The Laws of Thermodynamics Enrolment No.: 120350119534 Noble Group of Institutions - Junagadh.

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Presentation transcript:

The Laws of Thermodynamics Enrolment No.: Noble Group of Institutions - Junagadh

Principles of Thermodynamics  Energy is conserved oFIRST LAW OF THERMODYNAMICS oExamples: Engines (Internal -> Mechanical) Friction (Mechanical -> Internal)  All processes must increase entropy oSECOND LAW OF THERMODYNAMICS oEntropy is measure of disorder oEngines can not be 100% efficient

Converting Internal Energy to Mechanical Work done by expansion

Example A cylinder of radius 5 cm is kept at pressure with a piston of mass 75 kg. a) What is the pressure inside the cylinder? b) If the gas expands such that the cylinder rises 12.0 cm, what work was done by the gas? c) What amount of the work went into changing the gravitational PE of the piston? d) Where did the rest of the work go?

Solution Given: M =75, A =  ,  x=0.12, P atm = 1.013x10 5 Pa a) Find P gas = 1.950x10 5 Pa b) Find W gas = J c) Find W gravity = 88.3 J d) Where did the other work go? Compressing the outside air

Quiz Review A massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air. Pick the most correct statement? A. The pressure is higher inside the cylinder than outside. B. The temperature inside the cylinder is higher than outside the cylinder. C. If the gas is heated by applying a flame to cylinder, and the piston comes to a new equilibrium, the inside pressure will have increased. D. All of the above. E. A and C.

Some Vocabulary  Isobaric oP = constant  Isovolumetric oV = constant  Isothermal oT = constant  Adiabatic oQ = 0 P P P P V V V V

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates at room temperature (a) Weight is slowly added to the piston, isothermally compressing the gas to half its original volume (b) A. P b ( > < = ) P a B. T b ( > < = ) T a C. W ab ( > < = ) 0 D. U b ( > < = ) U a E. Q ab ( > < = ) 0 Outside Air: Room T, Atm. P Vocabulary: W ab is work done by gas between a and b Q ab is heat added to gas between a and b Example

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates at room temperature (a) Weight is slowly added to the piston, adiabatically compressing the gas to half its original volume (b) A. P b ( > < = ) P a B. W ab ( > < = ) 0 C. Q ab ( > < = ) 0 D. U b ( > < = ) U a E. T b ( > < = ) T a Outside Air: Room T, Atm. P Vocabulary: W ab is work done by gas between a and b Q ab is heat added to gas between a and b Example

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates at room temperature (a) The gas is cooled, isobarically compressing the gas to half its original volume (b) A. P b ( > < = ) P a B. W ab ( > < = ) 0 C. T b ( > < = ) T a D. U b ( > < = ) U a E. Q ab ( > < = ) 0 Outside Air: Room T, Atm. P Vocabulary: W ab is work done by gas between a and b Q ab is heat added to gas between a and b Example

Work from closed cycles Consider cycle A -> B -> A W A->B -W B->A

Work from closed cycles Consider cycle A -> B -> A W A->B->A = Area

Work from closed cycles Reverse the cycle, make it counter clockwise -W B->A W A->B

Example a) What amount of work is performed by the gas in the cycle IAFI? b) How much heat was inserted into the gas in the cycle IAFI? c) What amount of work is performed by the gas in the cycle IBFI? W IAFI = 3P atm = 3.04x10 5 J  U = 0 W = -3.04x10 5 J Q = 3.04x10 5 J V (m 3 )

One More Example Consider a monotonic ideal gas which expands according to the PV diagram. 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 beween 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) A B C

Solution a) Find W AB b) Find Q AB V (m 3 ) P (kPa) A B C W AB = Area= 20,000 J First find U A and U B U A = 22,500 J, U B = 22,500 J,  U = 0 Finally, solve for Q Q = 20,000 J

Solution c) Find W BC d) Find Q BC V (m 3 ) P (kPa) A B C W AB = -Area= -10,000 J First find U B and U C U B = 22,500 J, U C = 7,500 J,  U = -15,000 Finally, solve for Q Q = -25,000 J

Solution e) Find W CA f) Find Q CA V (m 3 ) P (kPa) A B C W AB = Area= 0 J First find U C and U A U C = 7,500 J, U A = 22,500 J,  U = 15,000 Finally, solve for Q Q = 15,000 J

Continued Example Take solutions from last problem and find: a) Net work done by gas in the cycle b) Amount of heat added to gas 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!

Entropy  Total Entropy always rises! (2nd Law of Thermodynamics)  Adding heat raises entropy Defines temperature in Kelvin!

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 changed by:  S = Q/T B - Q/T A  If T A > T B, then  S > 0  System will achieve more randomness by exchanging heat until T B = T A

Efficiencies of Engines  Consider a cycle described by: W, work done by engine Q hot, heat that flows into engine from source at T hot Q cold, heat exhausted from engine at lower temperature, T cold  Efficiency is defined: Q hot engine Q cold W Since ,

Carnot Engines  Idealized engine  Most efficient possible

Carnot Cycle

Example 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. Given, e=0.5 when T cold =293 ºK, Find e when T cold = 243 ºK = 586 ºK Solution First, find T hot Now, find e given T hot =586 ºK and T cold =243 ºK e = 0.585

Refrigerators Q hot engine Q cold W Given: Refrigerated region is at T cold Heat exhausted to region with T hot Find: Efficiency Since, Note: Highest efficiency for small T differences

Heat Pumps Q hot engine Q cold W Given: Inside is at T hot Outside is at T cold Find: Efficiency Since, Like Refrigerator: Highest efficiency for small  T

Example 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.

Solution Find T for which e = 3 for heat pump. Above this T: use a heat pump Below this T: use gas 50% of ideal heat pump