# PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 19. First Law of Thermodynamics Work done by/on a gas Last Lecture.

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PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 19

First Law of Thermodynamics Work done by/on a gas Last Lecture

Some Vocabulary Isobaric P = constant Isovolumetric V = constant W = 0 Isothermal T = constant  U = 0 (ideal gas) Adiabatic Q = 0 V V V V P P P P

P-V Diagrams P V Path moves to right: W by the gas = Area under curve P V Path moves to left: W by the gas = - Area under curve (W on the gas = - W by the gas )

Work from closed cycles W A->B->A = Area (work done by gas) W A->B->A = -Area Clockwise cycle: Counterclockwise cycle: in closed cycles

Example 12.8a Consider an ideal gas undergoing the trajectory through the PV diagram. In going from A to B to C, the work done BY the gas is _______ 0. P V A B a)> b)< c)= C

Example 12.8b In going from A to B to C, the change of the internal energy of the gas is _______ 0. P V A B a)> b)< c)= C

Example 12.8c In going from A to B to C, the amount of heat added to the gas is _______ 0. P V A B a)> b)< c)= C D

Example 12.8d In going from A to B to C to D to A, the work done BY the gas is _______ 0. P V A B a)> b)< c)= C D

Example 12.8e In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0. P V A B a)> b)< c)= C D

Example 12.8f In going from A to B to C to D to A, the heat added to the gas is _______ 0. P V A B a)> b)< c)= C D

Example 12.7 Consider a monotonic ideal gas. a) What work was done by the gas from A to B? b) What heat was added to the gas between A and B? c) What work was done by the gas from B to C? d) What heat was added to the gas between B and C? e) What work was done by the gas from C to A? f) What heat was added to the gas from C to A? V (m 3 ) P (kPa) 25 50 75 0.2 0.4 0.6 A B C 20,000 J -10,000 J -25,000 J 0 15,000 J

Example 12.7 (Continued) g) What was total work done by gas in cycle? h) What was total heat added to gas in cycle? V (m 3 ) P (kPa) A B C W AB + W BC + W CA = 10,000 J Q AB + Q BC + Q CA = 10,000 J This does NOT mean that the engine is 100% efficient! |Q in | = Q AB + Q CA = 35,000 J |Q out | = |Q BC | = 25,000 J W eng = |Q in |-|Q out | |Q in | |Q out | Exhaust!!!

Heat Engines Described by a cycle with: Q hot = heat that flows into engine from source at T hot Q cold = heat exhausted from engine at lower temperature, T cold W= work done by engine Efficiency is defined: Q hot engine Q cold W using

2 nd Law of Thermodynamics (version 1) The most efficient engine is the Carnot Engine (an idealized engine), for which: No heat engine can be 100% efficient In practice, we always have (T in Kelvin)

Carnot Cycle

Example 12.9 An ideal engine (Carnot) is rated at 50% efficiency when it is able to exhaust heat at a temperature of 20 ºC. If the exhaust temperature is lowered to -30 ºC, what is the new efficiency. e = 0.585

Refrigerators Q hot fridge Q cold W Just a heat engine run in reverse! Pull Q cold from fridge Exhaust Q hot to outside Most efficient is Carnot refrigerator: Note: Highest COP for small T differences Coefficient of Performance:

Heat Pumps Q hot heat pump Q cold W Like Refrigerator: Best performance for small  T Same as refrigerator, except Pull Q cold from environment Exhaust Q hot to inside of house Again, most efficient is Carnot: Coefficient of Performance:

Example 12.10 A modern gas furnace can work at practically 100% efficiency, i.e., 100% of the heat from burning the gas is converted into heat for the home. Assume that a heat pump works at 50% of the efficiency of an ideal heat pump. If electricity costs 3 times as much per kw-hr as gas, for what range of outside temperatures is it advantageous to use a heat pump? Assume T inside = 295 ºK.

Entropy Measure of Disorder of the system (randomness, ignorance) S = k B log(N) N = # of possible arrangements for fixed E and Q Relative probabilities for 12 molecules to arrange on two halves of container.

On a macroscopic level, one finds that adding heat raises entropy: Defines temperature in Kelvin! 2 nd Law of Thermodynamics (version 2) The Total Entropy of the Universe can never decrease.

Why does Q flow from hot to cold? Consider two systems, one with T A and one with T B Allow Q > 0 to flow from T A to T B Entropy changes by:  S = Q/T B - Q/T A This can only occur if  S > 0, requiring T A > T B. System will achieve more randomness by exchanging heat until T B = T A

Carnot Engine Carnot cycle is most efficient possible, because the total entropy change is zero. It is a “reversible process”. For real engines:

Example 12.11a An engine does an amount of work W, and exhausts heat at a temperature of 50 degrees C. The chemical energy contained in the fuel must be greater than, and not equal to, W. a) True b) False

Example 12.11b A locomotive is powered by a large engine that exhausts heat into a large heat exchanger that stays close to the temperature of the atmosphere. The engine should be more efficient on a very cold day than on a warm day. a) True b) False

Example 12.11c An air conditioner uses an amount of electrical energy U to cool a home. The amount of heat removed from the home must be less than or equal to U. a) True b) False

Example 12.11d A heat pump uses an amount of electrical energy U to heat a home. The amount of heat added to a home must be less than or equal to U. a) True b) False

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