{"id":31190,"date":"2021-03-28T13:00:51","date_gmt":"2021-03-28T13:00:51","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31190"},"modified":"2025-01-12T13:03:44","modified_gmt":"2025-01-12T13:03:44","slug":"problemas-de-combinatorias-em-termodinamica","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/","title":{"rendered":"Problemas de Combinat\u00f3rias em Termodin\u00e2mica"},"content":{"rendered":"<style>\n\tp, ul, ol {\n\t\ttext-align: justify;\n\t}\n\th1, h2 {\n\t\ttext-align: center;\n\t}\n<\/style>\n<h1>Problemas de Combinat\u00f3rias em Termodin\u00e2mica<\/h1>\n<p style=\"text-align: center;\"><em>Quantas maneiras existem de organizar um sistema f\u00edsico composto por milh\u00f5es de elementos? Nesta aula, abordaremos como a matem\u00e1tica permite responder a perguntas como essa no contexto da termodin\u00e2mica, desde a distribui\u00e7\u00e3o de quantas de energia em sistemas at\u00f4micos at\u00e9 o c\u00e1lculo de configura\u00e7\u00f5es poss\u00edveis em sistemas de grande escala. Utilizando ferramentas como combinat\u00f3rias, logaritmos e a f\u00f3rmula de Stirling, exploraremos como lidar com n\u00fameros extraordinariamente grandes e resolver problemas aparentemente inabord\u00e1veis.<\/em><\/p>\n<p style=\"text-align: center;\"><strong>Objetivos de Aprendizagem:<\/strong><br \/>\nAo final desta aula, o aluno ser\u00e1 capaz de:<\/p>\n<ol>\n<li><strong>Compreender<\/strong> como os problemas de combinat\u00f3ria s\u00e3o aplicados no contexto da termodin\u00e2mica, especificamente na organiza\u00e7\u00e3o de sistemas f\u00edsicos.<\/li>\n<li><strong>Calcular<\/strong> configura\u00e7\u00f5es poss\u00edveis de sistemas at\u00f4micos por meio de n\u00fameros combinat\u00f3rios.<\/li>\n<li><strong>Aplicar<\/strong> a f\u00f3rmula de Stirling para estimar a ordem de magnitude de configura\u00e7\u00f5es complexas.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><u>\u00cdNDICE DE CONTE\u00daDOS<\/u>:<br \/>\n<a href=\"#1\"><strong>Problemas de combinat\u00f3ria<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Problemas com grandes n\u00fameros<\/strong><\/a><br \/>\n<a href=\"#3\">Uso de logaritmos e da f\u00f3rmula de Stirling para o c\u00e1lculo da ordem de magnitude<\/a><br \/>\n<a href=\"#4\"><em>Desenvolvimento com a aproxima\u00e7\u00e3o simplificada<\/em><\/a><br \/>\n<a href=\"#5\"><em>Desenvolvimento com a aproxima\u00e7\u00e3o ordin\u00e1ria<\/em><\/a><br \/>\n<a href=\"#6\"><strong>Exemplos de c\u00e1lculos combinat\u00f3rios e de ordem de magnitude<\/strong><\/a><br \/>\n<a href=\"#7\">Caso 1: Grandes fatoriais<\/a><br \/>\n<a href=\"#8\">Caso 2: Grandes combinat\u00f3rias<\/a>\n<\/p>\n<p>Uma quest\u00e3o comum em certas situa\u00e7\u00f5es f\u00edsicas \u00e9: De quantas maneiras diferentes pode-se organizar um sistema dado? Esses problemas combinat\u00f3rios s\u00e3o frequentes na termodin\u00e2mica. Embora inicialmente pare\u00e7am simples, tornam-se complexos ao incorporar n\u00fameros extremamente grandes, como o <strong><a href=\"http:\/\/toposuranos.com\/material\/mol-y-masa-molar\/\" rel=\"noopener\" target=\"_blank\">n\u00famero de Avogadro<\/a><\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">N_A<\/span>, que exemplifica o qu\u00e3o esmagador pode ser trabalhar com magnitudes dessa escala.<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Problemas de Combinat\u00f3ria<\/h2>\n<p><center><br \/>\n\t<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/K7_De9wux4A?si=q93_T8xh15EdfJ6B\" \n\t        title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" \n\t        referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><br \/>\n<\/center><\/p>\n<p>Para compreender a magnitude dos problemas que envolvem combinat\u00f3ria na termodin\u00e2mica, consideremos o seguinte exemplo:<\/p>\n<h4>Exemplo: combina\u00e7\u00f5es sobre quantas de energia<\/h4>\n<p>Suponhamos um sistema composto por 10 \u00e1tomos. Cada \u00e1tomo pode armazenar apenas 1 ou 0 unidades de energia, chamadas <strong>quantas de energia<\/strong>. De quantas maneiras diferentes podemos distribuir essas quantas se tivermos (a) 10 quantas de energia e (b) 5 quantas de energia?