{"id":34650,"date":"2021-03-28T13:00:11","date_gmt":"2021-03-28T13:00:11","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=34650"},"modified":"2025-09-19T23:20:59","modified_gmt":"2025-09-19T23:20:59","slug":"problemata-combinatoria-in-thermodynamica","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/","title":{"rendered":"Problemata Combinatoria in Thermodynamica"},"content":{"rendered":"<style>\n\tp, ul, ol {\n\t\ttext-align: justify;\n\t}\n\th1, h2 {\n\ttext-align:center;\n\t}\n<\/style>\n<h1>Problemata Combinatoria in Thermodynamica<\/h1>\n<p style=\"text-align:center;\"><em>Quot modis systema physicum ex milionibus elementorum compositum ordinari potest? In hac lectione tractabimus quomodo mathematicae sinant quaestiones huiusmodi in contextu thermodynamicae respondere, a distributione quantorum energiae in systematibus atomicis usque ad computationem configurationum possibilium in systematibus magnae scalae. Uti instrumentis sicut combinatoriis, logarithmis et formula Stirling, explorabimus quomodo numeros extraordinarios tractare et problemata apparenter inaccessibilia solvere.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Proposita Discendi:<\/strong><br \/>\nPost hanc lectionem discipulus poterit<\/p>\n<ol>\n<li><strong>Intelligere<\/strong> quomodo problemata combinatoria ad contextum thermodynamicae applicentur, speciatim ad ordinationem systematum physicorum.<\/li>\n<li><strong>Computare<\/strong> configurationes possibiles systematum atomicorum per numeros combinatorios.<\/li>\n<li><strong>Adhibere<\/strong> formulam Stirling ad aestimandum ordinem magnitudinis configurationum complexarum.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><u>INDEX CONTENTORUM<\/u>:<br \/>\n<a href=\"#1\"><strong>Problemata combinatoria<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Problemata cum magnis numeris<\/strong><\/a><br \/>\n<a href=\"#3\">Usus logarithmorum et formulae Stirling ad computationem ordinis magnitudinis<\/a><br \/>\n<a href=\"#4\"><em>Evolutio per approximationem simplicem<\/em><\/a><br \/>\n<a href=\"#5\"><em>Evolutio per approximationem ordinariam<\/em><\/a><br \/>\n<a href=\"#6\"><strong>Exempla computationum combinatoriarum et ordinis magnitudinis<\/strong><\/a><br \/>\n<a href=\"#7\">Casu 1: Magna factorialia<\/a><br \/>\n<a href=\"#8\">Casu 2: Magna combinatoria<\/a>\n<\/p>\n<p>Communis quaestio in quibusdam condicionibus physicis est: Quot modis diversis systema datum ordinari potest? Haec problemata combinatoria in thermodynamica frequenter occurrunt. Quamvis initio simplicia videantur, fiunt tamen difficilia cum numeros eximie magnos includimus, sicut <strong><a href=\"http:\/\/toposuranos.com\/material\/mol-y-masa-molar\/\" rel=\"noopener\" target=\"_blank\">numerus Avogadro<\/a><\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">N_A<\/span>, qui exemplum est quantopere laboriosum sit cum magnitudinibus huiusmodi tractare.<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Problemata combinatoria<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/K7_De9wux4A?si=q93_T8xh15EdfJ6B\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Ad intellegendam magnitudinem problematum quae combinatoriam in thermodynamica implicant, consideremus hoc exemplum:<\/p>\n<h4>Exemplum: combinationes super quantos energiae<\/h4>\n<p>Ponamus systema ex 10 atomis compositum. Quisque atomus potest continere solum 1 vel 0 unitates energiae, appellatas <strong>quantos energiae<\/strong>. Quot modis diversis hi quanti distribui possunt si habemus (a) 10 quantos energiae et (b) 5 quantos energiae?<\/p>\n<h5>Solutio<\/h5>\n<p>Repraesentamus atomos ut spatia ad quantos energiae reponendos parata. Si spatium plenum est, significat atomo correspondenti iam suum quantum energiae inesse.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-zJa7H7iMOGY\/YF3QhG94SoI\/AAAAAAAAEuI\/olfdSHfeJOgKea3vHXEAyr5QCJ1ZjmcpgCLcBGAsYHQ\/s839\/cuantodeenergia.PNG\" \n               class=\"aligncenter lazyload\" alt=\"Problemata combinatoria in thermodynamica\" width=\"400\" height=\"200\" \/><\/center><\/p>\n<p>Ad numerandum modos quibus <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> quanti energiae inter <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> spatia distribuuntur, utimur numero combinatorio:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\binom{n}{k}=\\dfrac{n!}{k!(n-k)!}<\/span>\n<p>Haec ratio nobis praebet numerum <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> statuum possibilium.