Dr Roger Bennett Rm. 23 Xtn Lecture 8

Molar Heat Capacity- C v For ideal monatomic gas –U = 3/2 nkT –Per atom this is just 1/2kT per degree of freedom i.e. 3/2kT = K.E. x + K.E. y + K.E. z

Molar Heat Capacity- C v For ideal monatomic gas –U = 3/2 nkT –Per atom this is just 1/2kT per degree of freedom i.e. 3/2kT = K.E. x + K.E. y + K.E. z –Per mole C v = (đQ/  T) v = (dU/  T) v = 3/2R –Consider 1 st Law adiabat đQ = dU + PdV = 0 –dU + PdV = 0 –C v dT + PdV = 0

Molar Heat Capacity- C v For ideal gas

Molar Heat Capacity- C v For ideal diatomic gas –still non interacting but now dumbell shape –U translational per atom = 3/2kT as before –Molecule now can rotate –1/2kT per rotational degree of freedom –Molecule can vibrate –has potential energy –has vibrational kinetic energy U total = 3/2kT+ kT + kT = 7/2kT

Equipartition Theorem Classical mechanics predicts that: When a substance is in equilibrium, there is an average energy of ½kT per molecule (or ½RT per mole) associated with each degree of freedom.

Molar Heat Capacity- C v For ideal diatomic gas – classical physics predicts U total = 7/2RT per mole What do we observe? CvCv T 5R/2 3R/2 7R/2 No classical explanation Only explained by Q.M. Only discrete values of rotation and vibration allowed. Below some temperatures not Enough energy so some degrees of freedom “Frozen Out”

Properties of materials It is well known that most materials expand on heating. We define the volume thermal expansivity as:- Moduli of elasticity are the stress/strain or force/unit area divided by the fractional deformation:-

Cyclic Processes We have defined processes that can take us full circle in an indicator diagram. Consider two paths a and b between points 1 and 2. The A’s are the areas under paths a and b. P V a b 1 2

Cyclic Processes We end in the same place as we start so the state is the same  U=0 around loop.  U 12 = Q 1a2 + W 1a2  U 21 = Q 2b1 + W 2b1  U 12 +  U 21 = 0 (Q 1a2 + Q 2b1 ) = - (W 1a2 +W 2b1 ) Nett heat input = - (nett work done on the system) Nett heat input = A a - A b = work done by system We have transformed heat into work – an engine V P a b 1 2

Dr Roger Bennett Rm. 23 Xtn Lecture 9

Cyclic Processes We end in the same place as we start so the state is the same  U=0 around loop.  U 12 = Q 1a2 + W 1a2  U 21 = Q 2b1 + W 2b1  U 12 +  U 21 = 0 (Q 1a2 + Q 2b1 ) = - (W 1a2 +W 2b1 ) Nett heat input = - (nett work done on the system) Nett heat input = A a - A b = work done by system We have transformed heat into work – an engine V P a b 1 2

Carnot Cycle – Ideal Gas Need a cycle that we can solve Define path as two adiabats and two isotherms Isotherms a and c Adiabats b and d Where does heat flow? Work done by system = area enclosed V P a b c d

Carnot Cycle – Ideal Gas dU=đQ + đW On isotherm a dU = 0 đQ = -đW = PdV On isotherm c V P a b c d

Carnot Cycle – Ideal Gas dU=đQ + đW We know  U for cycle = 0 On adiabat b and d đQ = 0 V P a b c d

Carnot Cycle – Ideal Gas On adiabat b and d We know V P a b c d

Carnot Cycle – Ideal Gas We have calculated the work done by the system: Because we can find This is the definition of the Kelvin temperature scale. It is important as it is only defined by the temperatures of the reservoirs.

Carnot Cycle – Ideal Gas We can calculate the efficiency of the cycle. We turn heat from the hot reservoir into useful work and discard the remainder into the cold reservoir. Efficiency = Work out / Heat energy put in

Carnot Cycle – Ideal Gas The Carnot Cycle works between two reservoirs at differing temperatures. Its efficiency is uniquely defined by the two temperatures. It is reversible – we can go round the system backwards. It then uses work to extract heat from a cold reservoir to a hot reservoir. This may be considered a refrigerator!

Carnot Cycle – Ideal Gas In a refrigerator the coefficient of performance is given by how much heat can be removed from the cold reservoir per unit of work put in. Note this can be > 1

Engines and Refrigerators Cold T C Hot T H R Work QHQH QCQC Cold T C Hot T H E Work QHQH QCQC

Dr Roger Bennett Rm. 23 Xtn Lecture 10

Carnot Cycle We considered a cycle with 2 adiabats and 2 isotherms Found heat flows in and out on adiabats. Work output is = nett heat input. Efficiency is fundamental and fixed by temperature of two reservoirs. V P a b c d

Why is efficiency fixed? 0 th law gave us –“If two bodies A and B are in thermal equilibrium with a third body C then A and B are in thermal equilibrium with each other.” 1 st law gave us –Conservation of energy dU = đW + đQ None of these deals with the inherent direction of processes. Empirical evidence suggests that direction is important. Hot bodies cool over time, cold bodies heat up.

