Class Note for PH 101 at UA-Generl Physics II (1)
Class Note for PH 101 at UA-Generl Physics II (1)
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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Alabama - Tuscaloosa taught by a professor in Fall. Since its upload, it has received 18 views.
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Date Created: 02/06/15
Chapter 12 The Laws of Thermodynamics This chapter deals with the laws of thermodynamics Work in thermodynamics processes A AV 1 A gas in a cylinder can be compressed by pushing l F on a piston as shown to the right The work done V 39 I on the gas is given by W r FAx PA Ax eBAV 1 lt AX If the gas is compressed then AV is negative and the work done is positive Note the negative sign in the above equation If the gas expands then the work done on the gas is negative Work is done by the gas Isobaric process In an isobaric process the pressure is constant Thus the work done is W PAVPV Vf Isovolumeric process In an isovolumeric isochoric process the volume is constant Thus W PAV 0 If the pressure changes during the compression or expansion then the work done on the substance is the area under the P versus V curve Isothermal process It can be shown using calculus that the work done in an isothermal compression is W nRTaniVf isobaric isovolumetric isothermal First Law of Thermodynamics The rst law of thermodynamics relates the change in internal energy to the heat absorbed and the work done on a substance It is essentially a statement of conservation of energy AU Uf quotU Q W Q is positive if heat is absorbed and is negative if heat is lost by the system For a monatomic ideal gas U gNkBT gnRT Thus for an isothermal expansion or compression AU gnRAT 0 and Q W An adiabatic process is one for which no heat ows into or out of a system For an adiabatic process AU W Example P kPa An ideal monatomic gas goes through the A cyclic process A gtB gtC gtA as shown to 30 the right The temperature of the gas at A is 20 600K Calculate the work done on the gas the heat absorbed by the gas and the change in internal energy for each process and for 10 C B the total cycle 0 V L From the ideal gas equation PV nRT we 0 2 4 6 can calculate that TB 600K and Tc 200K Since this is a monatomic gas then we have U 32 nRT We keep in mind that 1 liter 10393 m3 A gtB Since the gas is expanding then the work done on the gas is negative and WAB area under PV curve 2x104 Pa4x10393 m3 80 J AUAB 32 nRATAB 0 TA TB From the lSt law QAB AUAB 7 WAB 0 7 80 J 80 J heat is absorbed B gtC Since the gas is being compressed the work done on the gas is positive and wBC area under PV curve 1110413a4XlO393 m3 40 J AUBC 32 HRATBC 32 nRTc 7 TB 32 Pch 7 32 PBVB 32 1x104 Pa2 7 6 x10393 m3 60 J QBC AUBC 7WBC 60 J J C gtA This is an isovolumeric process so WCA 0 AUCA HRATCA HRTA 7 Tc PAVA 7 Pch 32 3x104 Pa 7 leO4 Pa2X10393 m3 60 J QCA AUCA 7 WCA 60 J Summary results A gtB B gtC C gtA Total w J 80 40 0 40 Q J 80 100 60 40 AU J 0 60 60 0 Note that the total work done in the cycle is the enclosed area The negative value means that the gas does work on its environment during the cycle Since the gas returns to its original state the net change in internal energy is zero The net heat absorbed is equal to the net work done by the gas Heat is absorbed during the processes A gtB and C gtA and rejected during the process B gtC Heat Engines and the 2quot 1 Law of Thermodynamics The gas undergoing the cyclic process described in the above problem is an example of a heat engine It absorbs heat at high temperatures dumps heat at low temperatures and converts the difference into work P Qin Hot The ef ciency of a heat engine is the ratio of the net work done during the cycle to the heat absorbed 7 W 167 lQin Since W lQinl lQoutl then lQi llle e le i n 1 The QM law of thermodynamics states that no heat engine can have an ef ciency that is 100 e 1 In other words a heat engine cannot extract heat from a reservoir and convert it completely to work Some heat must be dumped at lower temperatures Example In the previous example of the ideal monatomic gas undergoing a cyclic process calculate the ef ciency Qquot QAB QCA 80 60 14o1 Wnet 40 done bythe gas W 40 e 7 7 0286 286 Qm 1401 Maximum possible ef ciency The maximum possible ef ciency of a heat engine that absorbs heat at That and dumps heat at Team is Ezekl This would apply if the heat engine were an ideal gas All other heat engines would have lower ef ciency Example A heat engine absorbs heat at 500 C and dumps heat at 25 C What is the maximum ef ciency e1 1 25273 0386 T h 500 273 If the engine takes in heat at the rate of 10 kW what is the power output W elQini Power K e 036810kW 368kW Ar Ar Entropy The entropy of a system is a measure of its disorder The higher the disorder the higher the entropy Speci cally if a system absorbs heat at a xed temperature then the change in entropy is given by units JK or calK Ifheat is absorbed then AS gt 0 Ifheat is lost then AS lt 0 Example 50 g of water melts at 0 C What is the change in entropy of the water mL 271 50g80wlg147 calK613JK T T 273K Example 100 cal of heat is transferred from a reservoir at 100 C to a reservoir at 0 C Assuming that the reservoirs are large enough so that their temperatures don t change what is the total change in entropy of the two reservoirs AS AShOt Assam Qhot led lOOcal lOOcal Th0 Twld 373K 273K 00980alK Note that AS gt 0 which means that the total disorder has increased If heat owed from the cold to the hot then AS would be negative This cannot occur spontaneously Another way of stating the 2quotd law of thermodynamics is to say that the total entropy of a closed system increases in all natural processes
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