gas from `color{green} {(P_4, V_4, T_2)}` to `color{green} {(P_1,V_1, T_1)}`.
Work done on the gas, [using Eq.(12.16)], is
`color{orange} { W_(4 ->1) = mu R ((T_1 - T_2 )/(gamma - 1 ) )}` ........................12.26
From Eqs. (12.23) to (12.26) total work done by the gas in one complete cycle is
`color{lime} { W = W_(1 → 2) + W_(2 → 3) – W_(3 → 4) – W_(4 → 1)}`
`color{blue} { = μ RT_1 ln (V_2/V_1) - μ RT_2 ln (V_3/V_4) }` ....................12.27
The efficiency η of the Carnot engine is
`eta = W/Q_1 = 1 - Q_2/Q_1`
`color{purple} {= 1 - (T_2/T_1 ) (In (V_3/V_4))/(In (V_2/V_1))}` ...................12.28
Now since step 2 → 3 is an adiabatic process,
`color{orange} {T_1 V_2^(gamma -1) = T_2 V_3^(gamma -1)}`
i.e. `V_2/V_3 = (T_2/T_1)^(1//(gamma -1))` ............12.29
Similarly, since step 4 → 1 is an adiabatic process
`color{navy} {T_2 V_4^(gamma -1) = T_1V_1^(gamma -1)}`
i.e. `color{purple} {V_1/V_4 = (T_2/T_1) ^(1//gamma -1)}` ......12.30
From Eqs. (12.29) and (12.30),
`color{green} {V_3/V_4 = V_2/V_1}` ...........12.31
Using Eq. (12.31) in Eq. (12.28), we get
`eta = 1 - T_2/T_1 ` (Carnot engine) ................12.32
We have already seen that a Carnot engine is a reversible engine. Indeed it is the only reversible engine possible that works between two reservoirs at different temperatures.
Each step of the Carnot cycle given in Fig. 12.11 can be reversed. This will amount to taking heat `Q_2` from the cold reservoir at `T_2`, doing work W on the system, and transferring heat `Q_1` to the hot reservoir. This will be a reversible refrigerator.
We next establish the important result (sometimes called Carnot’s theorem) that (a) working between two given temperatures `T_1` and `T_2` of the hot and cold reservoirs respectively, no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the Carnot engine is independent of the nature of the working substance.
To prove the result (a), imagine a reversible (Carnot) engine R and an irreversible engine I working between the same source (hot reservoir) and sink (cold reservoir).
Let us couple the engines, `I` and R, in such a way so that I acts like a heat engine and R acts as a refrigerator. Let I absorb heat `Q_1` from the source, deliver work W ′ and release the heat `Q_1- W′` to the sink. We arrange so that R returns the same heat `Q_1` to the source, taking heat `Q_2` from the sink and requiring work `W = Q_1 – Q_2` to be done on it.
Now suppose `η_R < η_I` i.e. if R were to act as an engine it would give less work output than that of I i.e. `W < W′` for a given `Q_1`. With R acting like a refrigerator, this would mean `Q_2 = Q_1 – W > Q_1 – W ′`.
Thus on the whole, the coupled I-R system extracts heat `(Q_1 – W) – (Q_1 – W′) = (W ′ – W)` from the cold reservoir and delivers the same amount of work in one cycle, without any change in the source or anywhere else. This is clearly against the Kelvin-Planck statement of the Second Law of Thermodynamics. Hence the assertion `η_I > η_R` is wrong.
No engine can have efficiency greater than that of the Carnot engine. A similar argument can be constructed to show that a reversible engine with one particular substance cannot be more efficient than the one using another substance.
The maximum efficiency of a Carnot engine given by Eq. (12.32) is independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of state, which allows us to readily calculate η, but the final result for `η`, [Eq. (12.32)], is true for any Carnot engine.