The Second Law of Thermodynamics Equivalent statements on handouts. The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”

The Second Law of Thermodynamics The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”

The Second Law of Thermodynamics The Kelvin-Planck Statement Cycle requires that the state of the working substances is the same at the start and the end of the process. An isothermal compression  U = 0 but V x dx F A Vacuum

The Second Law of Thermodynamics The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”

The Second Law of Thermodynamics The Kelvin-Planck Statement No effect other tells us that in addition to the rejection of heat to a body of lower temp. the only other effect on the surroundings is via the work delivered by the engine. This means that the bodies delivering and accepting the heat must do so without delivering any work. In other words they are sources of heat – reservoirs.

The Second Law of Thermodynamics The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”

The Second Law of Thermodynamics The Kelvin-Planck Statement Single Suppose Q 1 + Q 2 heat was supplied to an engine from bodies at T 1 and T 2 (T 1 > T 2 for instance). The cyclical engine could deliver work W = Q 1 + Q 2 which appears to violate the statement. However, Q 2 could be negative.

The Second Law of Thermodynamics The Kelvin-Planck Statement Single Suppose Q 1 + Q 2 heat was supplied to an engine from bodies at T 1 and T 2 (T 1 > T 2 for instance). The cyclical engine could deliver work W = Q 1 + Q 2 which appears to violate the statement. Hot T 1 E Work Q1Q1 Hot T 2 Q2Q2

The Second Law of Thermodynamics The Kelvin-Planck Statement Single Suppose Q 1 + Q 2 heat was supplied to an engine from bodies at T 1 and T 2 (T 1 > T 2 for instance). The cyclical engine could deliver work W = Q 1 + Q 2 which appears to violate the statement. However, Q 2 could be negative and there is no violation. This type of engine is excluded by specifying single.

The Second Law of Thermodynamics The Kelvin-Planck Statement “It is impossible for an engine, working in a cycle, to exchange heat with a single reservoir, produce an equal amount of work, and have no other effect.” Cold T C Hot T H E Work QHQH

The Second Law of Thermodynamics The Kelvin-Planck Statement “It is impossible for an engine, working in a cycle, to exchange heat with a single reservoir, produce an equal amount of work, and have no other effect.” We know that we can transfer energy by mechanical work (think stirring a liquid) to raise the internal energy or heat a system. This statement asserts that the reverse is not true. Turning work into heat may therefore be irreversible. We cannot transform heat directly into work. There is something different about these forms of energy!

The Second Law of Thermodynamics The Kelvin-Planck Statement “A process whose only effect is the complete conversion of heat into work is impossible.” Only here covers all the previously discussed eventualities.

The Second Law of Thermodynamics The Clausius Statement “It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body.” No effect is key here as the system must remain unchanged Cold T C Hot T H R QHQH QCQC

Equivalence of Statements R Work Q 2 + Q 1 Q2Q2 Cold T C Hot T H E Q1Q1 Assuming Kelvin statement is false

Equivalence of Statements R Work Q 2 + Q 1 Q2Q2 Cold T C Hot T H E Q1Q1 Composite Refrigerator Q2Q2 Q2Q2 Cold T C Hot T H Assuming Kelvin statement is false implies Clausius also false

Equivalence of Statements R Q1Q1 Q2Q2 Cold T C Hot T H E Q2Q2 Assuming Clausius statement is false Q2Q2

Equivalence of Statements R Work = Q 1 –Q 2 Q1Q1 Q2Q2 Cold T C Hot T H E Q2Q2 Composite Engine Q 1 -Q 2 Cold T C Hot T H Q2Q2 Assuming Clausius statement is false implies Kelvin also false

Dr Roger Bennett Rm. 23 Xtn Lecture 11

Maximum Efficiency – Carnot’s Theorem E Q3Q3 Q2Q2 Cold T C Hot T H C Q1Q1 Q4Q4 WEWE WcWc Assume an engine more efficient (  E ) than Carnot engine (  c). 1 st Law still applies. Q 2 =Q 1 -W E Q 4 =Q 3 -W c With matched W and a more efficient engine: W E /Q 1 > W c /Q 3 So Q 3 > Q 1 But Carnot engine is reversible….

