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2.5: Usambazaji wa kasi ya Masi

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    176473
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    Malengo ya kujifunza

    Mwishoni mwa sehemu hii, utaweza:

    • Eleza usambazaji wa kasi ya Masi katika gesi bora
    • Pata kasi ya wastani na inayowezekana zaidi ya Masi katika gesi bora

    Vipande katika gesi bora husafiri kwa kasi ya juu, lakini hawana kusafiri kwa kasi sawa. Kasi ya RMS ni aina moja ya wastani, lakini chembe nyingi huhamia kwa kasi na wengi huenda polepole. Usambazaji halisi wa kasi una athari kadhaa za kuvutia kwa maeneo mengine ya fizikia, kama tutakavyoona katika sura za baadaye.

    Usambazaji wa Maxwell-Boltzmann

    Mwendo wa molekuli katika gesi ni random kwa ukubwa na mwelekeo kwa molekuli ya mtu binafsi, lakini gesi ya molekuli nyingi ina usambazaji wa kutabirika wa kasi ya Masi. Hii usambazaji kutabirika ya kasi Masi inajulikana kama usambazaji Maxwell-Boltzmann, baada ya waanzilishi wake, ambao mahesabu yake kulingana na nadharia kinetic, na tangu hapo imethibitishwa majaribio (Kielelezo \(\PageIndex{1}\)).

    Ili kuelewa takwimu hii, tunapaswa kufafanua kazi ya usambazaji wa kasi ya Masi, kwa kuwa na idadi ya mwisho ya molekuli, uwezekano wa kuwa molekuli itakuwa na kasi ya kutolewa ni 0.

    takwimu ni graph ya uwezekano dhidi kasi v katika mita kwa sekunde ya gesi oksijeni katika 300 kelvin. graph ina kilele uwezekano katika kasi V p ya chini ya 400 mita kwa sekunde na mzizi-maana-mraba uwezekano katika kasi v r m s ya juu 500 mita kwa sekunde. Uwezekano ni sifuri katika asili na huelekea sifuri katika infinity. Grafu haipatikani, lakini badala ya mwinuko upande wa kushoto kuliko upande wa kulia wa kilele.
    Kielelezo\(\PageIndex{1}\): The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed \(v_p\) is less than the rms speed \(v_{rms}\). Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than \(v_{rms}\).

    We define the distribution function \(f(v)\) by saying that the expected number \(N(v_1, v_2)\) of particles with speeds between \(v_1\) and \(v_2\) is given by

    \[N(v_1,v_2) = N\int_{v_1}^{v_2} f(v)dv. \nonumber\]

    [Since N is dimensionless, the unit of f(v) is seconds per meter.] We can write this equation conveniently in differential form: \[dN = Nf(v)dv.\]

    In this form, we can understand the equation as saying that the number of molecules with speeds between v and \(v + dv\) is the total number of molecules in the sample times f(v) times dv. That is, the probability that a molecule’s speed is between v and \(v + dv\) is f(v)dv.

    We can now quote Maxwell’s result, although the proof is beyond our scope.

    Maxwell-Boltzmann Distribution of Speeds

    The distribution function for speeds of particles in an ideal gas at temperature \(T\) is

    \[f(v) = \dfrac{4}{\sqrt{\pi}}\left(\dfrac{m}{2k_BT}\right)^{3/2} v^2e^{-mv^2/2k_BT}.\]

    The factors before the \(v^2\) are a normalization constant; they make sure that \(N(0,\infty ) = N\) by making sure that \(\int_0^{|infty} f(v)dv = 1.\) Let’s focus on the dependence on v. The factor of \(v^2\) means that \(f(0) = 0\) and for small v, the curve looks like a parabola. The factor of \(e^{-m_0v^2/2k_BT}\) means that \(\lim_{v\rightarrow \infty} f(v) = 0\) and the graph has an exponential tail, which indicates that a few molecules may move at several times the rms speed. The interaction of these factors gives the function the single-peaked shape shown in the figure.

    Example \(\PageIndex{1}\): Calculating the Ratio of Numbers of Molecules Near Given Speeds

    In a sample of nitrogen \(N_2\) with a molar mass of 28.0 g/mol) at a temperature of \(273^oC\) find the ratio of the number of molecules with a speed very close to 300 m/s to the number with a speed very close to 100 m/s.

    Strategy

    Since we’re looking at a small range, we can approximate the number of molecules near 100 m/s as \(dN_{100} = f(100 \, m/s)dv\). Then the ratio we want is

    \[\dfrac{dN_{300}}{dN_{100}} = \dfrac{f(300 \, m/s)dv}{f(100 \, m/s)dv} = \dfrac{f(300 \, m/s)}{f(100 \, m/s)}. \nonumber\]

    All we have to do is take the ratio of the two f values.

