Problem Free Expansion of Ideal Gas

Transcription

1 Problem 4.3 Free Expanon o Ideal Ga In general: ds ds du P dv P dv NR V dn Snce U o deal ga ndependent on olume (du=), and N = cont n the proce: dv In a ere o nntemal ree expanon, entropy change by: S V V NR dv V V NR V Note that ree expanon alway rreerble S>.

2 Problem 4. 4 emperature and entropy change n ree expanon he equaton o tate: P Gen the ntal temperature, molar olume, and nal molar olume, nd the nal temperature and ncreae n molar entropy. Soluton: Snce equaton o tate nole and a ndependent extene arable, t conenent to nd the undamental relaton n the energy repreentaton: du du Pd du d d ntegratng u C

3 Problem 4. 4 emperature and entropy change n ree expanon u C In the ree expanon proce, no work done on the ytem and no heat tranerred to the ytem t adabatcally nulated orm the enronment. hereore, the energy o the ytem doe not change n the ree expanon proce: u u Note that not the ntal entropy, t jut a contant (th may be a ource o ome conuon). Otherwe we would get = n the rreerble ree expanon proce.

4 Problem 4. 4 emperature and entropy change n ree expanon

5 Problem 4.4 Heat Exchange between wo Identcal Object Coneraton o energy demand: U C d C d where the nal temperature o each body. hereore: A A A A Sog or 37.[ ] A K Note that not the aerage temperature nce the heat capacte depend on

6 Problem 4.4 Heat Exchange between wo Identcal Object he change o entropy o the two ytem n th proce can be ound rom the expreon below, where we hae taken nto account that no work done n the proce S Q Q Cd Cd A d A d Integratng, we obtan: S A A S A J K.6

7 Problem Maxmum work theorem P A adabat. he ytem deler a non-zero work durng the adabatc proce A : W A >.. Snce the proce adabatc, no heat low between the ytem and the reerble heat ource (RHS): Q A =. C ochore V obar 3. From the coneraton o energy and the aboe two tatement, the nternal energy o the ytem decreae n th proce U A <. 4. Snce energy a monotoncally ncreang uncton o temperature, temperature o the ytem decreae n the proce A : < A. 5. Snce temperature a uncton o the tate only, t alo decreae n the proce A C. 6. Snce W AC < W A and U A = U AC, we conclude rom the coneraton o energy that heat low rom the ytem to the reerble heat nk durng the A-C- proce Q AC >. 7. Entropy o the compote ytem (ytem + reerble heat ource) ncreae eery tme there heat traner between the ytem and the RHS and ytem RHS (rreerble proce). 8. For example, at pont on the obar cloe to pont, ytem < A RHS becaue pote heat ha been tranerred to the RHS durng the AC proce Q AC >. 9. Snce ytem RHS durng the - proce, the proce rreerble and thu the whole A-C- proce rreerble.

8 U Problem 4.5 Heat Flow between wo Derent Object Maxmum work can be obtaned rom the two bode n a reerble proce where the total change o entropy o the two bode zero: S he change o energy o the two bode only depend on the ntal and nal tate and gen by: S a d C d a d b d b C Q Q C C S d d a b In the aboe expreon we replaced Q wth du nce the proce wa carred out at contant olume o both bode, o no work wa done by the bode drectly. he bode act a hot and cold bode o a typcal heat engne. Sog or : a b a b Subttutng the expreon or U ab a b he maxmum work negate o U: Wmax ab a b

9 Problem 4.6 Ecency o the heat pump Ecency o a reerble heat pump: Q W h h c h 94. 6K c 83. 5K 6.5 When calculatng ecence (and many other quantte) n thermodynamc, one need to ue the abolute temperature cale (Ke cale). Indeed, there nothng pecal about zero Celu o Fahrenhet, but you ue Celu or Fahrenhet n the equaton aboe, your ecency alway turn to zero the hot ytem at zero degree ndependent on the temperature o the cold body. h unphycal becaue zero o Celu or Fahrenhet temperature cale to a large degree arbtrary. he Ke cale the abolute temperature cale (zero Ke the lowet temperature poble) and hould alway be ued n thermodynamc. Conert temperature to Ke beore ung t n a thermodynamc equaton.