This final remark shows that in a Carnot cycle, is a universal relation independent of the nature of the system. Here `Q_1` and `Q_2` are respectively, the heat absorbed and released isothermally (from the hot and to the cold reservoirs) in a Carnot engine. Equation (12.33), can, therefore, be used as a relation to define a truly universal thermodynamic temperature scale that is independent of any particular properties of the system used in the Carnot cycle. Of course, for an ideal gas as a working substance, this universal temperature is the same as the ideal gas temperature introduced in section 12.11.
`color{green} {Q_1/Q_2 = T_1/T_2}`
We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative effects, as remarked in the preceding section, and lowers efficiency.
A process is reversible if it is quasi-static and non-dissipative. We have seen that a process is not quasi-static if it involves finite temperature difference between the system and the reservoir. This implies that in a reversible heat engine operating between two temperatures, heat should be absorbed (from the hot reservoir) isothermally and released (to the cold reservoir) isothermally.
We thus have identified two steps of the reversible heat engine : isothermal process at temperature `T_1` absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature `T_2` releasing heat `Q_2` to the cold reservoir.
To complete a cycle, we need to take the system from temperature `T_1` to `T_2` and then back from temperature `T_2` to `T_1`. Which processes should we employ for this purpose that are reversible? A little reflection shows that we can only adopt reversible adiabatic processes for these purposes, which involve no heat flow from any reservoir.
If we employ any other process that is not adiabatic, say an isochoric process, to take the system from one temperature to another, we shall need a series of reservoirs in the temperature range `T_2` to `T_1` to ensure that at each stage the process is quasi-static.
(Remember again that for a process to be quasi-static and reversible, there should be no finite temperature difference between the system and the reservoir.) But we are considering a reversible engine that operates between only two temperatures. Thus adiabatic processes must bring about the temperature change in the system from `T_1` to `T_2` and `T_2` to `T_1` in this engine.
A reversible heat engine operating between two temperatures is called a Carnot engine. We have just argued that such an engine must have the following sequence of steps constituting one cycle, called the Carnot cycle, shown in Fig. 12.11. We have taken the working substance of the Carnot engine to be an ideal gas.
(a) `color{purple} "Step 1 → 2"`
Isothermal expansion of the gas taking its state from `color{green} {(P_1, V_1, T_1)}` to `color{green} {(P_2, V_2, T_1)}`.
The heat absorbed by the gas `(Q_1)` from the reservoir at temperature `T_1` is given by Eq. (12.12). This is also the work done `(W_(1 →2))` by the gas on the environment.
`color{orange} { W_(1 → 2) = Q_1 = μ R T_1 ln (V_2/V_1)}` ............12.23
(b) `color{purple} "Step 2 → 3"`
Adiabatic expansion of the gas from `color{green} {(P_2, V_2, T_1)}` to `color{green} {(P_3, V_3, T_2)}` .
Work done by the gas, using Eq. (12.16), is
`color{orange} {W_(2 ->3) = (mu R(T_1 - T_2))/(gamma -1)}` .............12.24
(c) `color{purple} "Step 3 → 4"`
Isothermal compression of the gas from `color{green} {(P_3, V_3, T_2)}` to `color{green} {(P_4, V_4, T_2)}`.
Heat released `(Q_2)` by the gas to the reservoir at temperature `T_2` is given by Eq. (12.12). This is also the work done `(W_(3 → 4))` on the gas by the environment.
`color{orange} { W_(3->4) = Q_2 = muRT_2 In (V_3/V_4)}` ...............12.25
(d) `color{purple} "Step 4 → 1"`
Adiabatic compression of the gas from `color{green} {(P_4, V_4, T_2)}` to `color{green} {(P_1,V_1, T_1)}`.