Maximum Efficiency – Carnot’s Theorem E Cold T C Hot T H C Q1Q1 Q 4 =Q 3 -W W E = W c With the Carnot refrigerator the system acts as a composite Heat flow out of the cold reservoir = Q 4 -Q 2 Q 4 -Q 2 = Q 3 -Q 1 But Q 1 < Q 3 ! So violates Clausius statements as it pumps heat from cold to hot with no external work being done.    c Q3Q3 Q 2 =Q 1 -W

Carnot Cycle We considered a cycle with 2 adiabats and 2 isotherms Found heat flows in and out on isotherms. V P a b c d

General Cycle V P a b c d We can build any closed reversible cycle out of a whole sequence of tiny reversible Carnot cycles. As the isotherms become very close the jagged contour approximates the actual path.

General Cycle Consider each tiny cycle individually - on the isotherms at T 1 and T 2 heat đQ 1 and đQ 2 flows. The total effect of each tiny cycle is therefore the sum of each cycle. In the limit that each individual heat flow becomes tiny the sum becomes an integral. The R remind us that this is for reversible paths.

General Cycle - Entropy The closed path integral around a reversible cycle is zero. It has all the hallmarks of a state function lets check…. P a b 1 2

Entropy Entropy is a state function because it does not matter which path the system takes between 1 and 2. Only differences in Entropy are defined – the reference point is usually taken at the lowest temperature that can be achieved. Ideally absolute zero. Therefore at finite temp T f the entropy is the integral of đQ R /T from the reference temp up to T f. All heat must be added reversibly. For tiny reversible changes.

Entropy and the 1 st Law The differential form of the first Law for reversible processes can be written as: Heat capacities can now be redefined: C v = (đQ/  T) v = (T  S/  T) v = T (  S/  T) v C p = (đQ/  T) p = T (  S/  T) P

Example Imagine heating a beaker of water slowly by putting it in contact with a reservoir whose temperature increases very slowly so that the system passes through a succession of equilibrium states at constant pressure. The heat flow in going from T to T+dT is đQ R = C p dT Hence entropy change: dS = đQ R /T = C p dT/T For 1kg of water with C p =4.2kJK -1 going from 20°C to 100°C  S = 1.01  JK -1

Dr Roger Bennett Rm. 23 Xtn Lecture 12

Real Engines and Irreversibility E Q3Q3 Q2Q2 Cold T C Hot T H C Q1Q1 Q4Q4 WEWE WcWc Assume an engine less efficient (  E ) than Carnot engine (  c). 1 st Law still applies. Q 2 =Q 1 -W E Q 4 =Q 3 -W c With matched W and a less efficient engine: W E /Q 1 < W c /Q 3 So Q 3 < Q 1 But Carnot engine is reversible….

Real Engines and Irreversibility E Cold T C Hot T H C Q1Q1 Q 4 =Q 3 -W W E = W c With the Carnot refrigerator the system acts as a composite Heat flow out of the cold reservoir = Q 4 -Q 2 Q 4 -Q 2 = Q 3 -Q 1 But Q 1 > Q 3 So doesn’t violate Clausius statements as it heat flows from hot to cold with no external work being done.  E   c (= for reversible) Q3Q3 Q 2 =Q 1 -W

Irreversible Cycle For Carnot Cycle For Irreversible engine  E <  c

In General Around entire closed path this must be true so: Clausius inequality < for irreversible = for reversible

General Cycle Remember this inequality is for irreversible processes involved in the cycle. Our expression previously defined for entropy is for reversible cycles. A reversible process is one that can be reversed by an infitessimal adjustment of the system coordinates (P,V,T etc) and takes place so slowly that the system can be considered to be passing through a succession of equilibrium states. I.E. Quasistatic and reversible by infitessimal adjustment.

General Cycle - Entropy Take a cycle in which path a is irreversible between points 1 and 2 which are both equilibrium states. Path b from 2 to 1 is reversible. P a b 1 2

The Entropy Statement The equality applies only for reversible processes. Entropy is created by irreversible processes! For a thermally isolated system đQ = 0 so any irreversible process occurring makes dS > 0. Entropy only ever increases!

The Entropy All changes in a thermally isolated system must lead to an increase in entropy (or stays the same if reversible). As such a system approaches equilibrium the entropy must increase therefore the final equilibrium configuration is the one with the maximum entropy. At the maximum of entropy there are no changes as entropy cannot decrease. The always increasing nature defines a natural direction to time. This all follows from the simple observation that heat flows from hot to cold!

Example- expansion of an ideal gas Imagine an ideal gas in volume V. A second empty vessel of the same volume is placed adjacent. A hole is formed between them and the gas expands to fill both. Gas in V Gas in 2V

Example- expansion of an ideal gas Process is clearly irreversible, the internal energy doesn’t change and the process is not quasistatic.