<\/p>\n<p><strong>(a)<\/strong> Si 10 quanti inter 10 spatia distribuuntur, una sola forma est id faciendi. Itaque <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega=1<\/span>:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\Omega = \\binom{10}{10}=\\dfrac{10!}{10!(10-10)!} = \\dfrac{10!}{10!0!} = 1 <\/span>\n<p><strong>(b)<\/strong> Pro 5 quantis inter 10 spatia distributis, computationem facimus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\Omega &amp;= \\displaystyle\\binom{10}{5} \\\\ \\\\\n\n&amp;=\\dfrac{10!}{5!(10-5)!} = \\dfrac{10!}{5!\\cdot 5!} \\\\ \\\\\n\n&amp;= \\dfrac{5! \\cdot 6\\cdot 7\\cdot 8 \\cdot 9\\cdot 10}{5! \\cdot 2\\cdot 3\\cdot 4\\cdot 5} \\\\ \\\\\n\n&amp;= \\dfrac{ 7\\cdot 8 \\cdot 9\\cdot 10}{ 4\\cdot 5} = 7\\cdot 2 \\cdot 9 \\cdot 2 = 252\n\n\\end{array}<\/span>\n<p>Itaque 252 configurationes possibiles sunt.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Problemata cum magnis numeris<\/h2>\n<p>Quae hucusque investigavimus tantum initium est. Si systema casus (b) ad 100 atomos et 50 quantos extendamus, obtinebimus <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega \\approx 10^{28}<\/span>. Nunc, imagina idem computum cum uno mole atomorum peractum; effectus inconceptibilis esset.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Usus logarithmorum et formulae Stirling ad computationem ordinis magnitudinis<\/h3>\n<p>Cum volumus aestimare magnitudinem formae <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega = \\binom{n}{k}<\/span> pro magnis valoribus <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, praesertim cum <span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span>, quod est casus quo valores maximi attinguntur, utile fit approximatione logarithmica Stirling uti.<\/p>\n<p>Ad numeros huius magnitudinis tractandos, possumus computus reformulare logarithmos sumendo, obtinentes:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\ln(\\Omega)=\\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!)<\/span>\n<p>Haec expressio tractari potest utens approximatione Stirling ad logarithmum factorialem; ad hoc habemus duas versiones possibiles, ordinariam et simplicem:<\/p>\n<ul>\n<li><strong>Approximatio ordinaria:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n<\/span> <\/li>\n<li><strong>Approximatio simplicata:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx  n\\ln(n) - n<\/span> <\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/p>\n<h4>Evolutio per approximationem simplicem<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/fnt99FeeHPM?si=H6r-5GhO2yqaeaXd\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Utendo approximatione simplici obtinentur sequentia resultata:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\ln(\\Omega) &amp; \\approx n\\ln(n) - n - (n-k)\\ln(n-k) + (n-k) - k\\ln(k) + k \\\\ \\\\\n\n&amp;= n\\ln(n) - (n-k)\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= n\\ln(n) - n\\ln(n-k) + k\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\left( \\dfrac{n}{n-k} \\right)^n \\right] + k\\ln\\left( \\dfrac{n-k}{k} \\right) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\dfrac{1}{\\left(1 - \\dfrac{k}{n} \\right)^n} \\right] + k\\ln\\left( \\dfrac{n}{k} - 1 \\right)\n\n\\end{array}<\/span>\n<p>Cum haec approximatio magnos valores <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> consideret, relationem applicamus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{n\\to\\infty} \\left(1-\\dfrac{k}{n} \\right)^n = e^{-k} <\/span>\n<p>Itaque:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(\\Omega) \\approx \\ln(e^k) + k\\ln\\left( \\dfrac{n}{k} -1 \\right) = k + k\\ln\\left( \\dfrac{n}{k} -1 \\right) <\/span>\n<p>Denique, adhibito mutatione basis pro logarithmis, obtinemus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right] <\/span>\n<p>Quod nos ducit ad resultatum:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\Omega \\approx 10^{k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right]}}<\/span>\n<p>Quamquam hic effectus non praebet valorem exactum <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span>, sinit tamen aestimationem numeri digitum necessariorum ad eum repraesentandum et melior fit cum <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> maius efficitur. Hoc methodo sufficit computare quod in exponente positum est, quod pleraeque calculatrices efficere possunt.