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P340 Lecture 5 (Second Law)

P340 Lecture 5 (Second Law). THE FUNDAMENTAL POSTULATE (OF THERMAL PHYSICS) Any isolated system in thermodynamic equilibrium is equally likely to be in any one of its available microstates THE SECOND LAW OF THERMODYNAMICS

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P340 Lecture 5 (Second Law)

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  1. P340 Lecture 5(Second Law) • THE FUNDAMENTAL POSTULATE • (OF THERMAL PHYSICS) • Any isolated system in thermodynamic equilibrium is equally likely to be in any one of its available microstates • THE SECOND LAW OF THERMODYNAMICS • Any isolated thermodynamic system will, with overwhelming probability, evolve into the macrostate of largest multiplicity (consistent with the constraints imposed on the system) and subsequently will remain in that state.

  2. Multiplicity and volume II I Fixed wall Thermal barrier Consider two volumes isolated from the rest of the universe by barriers to particle or heat exchange, and from each other by a thin membrane. Suppose initially one of them is filled with gas (at some temperature and pressure) and the other is empty. What happens to the number of microstates available to the system (i.e. the multiplicity) if the barrier is removed (or say it breaks)?

  3. Thermal interactions II I Wall with a thermal conductivity that is changed from zero to non-zero Thermal barrier Consider two systems that are isolated from the universe, and each other, but then we allow them to exchange heat (i.e. energy exchange with no work done by either one on the other). What does the 2nd law tell us about this situation, and how might we analyse the relevant physics quantitatively?

  4. Thermal interactions II I Wall with a thermal conductivity that is changed from zero to non-zero Thermal barrier Consider two systems that are isolated from the universe, and each other, but then we allow them to exchange heat (i.e. energy exchange with no work done by either one on the other). What does the 2nd law tell us about this situation, and how might we analyse the relevant physics quantitatively? The system evolves to the macrostate (i.e. division of energy between the two systems, since that is all we are allowing to change) of maximum multiplicity and stays there. (dWtot/dEI)V,E,VI= 0; what is the condition on the individual subsystems that assures this mathematical result?

  5. Lecture 6-- CALM • What is the microscopic significance of an object’s heat capacity? • Something to do with Heat Exchange (2 answers) • Something to do with multiplicity (8 answers) • This is not an easy question, but there are two key points to remember: • A number of answers appeared to confuse the idea of temperature (which directly measures the rate of entropy change with heat exchange) and heat capacity, which describes how that rate changes with heat exchange. • Knowing Cv(T) allows one to compute the change in entropy (i.e. the multiplicative change in number of microstates) by performing an integral! • It is closely related to the second derivative of the energy with respect to entropy (while holding external parameters fixed, e.g. V)

  6. Chapter 2 problems (examples)

  7. Lecture 8– The importance of being slow Po P1 Fo=PoA F1 Let the gas push the piston to make the large volume, larger V is compensated by a decrease in E to keep the entropy constant! S1=So If this is done slowly enough to allow one to assume that the gas always remains in thermodynamic equilibrium throughout the process, it is said to be a QUASI-STATIC process. In principle this means the changes must take infinite time, in practice this means that the changes must be very small on time scales of the order of the equilibration time for the system.

  8. Lecture 8– The importance of being slow Po P1 Fo F1 V1 Let the gas push the piston to make the large volume, larger V is compensated by a decrease in E to keep the entropy constant! S1=So Po P2 Vo V2 Alternatively, just let “Scotty” beam the barrier up to the enterprise and have the gas double its volume without doing work. S goes up by N ln(2)

  9. Lecture 6-- CALM • By what factor does the multiplicity of water change when you freeze an ice cube (say 1 mole of water; the latent heat of fusion is 334 kJ/kg for water)? • Does this not violate the second law (which seems to say that entropy always increases)? • Energy goes down, but entropy does not (3 answers) • Sidestep the issue (explain why entropy goes down. (1 answer) • It is not an isolated system, (8 answers; the 2nd law applies only to isolated systems, or, failing that, the universe).

  10. Lecture 6-- CALM • Explain as succinctly as you can, and in your own words, why, of all possible heat engines acting between two reservoirs of fixed temperature , a Carnot cycle has the greatest possible thermodynamic efficiency? • Most answers argued something about heat being extracted from one reservoir and dumped into another, using what sound like energy conservation arguments. • This is because there is no change in entropy in the Carnot cycle and it is reversible. This makes it so it can use all the energy available to it (I think). (about 5 people answered something like this).

  11. P340 Lecture 6(Second Law) • THE SECOND LAW OF THERMODYNAMICS • (Clausius Statement) • “No process is possible whose sole result is the transfer of heat from a reservoir at lower temperature to another at a higher temperature” • THE SECOND LAW OF THERMODYNAMICS • (Kelvin Statement) • “No process is possible whose SOLE result is the absorption of energy (as heat) from a thermal reservoir and the conversion of this energy into useful work.” (i.e. no perpetual motion machines of the second kind)

  12. Lecture 8– Typical Heat Engine ALL heat engines extract heat from a high-temperature source (reservoir), and perform useful work. The ratio of these two quantities is called the efficiency (h). The FIRST Law of thermodynamics tells us that the efficiency can never be greater than 1 (energy cannot be created), and the second law puts limits on how close to 1 the efficiency can get. h cannot be equal to 1, since doing so would decrease the entropy of the heat source without increasing the entropy of anything else.

  13. Lecture 8– Carnot Cycle DSH = -|QH|/TH DSC = |QC|/TC DS = DSH +DSC = 0 => |QH|/TH = |QC|/TC http://thermodynamicstudy.net/carnot.html The most efficient heat engine would be one that does not change the entropy of the universe (i.e. the decrease of the source’s entropy is exactly compensated by the increase in the sink’s entropy. Any overall increase in entropy would require additional heat extracted from the source but NOT put into work and therefore would have lower efficiency.

  14. An important engine to know about is the Stirling engine, which achieves the Carnot efficiency (at least in an idealized form, this is, not surprisingly, never achieved in practice). The diagrams below are from Zemansky’s book “Heat and Thermodynamics”. The wikipedia page on this has some nice applets that show how practical Stirling engines can be built. http://en.wikipedia.org/wiki/Stirling_engine In this style of engine, R is called the Regenerator, it accepts heat from the fluid and then gives it back. There are other variants of the same basic idea (a cycle with two “isochoric” processes linked by two “isothermal” processes).

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