1. MEL140 (Second Law: Continued)

2. Recap on Irreversible process
A system undergoes an irreversible process, when
second law demands that reversal of the process leaves a finite trace on the surroundings. Connection with the Kelvin-Planck statement (board).
OR equivalently.
Traceless reversal is prohibited by second law (Connection with Clausius statement)
Ways to look at “irreversibilities”:
Lack of thermodynamic equilibrium between system and surroundings renders (existence of “driving forces”) a process irreversible:
Lack of thermal equilibrium: finite temperature differences.
Lack of mechanical equilibrium: finite pressure differences (e.g. free expansion).
Chemical equilibrium: e.g. reactions/phase-transformations that complete, diffusion of dye/ink in water.
General sign of lack of equilibrium: if the system is isolated and observed instantaneously, processes (internal adjustments) will be found to occur.

3. Recap on Irreversible process
Ways to look at “irreversibilities”:
General sign of lack of equilibrium: if the system is isolated and observed instantaneously, processes (internal adjustments) will be found to occur.
During an irreversible process, “internal currents/fluxes” that lead to dissipation are present due to driving forces either between the system and surroundings or between parts of a system. Dissipation can be identified when during a process without thermal interaction with the surroundings, the sum of the macroscopic potential and macroscopic kinetic energy of the system decreases. (energy goes to “microscopic modes”, remember discussion on work energy theorem)
e.g. mechanical friction, shocks/”explosions”, plastic deformations, “Joule heating effects” (resistors differs from capacitor/inductor), eddy currents.
Moving from one equilibrium state to another is possible in finite time only by irreversible (fast) processes.

4. To show that heat transfer through a finite temperature difference is an irreversible process
Q1-Q
W=Q1-Q tH H Q Q1
W=Q1-Q Q tC
Violation of Kelvin Planck
statement

5. Example of irreversibility due to lack of equilibrium: unrestrained expansion of a gas
800 kPa
0 kPa B A
A membrane separates a gas in chamber A from vacuum
in chamber B. The membrane is ruptured and the gas expands
Into chamber B until pressure equilibrium is established. The
process is so fast and the container is insulated enough such
that negligible heat transfer takes place between
the gas and the surroundings during this process.
At the end of the unrestrained expansion process, the gas (system) has the same internal energy, as it had initially.

6. Some questions not yet answered
What kind of engines and refrigerators have the best possible performance?
What factor(s) affect the performance of a heat engine and a refrigerator?
What is the best possible performance of a heat engine and a refrigerator?

7. The Carnot principles
First Carnot principle: The efficiency of an irreversible heat engine is always less than the efficiency of a reversible heat engine operating between the same two reservoirs.
(Irr<rev)
Second Carnot principle: All reversible engines operating between the same two reservoirs have the same efficiency.
 Rev1 =Rev2

8. Proof of First Carnot principle
Proof by contradiction: Assume Irr>Rev tH
Irr
QIrr Q
WIrr=Q-QIrr
Rev
QRev>QIrr Q
WRev<Wirr
Rev
QRev>QIrr Q
WRev<Wirr tL
Rev+Irr
QRev-QIrr
Wirrev-Wrev tL TC
Conclusion: Assumption Irr>Rev is incorrect. Efficiency of a reversible engine is
higher than that of an irreversible engine.

9. Proof of Second Carnot principle
Proof by contradiction: Assume Rev1>Rev2. tH
QRev1
Rev1 Q
WRev1=Q-QRev1
Rev2
QRev2>QRev1 Q
WRev2<WRev1
Rev2
QRev2>QRev1 Q
WRev2<WRev1 tL
Rev1+Rev2
QRev2-QRev1
WRev1-WRev2 tL TC
Conclusion so far: Assumption Rev1>Rev2 is incorrect.

10. Proof of Second Carnot principle (continued)
Proof by contradiction (continued): Assume Rev1<Rev2. tH
Rev1
QRev1>QRev1 Q
WRev1<WRev2
QRev1
Rev1 Q
WRev1<WRev2
Rev2
QRev2 Q
WRev2=Q-QRev2 tL
Rev1+Rev2
QRev1-QRev2
WRev2-WRev1 TC tL
Final conclusion: Rev1=Rev2. All heat engines working between the
same reservoirs have the same efficiency.

11. An important implication of the second Carnot principle
The efficiency of a reversible heat engine does not depend on its working fluid, method of execution of cycle, type of reversible engine used, amount of heat drawn from or rejected by the engine etc. It may however depend on a characteristic of the reservoirs.
By what characteristic is a reservoir specified?
Ans.: Temperature.
The only factors that could affect the efficiency of a reversible engine is, therefore, the temperatures (tH,tL) of the reservoirs it is connected to.

12. Uses of the second Carnot principle
To develop a thermodynamic temperature scale, which is a temperature scale that does not depend on the properties of a particular substance.
To calculate the maximum efficiency of a heat engine (or maximum COP of a refrigerator/heat pump).

13. Empirical and thermodynamic temperature scales
Empirical temperature scale
A scale that is based on the measurement of a temperature-sensitive property of a certain substance (e.g. pressure exerted by a constant volume of helium gas, thermal expansion of a enclosed mass of mercury/alcohol etc.).
A thermometer reads the “empirical temperature”.
Notation for empirical temperature: (t)
Thermodynamic temparature scale
A scale that is independent of the properties of any substance.
Lord Kelvin was aware of the second Carnot principle and suggested in1848, that a thermodynamic temperature scale could be based on the theoretical consideration that, during the operation of a reversible heat engine, the amounts of heat exchanged between system and the reservoirs depend only on the temperature of the reservoirs and not on the properties of any substance.
Notation for the
thermodynamic
temperature scale
to be developed : (T) 13 13

14. Empirical temperature scales (t)
Empirical scales are determined through experimentation with “thermometric substances”.
Single fixed point scale (>1954)
Constant volume gas thermometer and the ideal gas temperature scale.

15. Perform this experiment with various gases (A,B,C)
The ideal gas temperature scale: a very accurate empirical temperature scale developed from experiments using the constant volume gas thermometer
Step 1: Bring the thermometer in contact with water at triple point (tp). Measure the pressure ptp.
Step 2: Bring the thermometer in contact with the body at temperature T. Measure the pressure p. Calculate
Now, redo steps 1 and 2, at each instance reducing the number of moles (mass) of gas used in step 1 (and 2), such that pt=2n (n=10,9,8,.,3 etc.) mm Hg.
Perform this experiment with various gases (A,B,C)
C Capillary tube
L Lip
M Mercury manometer
Gas A
Gas B
Gas C
At low pressures, ideal gas
behavior is approached by all gases.
tmeasured
ptp

16. A temperature scale based on the second Carnot principle
Since
But

17. Properties of according to second lawQ3

18. Constructing a single-fixed-point thermodynamic temperature scale
for t<ttp
Place one of the reservoirs in thermal equilibrium with water at triple point (tp); prescribe the value to the constant
for t>ttp
Procedure for calculating thermodynamic temperature
The reversible engine is operated with a fixed Qtp ; then Q(t) is measured to obtain T(t) using:
(in either case)
T is known as the Kelvin/absolute temperature scale

19. The Kelvin scale is not the only thermodynamic temperature scale.
The Kelvin scale calculates the thermodynamic temperature using :
Alternatively any monotonic function T’(t) of T(t)
can also be chosen to define a new thermodynamic temperature scale.