    Solution

    1. Identify the knowns and convert to SI units if necessary. \[T = 300 \, K, \, k_B = 1.38 \times 10^{-23} J/K\] \[M = 0.0280 \, kg/mol \, so \, m = 4.65 \times 10^{-26} \, kg\]
    2. Substitute the values and solve. \[\begin{align*} \dfrac{f(300 \, m/s)}{f(100 \, m/s)} &= \dfrac{\frac{4}{\sqrt{\pi}}\left(\frac{m}{2k_BT}\right)^{3/2} (300 \, m/s)^2 exp[-m(300 \, m/s)^2/2k_BT]}{\frac{4}{\sqrt{\pi}}\left(\frac{m}{2k_BT}\right)^{3/2} (100 \, m/s)^2 exp[-m(100 \, m/s)^2 /2k_BT]} \\[4pt] &= \dfrac{(300 \, m/s)^2 exp [-(4.65 \times 10^{-26} \, kg)(300 \, m/s)^2 /2(1.38 \times 10^{-23} J/K) (300 \, K)]}{(100 \, m/s)^2 exp [-(4.65 \times 10^{-26} \, kg)(100 \, m/s)^2 /2(1.38 \times 10^{-23} J/K) (300 \, K)]} \\[4pt] &= 3^2 exp \left[-\dfrac{(4.65 \times 10^{-26} kg)[(300 \, m/s)^2 - (100 \, m/s)^2]}{2(1.38 \times 10^{-23} \, J/K)(300 \, K)} \right] = 5.74 \end{align*}\]

    Figure \(\PageIndex{2}\) shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds.

    Two distributions of probability versus velocity v in meters per second at two different temperatures, T one and T two, are plotted on the same graph. Temperature two is greater than Temperature one. The distribution for T two has a broader peak with a maximum at a higher velocity and lower probability than the distribution for Temperature one.
    Kielelezo\(\PageIndex{2}\): Usambazaji wa Maxwell-Boltzmann umebadilishwa kwa kasi ya juu na kupanuliwa kwa joto la juu.

    Kwa idadi ndogo tu ya molekuli, usambazaji wa kasi hubadilika karibu na usambazaji wa Maxwell-Boltzmann . Hata hivyo, unaweza kuona simulation hii kuona vipengele muhimu kwamba molekuli kubwa zaidi hoja polepole na kuwa na usambazaji nyembamba. Tumia kuweka-up “Gesi 2, kasi ya Random”. Kumbuka kuonyesha chini kulinganisha histograms ya mgawanyo wa kasi na curves kinadharia.

    Tunaweza kutumia usambazaji uwezekano wa kuhesabu maadili ya wastani kwa kuzidisha kazi ya usambazaji kwa wingi kuwa wastani na kuunganisha bidhaa juu ya kasi zote iwezekanavyo. (Hii ni sawa na kuhesabu wastani wa mgawanyo wa kipekee, ambapo huzidisha kila thamani kwa idadi ya mara hutokea, kuongeza matokeo, na ugawanye na idadi ya maadili. Muhimu ni sawa na hatua mbili za kwanza, na kuimarisha ni sawa na kugawa kwa idadi ya maadili.) Hivyo kasi ya wastani ni

    \ [overline {v} =\ int_0^ {\ infty} vf (v) dv =\ sqrt {\ dfrac {8} {\ pi} \ dfrac {k_bt} {m}} =\ sqrt {\ dfrac {8} {\ pi}\ dfrac {RT} {M}}.\]

    Vile vile,

    \ [v_ {rms} =\ sqrt {\ overline {v} ^2} =\ sqrt {\ int_0^ {\ infty} v ^ 2 f (v) dv} =\ sqrt {\ dfrac {3K_BT} {m}} =\ sqrt {3RT} {M}}\]

    kama katika Shinikizo, Joto, na RMS Speed. Kasi inayowezekana zaidi, pia huitwa kasi ya kilele\(v_p\) ni kasi kwenye kilele cha usambazaji wa kasi. (Katika takwimu itakuwa kuitwa mode.) Ni chini ya kasi ya RMS\(v_{rms}\). Kasi inayowezekana zaidi inaweza kuhesabiwa kwa njia inayojulikana zaidi ya kuweka derivative ya kazi ya usambazaji, kuhusiana na v, sawa na 0. Matokeo yake ni

    \[v_p = \sqrt{\dfrac{2k_BT}{m}} = \sqrt{\dfrac{2RT}{M}},\]

    ambayo ni chini ya\(v_{rms}\). Kwa kweli, kasi ya RMS ni kubwa zaidi kuliko kasi inayowezekana zaidi na kasi ya wastani.