<\/p>\n<h5>Solu\u00e7\u00e3o<\/h5>\n<p>Representamos os \u00e1tomos como espa\u00e7os dispon\u00edveis para armazenar uma quanta de energia. Se um espa\u00e7o est\u00e1 preenchido, isso significa que o \u00e1tomo correspondente j\u00e1 possui sua quanta de energia.<\/p>\n<p><center><br \/>\n\t<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\t     class=\"aligncenter lazyload\" alt=\"Problemas de combinat\u00f3ria na termodin\u00e2mica\" width=\"400\" height=\"200\" \/><br \/>\n<\/center><\/p>\n<p>Para contar as maneiras como <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> quantas de energia podem ser distribu\u00eddas entre <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> espa\u00e7os, usamos o n\u00famero combinat\u00f3rio:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\binom{n}{k}=\\dfrac{n!}{k!(n-k)!}<\/span><\/span>\n<\/p>\n<p>Esse c\u00e1lculo nos fornece o n\u00famero <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> de estados poss\u00edveis.<\/p>\n<p><strong>(a)<\/strong> Se houver 10 quantas distribu\u00eddas entre 10 espa\u00e7os, s\u00f3 existe uma maneira de faz\u00ea-lo. Portanto, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega=1<\/span><\/span>:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\Omega = \\binom{10}{10}=\\dfrac{10!}{10!(10-10)!} = \\dfrac{10!}{10!0!} = 1 <\/span><\/span>\n<\/p>\n<p><strong>(b)<\/strong> Para 5 quantas distribu\u00eddas entre 10 espa\u00e7os, realizamos o c\u00e1lculo:<\/p>\n<p style=\"text-align:center;\">\n<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>\n<p>Portanto, existem 252 configura\u00e7\u00f5es poss\u00edveis.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Problemas com Grandes N\u00fameros<\/h2>\n<p>O que analisamos at\u00e9 agora \u00e9 apenas o come\u00e7o. Se ampliarmos o sistema do caso (b) para 100 \u00e1tomos e 50 quantas, obteremos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega \\approx 10^{28}<\/span><\/span>. Agora, imagine realizar o mesmo c\u00e1lculo com um mol de \u00e1tomos; o resultado seria inconceb\u00edvel.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Uso de Logaritmos e da F\u00f3rmula de Stirling para o C\u00e1lculo da Ordem de Magnitude<\/h3>\n<p>Quando desejamos estimar uma magnitude da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega = \\binom{n}{k}<\/span><\/span> para valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, especialmente quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, que \u00e9 o caso em que se alcan\u00e7am os valores m\u00e1ximos, torna-se \u00fatil empregar a aproxima\u00e7\u00e3o logar\u00edtmica de Stirling.<\/p>\n<p>Para lidar com n\u00fameros dessa magnitude, podemos reformular os c\u00e1lculos tomando logaritmos, obtendo:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p>Essa express\u00e3o pode ser trabalhada utilizando a aproxima\u00e7\u00e3o de Stirling para o logaritmo fatorial, para isso temos duas vers\u00f5es poss\u00edveis, a ordin\u00e1ria e a simplificada:<\/p>\n<ul>\n<li><strong>Aproxima\u00e7\u00e3o ordin\u00e1ria:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n<\/span><\/span><\/li>\n<li><strong>Aproxima\u00e7\u00e3o simplificada:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx  n\\ln(n) - n<\/span><\/span><\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/p>\n<h4>Desenvolvimento com a Aproxima\u00e7\u00e3o Simplificada<\/h4>\n<p><center><br \/>\n\t<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/fnt99FeeHPM?si=H6r-5GhO2yqaeaXd\" \n\t        title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" \n\t        referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><br \/>\n<\/center><\/p>\n<p>Usando a aproxima\u00e7\u00e3o simplificada, obt\u00eam-se os seguintes resultados:<\/p>\n<p style=\"text-align:center;\">\n<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>\n<p>Como essa aproxima\u00e7\u00e3o considera valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, aplicamos a rela\u00e7\u00e3o:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{n\\to\\infty} \\left(1-\\dfrac{k}{n} \\right)^n = e^{-k} <\/span><\/span>\n<\/p>\n<p>Portanto:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p>Finalmente, ao empregar a mudan\u00e7a de base para logaritmos, obtemos:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p>O que nos leva ao resultado:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p>Ainda que esse resultado n\u00e3o forne\u00e7a o valor exato de <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span>, permite obter uma estimativa da quantidade de d\u00edgitos necess\u00e1rios para represent\u00e1-lo e que melhora conforme <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> aumenta. Com esse m\u00e9todo, basta calcular o que aparece no expoente, algo que a maioria das calculadoras pode realizar.