UIERSIY OF CALIFORIA - SAA CRUZ DEPARME OF PHYSICS PHYS Homework #4 Benjamin Stahl February, 05 PROBLEM It is given that the heat absorbed by a mole o ideal gas in a uasi-static process in which both its temperature and volume change is: Q c d + Pd. Where c is the molar speciic heat at constant volume. An expression or the change in entropy o the gas in a uasi-static process that takes it rom the initial volume, i, and temperature, i, to the inal volume,, and temperature,. ow the above euation is divided through everywhere by the temperature, and thereby redeined using the deinition o entropy change or a uasi-static process: ds Q c d + Pd. ow the pressure is related to the volume using the ideal gas law and then substituted into the above result: P k P k ds c d d + k.3 Integrating this result leads to the desired expression or the change in entropy: S S i ds c i d + k i d S c ln + k ln i i.4 Clearly, this result depends only on the initial and inal values o the temperature and volume. hus it is a state unction and thereore the result does not depend on the process used to go between the initial and inal states. PROBLEM For an Einstein solid with large and, the multiplicity is approximately given by: + + Ω,.
a. From this, the corresponding relationship or the chemical potential in terms o,, and will be derived. First, the entropy, S, as a unction o and is ound. Starting rom the deinition o entropy and then using properties o logarithms, the desired result can be ound: [ + ] + ] + + S k lnω k ln [ k ln + ln. ow, the chemical potential, µ, is ound using the deinition: [ S µ k U, + + ln + b. ow the high temperature limit >> will be investigated: + µ k ln k ln + [ k ln + + ] + k ln + ] k ln k ln +.3.4 In the high temperature limit, the argument o the second logarithm goes to unity and thus the second term in the above result goes to zero and can be saely ignored. hereore, the chemical potential in the high temperature limit is: µ k ln.5 ow, the low temperature limit << will be investigated: + µ k ln k ln + Expanding this in power series gives the result or low temperatures: µ k.6.7 From the results ound above it is clear that in the high temperature limit, the eect o adding another particle is much more substantial on the entropy than in the low temperature limit. 3 PROBLEM 3 It is given that one mole o a mono-atomic ideal gas is initially at a temperature o 300 K and a pressure o one atmosphere. a. he heat absorbed when the gas undergoes a uasi-static isothermal expansion to twice the initial volume will be calculated. First, the irst law o thermodynamics is stated: U Q +W 3. Since the internal energy depends on temperature, and no temperature change occurs in an isothermal process, this can be used to ind the heat, Q: 0 Q +W Q W 3. Solving the ideal gas law or the pressure, P, and substituting into the deinition o work leads to: P nr P nr W Pd nr i i d nr ln ow, the heat absorbed can be calculated: i Q nr ln mole8.3 J/mole K300 Kln 78 J 3.4 i 3.3
ow, the entropy can be calculated or this process as ollows: ds S Q 78 J 5.76 J/K 3.5 300 K b. ow, the temperature will be ound ater the gas undergoes an adiabatic and reversible isentropic expansion to our times the initial volume. Euation.39 o Schroeder gives the ollowing relationship between the degrees o reedom per molecule,, and the temperature and volume o an ideal gas undergoing an adiabatic expansion, which can be used to obtain an expression or the inal temperature: constant i i i i 3.6 Evaluating this result or the given scenario, where or an ideal gas 3, will give the desired result: 3 300 K 9 K 3.7 4 c. ow the inal temperature will be calculated i instead o the above processes, the gas had simply undergone a ree expansion to our times its initial volume. In a ree expansion, no mechanical work is done and no heat lows in or out o the gas, thus by the irst law o thermodynamics there will be no change in temperature and the inal temperature will be the same as the initial temperature: ow the increase in entropy is calculated using Euation.5 o Schroeder: 300 K 3.8 S k ln 6.0 0 3.38 0 3 J/K ln4.5 J/K 3.9 i 4 PROBLEM 4 Using a mono-atomic ideal gas, the amount o heat absorbed by the gas in one cycle, Q h, and the work done by the gas in one cycle, W, will be ound or the Carnot engine cycle. he ollowing igure depicts the Carnot cycle: Figure 4.: he Carnot cycle is exempliied in the above P diagram. From a to b, the gas expands isothermally at a temperature, H. hen rom b to c, the gas expands adiabatically and reaches a new temperature, L. c to d takes the gas through an isothermal compression, and then d to a takes the gas through an adiabatic compression, reaching the starting temperature o the entire cycle, H. Source: http://www.oberlin.edu/physics/dstyer/p/carnot.pd 3
he total heat absorbed by the gas will come rom the isothermal expansion: Q ab 4. ow, using the heat or isothermal processes as was discussed in Problem 3, the above result can become: nr H ln b 4. ow the total work done by the gas in one cycle can be ound as the sum o the work rom each step in the cycle: W W ab +W bc +W cd +W da 4.3 he work done in an isothermal process was discussed in Problem 3, now the work done in an adiabatic process will be developed. Euation.40 o Schroeder gives the ollowing relationship between volume and pressure or an adiabatic process involving an ideal gas: P constant K P K 4.4 Where is the adiabatic exponent deined by: +, and denotes the degrees o reedom; it is also deined as the ratio o speciic heats c P c. ote also the the constant has been arbitrarily assigned to K. ow, the deinition o work is employed to derive the work or an adiabatic process: W Pd K i i d K i K i 4.5 ow the given euation or total work can be illed in with the pertinent terms: W nr H ln b + K b c c + nr L ln d + K 4 d a 4.6 Where: Using this act allows the expression or the work to be simpliied: K b P c c & K 4 P d d P a a 4.7 W nr H ln nr H ln b b + b P b c c c + b P c c + nr L ln Applying the ideal gas law to this result gives: W nr H ln b hus, the eiciency, e, will be: c + nr L ln c d d + P d d P a a + nr H L + nr c L ln + nr L H d nr H ln e W Q H nr H ln + nr b L ln c nr H ln b d + P d d P d a a a b L ln c H ln d b c + nr L ln d 4.8 4.9 4.0 ow, starting again rom Euation.40 o Schroeder and using the ideal gas law gives: P c c b P c c c b b nr L c nr H b H c 4. L b 4
