Jednadžba idealnog plina i kinetička teorija
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1 Jednadžba idealnog plina i kinetička teorija FIZIKA PSS-GRAD 9. studenog 017.
2 14.1 Molekulska masa, mol i Avogadrov broj To facilitate comparison of the mass of one atom with another, a mass scale know as the atomic mass scale has been established. The unit is called the atomic mass unit (symbol u). The reference element is chosen to be the most abundant isotope of carbon, which is called carbon-1. 1 u kg The atomic mass is given in atomic mass units. For example, a Li atom has a mass of 6.941u.
3 14.1 Molekulska masa, mol i Avogadrov broj One mole of a substance contains as many particles as there are atoms in 1 grams of the isotope carbon-1. The number of atoms per mole is known as Avogadro s number, NA. N A mol 1 N n NA number of moles number of atoms
4 14.1 Molekulska masa, mol i Avogadrov broj mparticle N m n mparticle N A Mass per mole The mass per mole (in g/mol) of a substance has the same numerical value as the atomic or molecular mass of the substance (in atomic mass units). For example Hydrogen has an atomic mass of g/mol, while the mass of a single hydrogen atom is u.
5 14.1 Molekulska masa, mol i Avogadrov broj Example 1 The Hope Diamond and the Rosser Reeves Ruby The Hope diamond (44.5 carats) is almost pure carbon. The Rosser Reeves ruby (138 carats) is primarily aluminum oxide (Al O3). One carat is equivalent to a mass of 0.00 g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of AlO3 molecules in the ruby.
6 14.1 Molekulska masa, mol i Avogadrov broj (a) n 44.5 carats 0.00 g 1 carat mol m Mass per mole g mol N nn A mol mol atoms (b) n 138 carats 0.00 g 1 carat 0.71 molecules m Mass per mole g mol N nn A 0.71 mol mol atoms
7 14. Jednadžba stanja idealnog plina An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition of low density means that the molecules are so far apart that they do not interact except during collisions, which are effectively elastic. At constant volume the pressure is proportional to the temperature. P T
8 14. Jednadžba stanja idealnog plina At constant temperature, the pressure is inversely proportional to the volume. P 1 V The pressure is also proportional to the amount of gas. P n
9 14. Jednadžba stanja idealnog plina THE IDEAL GAS LAW The absolute pressure of an ideal gas is directly proportional to the Kelvin temperature and the number of moles of the gas and is inversely proportional to the volume of the gas. nrt P V PV nrt R 8.31 J mol K
10 14. Jednadžba stanja idealnog plina n N NA R T NkT PV nrt N NA k R 8.31 J mol K J K 3 1 N A mol
11 14. Jednadžba stanja idealnog plina Example Oxygen in the Lungs In the lungs, the respiratory membrane separates tiny sacs of air (pressure 1.00x105Pa) from the blood in the capillaries. These sacs are called alveoli. The average radius of the alveoli is 0.15 mm, and the air inside contains 14% oxygen. Assuming that the air behaves as an ideal gas at 310K, find the number of oxygen molecules in one of these sacs. PV NkT
12 14. Jednadžba stanja idealnog plina Pa m PV N J K 310 K kt molecules of air Number of molecules of oxygen in one sac molecules
13 14. Jednadžba stanja idealnog plina Conceptual Example 3 Beer Bubbles on the Rise Watch the bubbles rise in a glass of beer. If you look carefully, you ll see them grow in size as they move upward, often doubling in volume by the time they reach the surface. Why does the bubble grow as it ascends?
14 14. Jednadžba stanja idealnog plina Consider a sample of an ideal gas that is taken from an initial to a final state, with the amount of the gas remaining constant. PV nrt PV nr constant T Pf V f Tf PiVi Ti
15 14. Jednadžba stanja idealnog plina Pf V f Tf Constant T, constant n: Constant P, constant n: PiVi Ti Pf V f PiVi Boyle s law Vf Vi T f Ti Charles law
16 14.3 Kinetička teorija plinova Čestice se u neprekidnom, nasumičnom gibanju sudaraju jedna s drugom te sa stijenkama posude. U svakom se sudaru brzina čestice mijenja. Kao rezultat toga, čestice (atomi ili molekule) imaju različite brzine.
17 14.3 Kinetička teorija plinova postotak molekula u intervalu brzina RASPODJELA BRZINA MOLEKULA najvjerojatnija brzina je oko 400 m/s najvjerojatnija brzina je oko 800 m/s brzina molekule, m/s
18 14.3 Kinetička teorija plinova KINETIČKA TEORIJA Prosječna sila za jednu molekulu: Δ (m v ) Δv F = ma = m = Δt Δt ( +m v ) ( m v ) m v F= = L/ v L
19 14.3 Kinetička teorija plinova Prosječna sila za jednu molekulu: mv F= L Prosječna sila za N molekula: N mv F= 3 L rms (root-mean-square) brzina v rms = v F F N mv p= = = A 3 L3 L v rms = v volumen
20 14.3 Kinetička teorija plinova N mv p= 3 V Ek NkT rms mv 1 pv = N ( m v rms ) = N ( 3 3 ) N k T = N Ek 3 3 Ek = k T
21 14.3 Kinetička teorija plinova Konceptualni primjer 5 Ima li pojedinačna čestica temperaturu? Svaka čestica plina ima kinetičku energiju. Prosječnu kinetičku energiju povezali smo s temperaturom idealnog plina. Možemo li onda zaključiti da i pojedinačna čestica ima temperaturu?
