Ch. 5

1
Ch. 5 2nd Law of Thermodynamics

• Entropy
• The 1st Law is a statement of energy conservation
  for processes we could consider.
• However, it does not say whether or not any
  process could actually occur.
• Thermodynamic processes fall into 1 of 3
  categories
• Natural
• Impossible (unnatural or anti-natural)
• Reversible

2
Ch. 5 2nd Law of Thermodynamics

• Entropy
• Natural thermodynamic processes are, to one
  degree or another, irreversible.
• Such processes move toward equilibrium.
• Gas expanding into a vacuum
• Heat conduction through a finite temperature
  gradient
• Diffusion of a gas into another through a finite
  concentration gradient
• Freezing of supercooled water

3
Ch. 5 2nd Law of Thermodynamics

• Entropy
• Reverse processes are not excluded by 1st Law.
• Compression of gas under no external pressure
• Heat flowing from a colder body to a warmer one
• We know that these processes are impossible in
  nature.
• Since they are not excluded by 1st Law, we need
  something else.
• The 2nd Law of Thermodynamics takes care of this.

4
Ch. 5 2nd Law of Thermodynamics

• Entropy
• Depending on external conditions, natural
  processes can produce opposite changes in a
  system.
• Irreversibility of such processes can be reduced
  by modifying their paths.
• Attempt to change path so difference between
  actual values of state variables and values
  corresponding to equilibrium is reduced at all
  stages

5
Ch. 5 2nd Law of Thermodynamics

• Entropy
• Limit is a reversible process.
• Ideal limit cannot be realized
• Can approximate, however
• Can be defined as
• A series of states that differ infinitesimally
  from equilibrium
• States follow each other infinitely slowly
• Variables change in continuous way
• Underlying assumption rate of process tends to
  zero as equilibrium approached

6
Ch. 5 2nd Law of Thermodynamics

• Entropy
• We can reframe our earlier examples to find a
  reversible analogue.
• Expansion of gas with pressure p against external
  pressure p ?p
• Heat conduction along an infinitesimal
  temperature gradient
• By reversing the sign of infinitesimal difference
  from equilibrium, reversible process of opposite
  sense occurs (replace dp by dp, or dT/dx by
  dT/dx.
• Key idea, slow changes departing very little from
  equilibrium.

7
Ch. 5 2nd Law of Thermodynamics

• Entropy
• For every reversible process there is a positive
  integrating factor 1/? of the differential
  expression ?Q, which is only dependent on the
  thermal state and which is equal for all
  thermodynamic systems.
• This defines a state function S, the entropy,
  according to the exact differential,

8
Ch. 5 2nd Law of Thermodynamics

• Entropy
• The following is found to be true for
  irreversible processes
• We can combine the two expressions to generalize
  for any process
• The inequality and equality symbols correspond to
  irreversible and reversible processes,
  respectively.
• MATHEMATICAL EXPRESSION OF 2nd LAW.

9
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle (thermal engine)
• All processes are reversible.
• Contains 2 isotherms, T1 and T2 lt T1, 2 adiabats
  ?2 lt ?1.
• Steps are
• Isothermal expansion T1 const
• Adiabatic expansion ?1 const
• Isothermal compression T2 const
• Adiabatic compression ?2 const

10
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle (ideal gas)
• Calculate work done and heat absorbed.
• Step 1 (isothermal)
• Step 2 (adiabatic)

11
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle (ideal gas)
• Step 3 (isothermal)
• Step 4 (adiabatic)
• Total work done is (shaded)

12
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle
• Total internal energy change is zero (as
  expected).
• The gas absorbs heat Q1 gt 0 from the hotter
  reservoir.
• Transfers (rejects) heat Q2 lt 0 to the colder
  reservoir.
• For the adiabatic paths we have

13
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle
• That leaves us with
• Denote the heat absorbed by the gas at T1 as Q1
  and heat given up by gas at T2 as Q2, and since
  Q1 W12 and Q2 W34, we have
• Since W gt 0, Q gt 0.

14
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle
• We know that W Q1 Q2 and Q2 lt 0. From this
• Part of heat absorbed from reservoir at T1
  becomes work.
• The rest is given up to colder source at T2.
• Efficiency, ? defined as ratio of work done to
  heat absorbed (at T1)
• We know that Q1 gt 0 and Q2 lt 0 and
  also

15
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• Thermodynamic efficiency is function only of the
  temperature of the reservoirs.
• Cant do work with only one heat reservoir, i.e.,
  during the performance of work, some heat must be
  rejected to the a lower temperature reservoir
• The greater the difference in temperature between
  the two sources, the more efficient the thermal
  engine
• Example given of stronger winter circulations

16
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• A transformation whose only final result is to
  transfer heat from a body at a given temperature
  to an body at a higher temperature is
  impossible.
• Transfer of heat from cold body to warm body is
  impossible without external work being done.
• We know that refrigerators and air
  conditioners/heat pumps exist, so such
  transformations can take place
• Key point here is that external work must be done
  to drive the transformation (use electricity to
  run a compressor).

17
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• Carnot Cycle allows definition of absolute
  temperature.
• From the expressions for efficiency we get
• recalling that Q1 gt 0 and Q2 lt 0.
• Choose a reference temperature (T1 273.15 K)
• Define all other colder temperatures by a Carnot
  Cycle operating from the reference temperature
  source (melting ice)
• Measure heat transfers at each temperature to
  solve for T2
• 0 K is where Q2 0, cold source absorbs no heat
  (perfect engine)

18
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• Efficiency values vary between 0 and a maximum.
• Perfectly irreversible engine has ? 0
• Thermal losses due to friction and heat leakage
  maximize
• No work is done (all heat lost)
• Perfectly reversible engine has ? gt 0
• Friction and heat leakage are zero
• Maximum work is done
• Carnot Cycle (engine) is the most efficient
  engine (perfectly reversible), has maximum
  efficiency.

19
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• Carnot Theorem It is impossible to construct
  an engine operating between two heat sources that
  would have an efficiency greater than the
  efficiency of a Carnot cycle operating between
  the same heat sources.
• Carnot engine is an ideal engine cannot be
  realized.
• Note also that ? lt 0, i.e., T2 0 K cannot be
  reached.

20
Ch. 5 2nd Law of Thermodynamics

• Carnot Cycle Key Points
• So for cyclic processes we can write.
• which leads to the combined form for any process
• We also have, for the finite form, for
  completeness