Applying this methodology to the other two states yields: ow the above two uantities are related: H L d a 4. H c L b d a c d b a 4.3 Substituting this into the relationship or the eiciency gives: L ln c e d b H ln + L L 4.4 H H a 5 PROBLEM 5 It is given that gas turbine engines operate on the Brayton thermodynamic cycle, which consists o two adiabatic processes the compression and expansion o gas through the turbine and two isobaric processes. It is also given that heat hermodynamics is transerredand to the Statistical gas during Mechanics combustion in the irst isobaric Winter 05 process and removed via a heat exchanger during the last second isobaric process. he ollowing igure depicts the Brayton cycle: turbine engines, such as those powering an rat, operate on a thermodynamic cycle called Brayton cycle, illustrated in the igure. It consists two adiabatic processes, the compression and ansion o gas through the turbine, and two aric processes. Heat is transerred to the gas ing combustion in an isobaric process point to t 3 and removed in a heat exchanger during ther isobaric process point 4 to point. a. Derive an expression or the maximum eiciency o the gas turbine as a unction o Figure 5.: he Carnot cycle. the pressure ratio max min and the ratio o speciic heats. b. Calculate the a. maximum ow, the maximum eiciency eiciency o a Brayton will beengine ound. that First the uses heat a diatomic absorbedgas in the ignition step will be ound. hose, 7 5 and the has euation a pressure oratio determining o 0. the heat in an isobaric process must be derived. Starting rom the irst law o eat pump is identical thermodynamics to an air conditioner, and applying except the that euipartition it is used to and move the deinition heat rom othe work yields: outdoor air or rom water circulated through U the Q ground +W into the Q indoors, U W in order nc to P P 5. m the building. a. I the outdoor or Using ground the ideal temperature gas law thisis becomes: and the indoor temperature is, then what is the maximum power that could be supplied Q ncin P heat + nr to the building interior Q nc 5. or every Watt o power used to operate the heat pump? Is this advantageous compared to using Where ohmic the act electrical that c heating, c P + Rwhere has been all utilized. o the work hereore is turned the heat into absorbed the by the gas will be: same amount o heat? Why or why not? nc 3 5.3 b. Give a numerical answer or the case that the outdoor temperature is 0 C and the Using Euation 4.3 o Schroeder, the eiciency can be expressed as: indoor temperature is 3 C. c. Repeat or an outdoor temperature o 0 C. What would you conclude about the e Q C 5.4 relative suitability o a heat pump in very cold climates? his is also why in that case it is advantageous, although initially more expensive, to use the underground water circulation as the heat reservoir. 5
hus, the the only remaining uantity is the heat removed which is also in an isobaric process. hus: hus the eiciency will be: Q C nc 4 5.5 e nc 3 nc 4 3 4 5.6 he two isentropic adiabatic processes: & 3 4 obey euation.40 o Schroeder, which can be used along with ideal gas law to obtain the ollowing: P a a b P a b a nr b nr a P a Pa b a b a Pa Where a and b have been used to preserve generality so long as they are the starting and end o an adiabatic process. ow, this relationship will be used to replace and 3 in the expression or the eiciency: ow, given that P P 3 and P 4 P, the pressure ratio, r P, is deined as: his allows or the eiciency to be urther simpliied: 5.7 P P 3 P4 4 P e 5.8 4 r P P max P min 5.9 e 4 r e r 5.0 P P 4 b. ow the maximum eiciency will be calculated or a Brayton engine using a diatomic gas 7 5 with pressure ratio o 0: 5 7 7 e 0 5 0 7 0.48 5. 6 PROBLEM 6 It is given that a heat pump is in principle the same as an air conditioner, except that it is used to move heat rom a cold reservoir to a warmer reservoir. a. It is given that the outdoor or ground temperature is 0 and that the indoor temperature is i. he maximum power that could be supplied in heat to the building interior or every Watt o power used to operate the heat pump will be determined. his reuires irst that the coeicient o perormance, COP, be ound. In this case, this can be generally expressed as: COP W Applying the irst law to this case allows or the work to be re-expressed: COP Q C he entropy extracted rom the hot reservoir is: H, thus the second law will reuire: 6. Q C 6. H Q C C Q C C H 6.3 6
his allows or an upper bound to be placed on the coeicient o perormance: COP Putting the appropriate temperatures into this gives the desired result: C H 6.4 COP 0 i 6.5 Clearly it is possible or the coeicient o perormance to be greater than unity while ohmic heating has an eiciency o one, so a heat pump can be advantageous in situations where it delivers more heat than it reuires work. b. ow a numerical result will be be ound or the case where the outdoor temperature is 0 C 73 K and the indoor temperature is 3 C 96 K: COP 73 K 96 K COP.9 6.6 hus or every Watt o power used to operate the heat pump between the speciied temperatures,.9 Watts o heat could be supplied to the indoors. c. ow this calculation will be repeated or the case where the outdoor temperature is 0 C 53 K: COP 53 K 96 K COP 6.9 6.7 It is thus apparent and also rom the unction itsel that the larger the temperature dierence between the hot and cold reservoirs, the smaller the coeicient o perormance o the heat pump. In very cold climates the heat pump is less practical because the coeicient o perormance is not as large, this is why it is more advantageous to use underwater circulation which is closer to the desired temperature as the heat reservoir. 7