<\/p>\n<p>Praeterea, hic modus sinit celeriter aestimare valorem maximum <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> pro magno valore <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. Considerato casu quo <span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span>, obtinemus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Max}\\left(\\Omega\\right) \\approx 10^{\\dfrac{n}{2}\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{n\/2} - 1 \\right) \\right]} = 10^{ n\\log(e)\/2 } <\/span>\n<p><a name=\"5\"><\/a><\/p>\n<h4>Evolutio per approximationem ordinariam<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/i3OO4lHV24Q?si=jEPCafXqgVxNYqH-\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Quamquam evolutio per approximationem ordinariam effectum accuratiorem praebebit, implicabit tamen nonnullos computus additos qui ad eventus fere aequivalentes pro magnis valoribus <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> ducent. Evolutio huius approximationis plures ex computationibus iam factis in approximatione simplici recyclet, ut in sequenti ratiocinio ostenditur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; = &amp; \\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!) \\\\ \\\\\n\n&amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln(2n\\pi)\\color{black} + n\\ln(n) - n \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2(n-k)\\pi)\\color{black} - (n-k)\\ln(n-k) + (n-k) \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2k\\pi)\\color{black} - k\\ln(k) + k\n\n\\end{array}<\/span>\n<p>Pars rubro signata respondet elementis additis in approximatione ordinaria consideratis, dum cetera sunt quae iam in approximatione simplici obtenta sunt. Ex hoc habetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln\\left( \\dfrac{2n\\pi}{2(n-k)\\pi \\cdot 2k\\pi} \\right)\\color{black} + k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) \\\\ \\\\\n\n&amp; = &amp;  k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right)\n\n\\end{array}<\/span>\n<p>Deinde, adhibito mutatione basis logarithmorum, habetur<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx \\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right] <\/span>\n<p>Denique, sumendo exponentialem basis 10 obtinetur<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\Omega \\approx 10^{\\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right]}\n\n<\/span>\n<p>Nunc, similiter ac prius, possumus invenire valorem maximum huius numeri evaluando cum <span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span>, quod hoc in casu sequentem effectum dabit:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Max}(\\Omega) &amp;\\approx &amp; 10^{\\log(e) \\left[ \\dfrac{n}{2} + \\dfrac{n}{2}\\ln\\left(\\dfrac{n}{(n\/2)} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2(n\/2)\\pi(n-n\/2)}{n}\\right) \\right]} \\\\ \\\\\n\n&amp; = &amp; 10^{\\log(e) \\left[\\dfrac{n}{2} - \\dfrac{1}{2}\\ln\\left(\\dfrac{n\\pi}{2} \\right) \\right]} = 10^{\\log(e)(n-\\ln(n\\pi\/2))\/2}\n\n\\end{array}<\/span>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Exempla computationum combinatoriarum et ordinis magnitudinis<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/M7NrtICrSzU?si=6UT34333Wp5YwFU1\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><br \/>\n<a name=\"7\"><\/a><\/p>\n<h3>Casus 1: Magna factorialia<\/h3>\n<p>Aestimemus ordinem magnitudinis <span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span>, id est, quantitatem digitum necessariorum ad hunc numerum scribendum.<\/p>\n<h5>Solutio<\/h5>\n<p>Ad hunc computum perficiendum, formulam Stirling hoc modo adhibemus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\ln\\left[ \\left(10^{50}\\right)! \\right] &amp;\\approx 10^{50}\\ln\\left(10^{50}\\right) - 10^{50}\\\\ \\\\\n\n&amp;= \\left[\\ln\\left(10^{50}\\right) -1\\right]10^{50}  \\\\ \\\\\n\n&amp;= \\left[50\\ln(10)-1 \\right]10^{50} \\\\ \\\\\n\n\\end{array}\n\n<\/span>\n<p>Deinde, mutationem basis logarithmorum applicamus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\ln\\left[ \\left(10^{50}\\right)! \\right] = \\dfrac{\\log\\left[\\left(10^{50}\\right)!\\right]}{\\log{e}}<\/span>\n<p>Itaque:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\log\\left[ \\left(10^{50}\\right)! \\right] \\approx \\log(e)\\left[50\\ln(10)-1 \\right]10^{50}\n\n<\/span>\n<p>Denique, applicata exponentiali basis 10, obtinemus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\left(10^{50}\\right)!  \\approx  10^{\\log(e)\\left[50\\ln(10)-1 \\right]10^{50}} = 10^{49,5657 \\cdot 10^{50}}\n\n<\/span>\n<p>Exponent super 10 ordinem magnitudinis repraesentat, praebens aestimationem numeri digitum quos numerus <span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span> habet.<\/p>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Casus 2: Magna combinatoria<\/h3>\n<p>Domus mediocris circiter 12 interruptoria lucis habet, quae accensa vel exstincta esse possunt. In mediocris, unaquaeque domus 4 personas alit. Si civitas 5 miliones incolarum habet, quot modis possibilibus dimidia pars interruptorum civitatis accensa esse potest?<\/p>\n<h5>Solutio<\/h5>\n<p>Numerus <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> interruptorum totorum in urbe est:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\nn &amp;=&amp;\\dfrac{\\text{incolae in urbe}}{\\text{personae per domum}} \\times \\text{interruptoria per domum} \\\\ \\\\\n\n&amp;=&amp; \\dfrac{5\\cdot 10^6}{4}\\cdot 12 = 15\\cdot 10^6\n\n\\end{array}\n\n<\/span>\n<p>Macrostatus ex omnibus microstatibus constitutus, in quibus dimidia pars interruptorum accensa est, congruit cum macrostatu qui maximum numerum configurationum possibilium habet. Hoc numerum maximum, <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max}<\/span> denotatum, aestimationes sequentes praebere possumus secundum quemque modum:<\/p>\n<ul>\n<ol><strong>Aestimatio ordinaria:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 - \\ln\\left(15\\pi\\cdot10^6 \/ 2 \\right) \\right]\/2} \\approx 10^{6.514.413,542}<\/span><\/ol>\n<ol><strong>Aestimatio simplicata:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 \\right]\/2} \\approx 10^{6.514.417,229}<\/span><\/ol>\n<\/ul>\n<p>Quamvis inter utrasque approximationes differentia prope 4 ordines magnitudinis (quae magna videri potest) exstet, re vera hoc parum momenti est respectu amplius 6 milionum et dimidii ordinum magnitudinis.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problemata Combinatoria in Thermodynamica Quot modis systema physicum ex milionibus elementorum compositum ordinari potest? In hac lectione tractabimus quomodo mathematicae sinant quaestiones huiusmodi in contextu thermodynamicae respondere, a distributione quantorum energiae in systematibus atomicis usque ad computationem configurationum possibilium in systematibus magnae scalae. Uti instrumentis sicut combinatoriis, logarithmis et formula Stirling, explorabimus quomodo numeros extraordinarios [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":30534,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":6,"footnotes":""},"categories":[1250,1292],"tags":[],"class_list":["post-34650","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-physica","category-thermodynamica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Problemata Combinatoria in Thermodynamica - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"In hac lectione pervidebimus Applicationem combinatoriae et formulae Stirling in problematibus thermodynamicis cum magnis numeris\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Problemata Combinatoria in Thermodynamica\" \/>\n<meta property=\"og:description\" content=\"In hac lectione pervidebimus Applicationem combinatoriae et formulae Stirling in problematibus thermodynamicis cum magnis numeris\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-03-28T13:00:11+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-09-19T23:20:59+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/12\/PROBLEMASDECOMBINATORIA-1024x585.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Problemata Combinatoria in Thermodynamica\" \/>\n<meta name=\"twitter:description\" content=\"In hac lectione pervidebimus Applicationem combinatoriae et formulae Stirling in problematibus thermodynamicis cum magnis numeris\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/12\/PROBLEMASDECOMBINATORIA.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Problemata Combinatoria in Thermodynamica\",\"datePublished\":\"2021-03-28T13:00:11+00:00\",\"dateModified\":\"2025-09-19T23:20:59+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\"},\"wordCount\":1448,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/12\/PROBLEMASDECOMBINATORIA.jpg\",\"articleSection\":[\"Physica\",\"Thermodynamica\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\",\"url\":\"http:\/\/toposuranos.com\/material\/la\/problemata-combinatoria-in-thermodynamica\/\",\"name\":\"Problemata Combinatoria in Thermodynamica - 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