22 14.3 Kinetička teorija plinova Primjer 6 Brzina molekula u zraku Zrak je smjesa molekula dušika N (molekulske mase 8,0 u) i molekula kisika O (molekulske mase 3,0 u). Uz pretpostavku da se zrak ponaša kao idealni plin, odredite prosječnu (rms) brzinu molekula kisika i dušika, ako je temperatura zraka 93 K. m v rms 3 = kt 3 kt v rms = m
23 14.3 Kinetička teorija plinova Za dušik, masa jedne (N =1) molekule: m N 1 n= = = M NA NA M 8,0 g/mol 6 m= = = 4,65 10 kg 3 1 NA 6,0 10 mol 3 kt 3 1, J/K 93 K v rms = = = 511 m/s 6 m 4,65 10 kg
24 14.3 Kinetička teorija plinova UNUTRAŠNJA ENERGIJA JEDNOATOMNOG IDEALNOG PLINA m v rms 3 Ek = = kt R = kna 3 3 U = N E k = N k T = n R T
25 14.4 Difuzija Proces u kojem se molekule gibaju iz područja više koncentracije u područje niže koncentracije nazivamo difuzijom. NA POČETKU KASNIJE
26 14.4 Difuzija Konceptualni primjer 7 Zašto je difuzija relativno spora? Molekula plina, na sobnoj temperaturi, ima prosječnu (rms) translacijsku brzinu od nekoliko stotina metara po sekundi. S takvom brzinom molekula bi mogla stići s kraja na kraj sobe u djeliću sekunde. No, obično treba nekoliko sekundi, ili čak minuta, da se miris iz otvorene bočice parfema proširi po sobi. Zašto treba toliko puno vremena?
27 14.4 Difuzija viša koncentracija C poprečni presjek = A niža koncentracija C1 tok topive tvari poprečni presjek = A viša temperatura T tok topline niža temperatura T
28 14.4 Difuzija FICKOV ZAKON DIFUZIJE Masa m otopine koja difundira u vremenu t kroz stupac otapala duljine L i poprečnog presjeka A je difuzijska konstanta DAΔC t m L gradijent koncentracije Jedinica SI za difuzijsku konstantu: m/s
29 ZADACI ZA VJEŽBU 1. Masa od 135 g nekog elementa sadrži 30,1 103 atoma. Koji je to element? RJEŠENJE: aluminij. Valjkasta čaša vode ima polumjer 4,50 cm i visinu 1,0 cm. Gustoća vode je 1,00 g/cm3. Kolika je količina vode u toj čaši? RJEŠENJE: 4,4 mol 3. Da bi se napunio balon potrebno je 0,16 g helija. Koliko bi, za isti balon, trebalo grama dušika (istog tlaka, volumena i temperature)? RJEŠENJE: 1,1 g 4. Na sunčanoj strani Venere atmosferski tlak je 9,0 106 Pa, a temperatura 740 K. Na površini Zemlje tlak je 1,0 105 Pa, a temperatura može doseći 30 K. Ovi podaci pokazuju da Venera ima "deblju" atmosferu tj. da je broj molekula po jediničnom volumenu (N/V) na Veneri veći nego na Zemlji. Koliko je puta veći? RJEŠENJE: 39
30 ZADACI ZA VJEŽBU 5. Soba je široka 4,0 m, dugačka 5,0 m i visoka,5 m. Zrak u sobi sadrži 79% dušika (N) i 1% kisika (O). Kolika je masa zraka pri temperaturi od oc i tlaku od 1, Pa? RJEŠENJE: 59 kg 6. Dva mola idealnog plina nalaze se u posudi volumena 8, m3 pri tlaku 4,5 105 Pa. Odredite prosječnu translacijsku kinetičku energiju molekule plina. RJEŠENJE: 4, J 7. Prije početka vožnje tlak u automobilskoj gumi je, Pa. Vanjska temperatura je 84 K. Nakon vožnje tlak je 3, Pa. Kolika je temperatura zraka u gumi? Zanemarite rastezanje gume. RJEŠENJE: 304 K
31 ZADACI ZA VJEŽBU 8. Kisik za bolničke pacijente drži se u posebnom rezervoaru pod tlakom od 65 atmosfera i na temperaturi od 88 K. Rezervoar se nalazi u posebnoj sobi, a kisik se dovodi u pacijentovu sobu pod tlakom od 1,00 atmosfera i na temperaturi od 97 K. Koji volumen u pacijentovoj sobi zauzima kisik koju u rezervoaru zauzima 1,00 m3? RJEŠENJE: 67 m3 9. Tlak sumporovog dioksida (SO) je,1 104 Pa. U 50,0 m3 nalazi se 41 mol toga plina. Odredite prosječnu (rms) translacijsku brzinu molekula SO. RJEŠENJE: 343 m/s 10. Helij se nalazi u posudi volumena 0,010 m3 pod tlakom od 6, 105 Pa. Koliko dugo mora raditi stroj snage 0,5 konjskih snaga (jedna konjska snaga je 746 W) da bi energija koju "proizvede" odgovarala unutrašnjoj energiji spomenutog helija? RJEŠENJE: 50 s
32 PITANJA ZA PONAVLJANJE 1. Idealni plin. Atomska jedinica mase 3. Količina tvari 4. Avogadrova konstanta 5. Boyle-Mariotteov zakon 6. Charlesov zakon 7. Gay-Lussacov zakon 8. Jednadžba idealnog plina 9. Boltzmannova konstanta 10. Unutrašnja energija PITANJA ZA PONAVLJANJE
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