Work done on the gas, [using Eq.(12.16)], is
`color{orange} { W_(4 ->1) = mu R ((T_1 - T_2 )/(gamma - 1 ) )}` ........................12.26
From Eqs. (12.23) to (12.26) total work done by the gas in one complete cycle is
`color{lime} { W = W_(1 → 2) + W_(2 → 3) – W_(3 → 4) – W_(4 → 1)}`
`color{blue} { = μ RT_1 ln (V_2/V_1) - μ RT_2 ln (V_3/V_4) }` ....................12.27
The efficiency η of the Carnot engine is
`eta = W/Q_1 = 1 - Q_2/Q_1`
`color{purple} {= 1 - (T_2/T_1 ) (In (V_3/V_4))/(In (V_2/V_1))}` ...................12.28
Now since step 2 → 3 is an adiabatic process,
`color{orange} {T_1 V_2^(gamma -1) = T_2 V_3^(gamma -1)}`
i.e. `V_2/V_3 = (T_2/T_1)^(1//(gamma -1))` ............12.29
Similarly, since step 4 → 1 is an adiabatic process
`color{navy} {T_2 V_4^(gamma -1) = T_1V_1^(gamma -1)}`
i.e. `color{purple} {V_1/V_4 = (T_2/T_1) ^(1//gamma -1)}` ......12.30
From Eqs. (12.29) and (12.30),
`color{green} {V_3/V_4 = V_2/V_1}` ...........12.31
Using Eq. (12.31) in Eq. (12.28), we get
`eta = 1 - T_2/T_1 ` (Carnot engine) ................12.32
We have already seen that a Carnot engine is a reversible engine. Indeed it is the only reversible engine possible that works between two reservoirs at different temperatures.
Each step of the Carnot cycle given in Fig. 12.11 can be reversed. This will amount to taking heat `Q_2` from the cold reservoir at `T_2`, doing work W on the system, and transferring heat `Q_1` to the hot reservoir. This will be a reversible refrigerator.
We next establish the important result (sometimes called Carnot’s theorem) that (a) working between two given temperatures `T_1` and `T_2` of the hot and cold reservoirs respectively, no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the Carnot engine is independent of the nature of the working substance.
To prove the result (a), imagine a reversible (Carnot) engine R and an irreversible engine I working between the same source (hot reservoir) and sink (cold reservoir).
Let us couple the engines, `I` and R, in such a way so that I acts like a heat engine and R acts as a refrigerator. Let I absorb heat `Q_1` from the source, deliver work W ′ and release the heat `Q_1- W′` to the sink. We arrange so that R returns the same heat `Q_1` to the source, taking heat `Q_2` from the sink and requiring work `W = Q_1 – Q_2` to be done on it.
Now suppose `η_R < η_I` i.e. if R were to act as an engine it would give less work output than that of I i.e. `W < W′` for a given `Q_1`. With R acting like a refrigerator, this would mean `Q_2 = Q_1 – W > Q_1 – W ′`.
Thus on the whole, the coupled I-R system extracts heat `(Q_1 – W) – (Q_1 – W′) = (W ′ – W)` from the cold reservoir and delivers the same amount of work in one cycle, without any change in the source or anywhere else. This is clearly against the Kelvin-Planck statement of the Second Law of Thermodynamics. Hence the assertion `η_I > η_R` is wrong.
No engine can have efficiency greater than that of the Carnot engine. A similar argument can be constructed to show that a reversible engine with one particular substance cannot be more efficient than the one using another substance.
The maximum efficiency of a Carnot engine given by Eq. (12.32) is independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of state, which allows us to readily calculate η, but the final result for `η`, [Eq. (12.32)], is true for any Carnot engine.
This final remark shows that in a Carnot cycle, is a universal relation independent of the nature of the system. Here `Q_1` and `Q_2` are respectively, the heat absorbed and released isothermally (from the hot and to the cold reservoirs) in a Carnot engine. Equation (12.33), can, therefore, be used as a relation to define a truly universal thermodynamic temperature scale that is independent of any particular properties of the system used in the Carnot cycle. Of course, for an ideal gas as a working substance, this universal temperature is the same as the ideal gas temperature introduced in section 12.11.
`color{green} {Q_1/Q_2 = T_1/T_2}`