    Kasi ya kilele hutoa njia rahisi zaidi ya kuandika kazi ya usambazaji wa Maxwell-Boltzmann:

    \[f(v) = \dfrac{4v^2}{\sqrt{\pi}v_p^3}e^{-v^2/v_p^2}\]

    Kwa sababu hiyo\(e^{-mv^2/2k_BT}\), ni rahisi kutambua nishati ya kutafsiri kinetic. Hivyo, maneno hayo ni sawa na \(e^{-K/k_BT}\). Usambazaji f (v) unaweza kubadilishwa kuwa usambazaji wa nishati ya kinetic kwa kuhitaji hilo \(f(K)dK = f(v)dv\). Boltzmann alionyesha kuwa formula kusababisha ni zaidi kwa ujumla husika kama sisi kuchukua nafasi ya nishati kinetic ya tafsiri na jumla ya nishati mitambo E. Matokeo ya Boltzmann ni

    \ [f (E) =\ dfrac {2} {\ sqrt {\ pi}} (k_bt) ^ {-3/2}\ sqrt {E} e^ {-E/k_BT} = \ dfrac {2} {\ sqrt {\ pi} (k_bt) ^ {3/2}} \ dfrac {E} {e} {e^ {e/kt} {e^ _bt}}.\]

    Sehemu ya kwanza ya equation hii, pamoja na ufafanuzi hasi, ni njia ya kawaida ya kuandika. Tunatoa sehemu ya pili tu kwa kusema kwamba \(e^{E/k_BT}\) katika denominator ni ubiquitous katika quantum pamoja na mechanics classical takwimu.

    Mkakati wa Kutatua matatizo: Usambazaji wa kasi

    • Hatua ya 1. Kuchunguza hali ili kuamua kwamba inahusiana na usambazaji wa kasi ya Masi.
    • Hatua ya 2. Fanya orodha ya kiasi gani kinachopewa au kinaweza kuhitimishwa kutokana na tatizo kama ilivyoelezwa (kutambua kiasi kinachojulikana).
    • Hatua ya 3. Tambua hasa kile kinachohitajika kuamua katika tatizo (kutambua kiasi haijulikani). Orodha iliyoandikwa ni muhimu.
    • Hatua ya 4. Badilisha maadili inayojulikana katika vitengo sahihi vya SI (K kwa joto, Pa kwa shinikizo,\(m^3\) kwa kiasi, molekuli kwa N, na moles kwa n). Mara nyingi, ingawa, kutumia R na molekuli ya molar itakuwa rahisi zaidi kuliko kutumia\(k_B\) na molekuli ya Masi.
    • Hatua ya 5. Kuamua kama unahitaji kazi ya usambazaji kwa kasi au moja kwa nishati, na kama unatumia formula kwa moja ya kasi ya tabia (wastani, pengine, au RMS), kutafuta uwiano wa maadili ya kazi ya usambazaji, au kukadiria muhimu.
    • Hatua ya 6. Tatua equation sahihi kwa sheria bora ya gesi kwa wingi kuamua (kiasi haijulikani ). Kumbuka kwamba ikiwa unachukua uwiano wa maadili ya kazi ya usambazaji, mambo ya kuhalalisha hugawanyika. Au kama makadirio muhimu, kutumia njia aliuliza kwa tatizo.
    • Hatua ya 7. Badilisha kiasi kinachojulikana, pamoja na vitengo vyao, katika equation sahihi na kupata ufumbuzi wa namba kamili na vitengo.

    Sasa tunaweza kupata ufahamu wa ubora wa puzzle kuhusu muundo wa anga ya dunia. Hidrojeni ni kwa mbali elementi ya kawaida katika ulimwengu, na heliamu ni kwa mbali ya pili ya kawaida. Aidha, heliamu huzalishwa daima duniani na kuoza kwa mionzi. Kwa nini mambo hayo ni nadra sana katika anga yetu? Jibu ni kwamba molekuli za gesi zinazofikia kasi juu ya kasi ya kutoroka duniani, takriban 11 km/s, zinaweza kutoroka kutoka angahewa hadi angani. Kwa sababu ya masi ya chini ya molekuli ya hidrojeni na heliamu, huhamia kwa kasi ya juu kuliko molekuli nyingine za gesi, kama vile nitrojeni na oksijeni. Wachache tu kisichozidi kutoroka kasi, lakini molekuli mbali wachache nzito kufanya. Kwa hiyo, zaidi ya mabilioni ya miaka ambayo Dunia imekuwepo, molekuli nyingi za hidrojeni na heliamu zimetoroka kutoka angahewa kuliko molekuli nyingine, na hakuna hata mojawapo iko sasa.

    Tunaweza pia kuangalia nyingine kwenye baridi ya uvukizi, ambayo tulijadiliwa katika sura ya joto na joto. Liquids, kama gesi, zina usambazaji wa nguvu za Masi. Molekuli ya juu ya nishati ni wale ambao wanaweza kutoroka kutoka vivutio vya intermolecular ya kioevu. Kwa hiyo, wakati kioevu kinachopuka, molekuli zilizoachwa nyuma zina nishati ya chini, na kioevu kina joto la chini.