<\/p>\n<p>Al\u00e9m disso, essa abordagem permite estimar rapidamente o valor m\u00e1ximo de <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> para um valor grande de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. Considerando o caso em que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, obtemos:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h4>Desenvolvimento com a Aproxima\u00e7\u00e3o Ordin\u00e1ria<\/h4>\n<p><center><br \/>\n\t<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/i3OO4lHV24Q?si=jEPCafXqgVxNYqH-\" \n\t        title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" \n\t        referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><br \/>\n<\/center><\/p>\n<p>Ainda que o desenvolvimento com a aproxima\u00e7\u00e3o ordin\u00e1ria ofere\u00e7a um resultado mais preciso, isso implicar\u00e1 alguns c\u00e1lculos adicionais que conduzir\u00e3o a resultados aproximadamente equivalentes para grandes valores de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. O desenvolvimento dessa aproxima\u00e7\u00e3o reaproveitar\u00e1 v\u00e1rios dos c\u00e1lculos j\u00e1 realizados na aproxima\u00e7\u00e3o simplificada, ficando como mostrado no racioc\u00ednio a seguir:<\/p>\n<p style=\"text-align:center;\">\n<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>\n<p>A parte destacada em vermelho corresponde aos elementos adicionais considerados na aproxima\u00e7\u00e3o ordin\u00e1ria, enquanto todo o restante \u00e9 o que j\u00e1 foi obtido na aproxima\u00e7\u00e3o simplificada. A partir disso, temos:<\/p>\n<p style=\"text-align:center;\">\n<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>\n<p>Depois, utilizando a mudan\u00e7a de base para logaritmos, obtemos:<\/p>\n<p style=\"text-align:center;\">\n\t<span dir=\"ltr\"><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><\/span>\n<\/p>\n<p>Finalmente, tomando a exponencial de base 10, chegamos a:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\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]}<\/span>\n<\/p>\n<p>Agora, de forma an\u00e1loga \u00e0 anterior, podemos encontrar o valor m\u00e1ximo deste n\u00famero avaliando com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, que neste caso fornecer\u00e1 o seguinte resultado:<\/p>\n<p style=\"text-align:center;\">\n<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>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Exemplos de C\u00e1lculos Combinat\u00f3rios e de Ordem de Magnitude<\/h2>\n<p><center><br \/>\n\t<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/M7NrtICrSzU?si=6UT34333Wp5YwFU1\" \n\t        title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" \n\t        referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><br \/>\n<\/center><\/p>\n<p><a name=\"7\"><\/a><\/p>\n<h3>Caso 1: Grandes Fatoriais<\/h3>\n<p>Estimemos a ordem de magnitude de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span>, ou seja, a quantidade de d\u00edgitos necess\u00e1ria para escrever esse n\u00famero.<\/p>\n<h5>Solu\u00e7\u00e3o<\/h5>\n<p>Para realizar este c\u00e1lculo, utilizamos a f\u00f3rmula de Stirling da seguinte maneira:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\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}<\/span>\n<\/p>\n<p>A seguir, aplicamos a mudan\u00e7a de base dos logaritmos:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left[ \\left(10^{50}\\right)! \\right] = \\dfrac{\\log\\left[\\left(10^{50}\\right)!\\right]}{\\log{e}}<\/span>\n<\/p>\n<p>Portanto:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\log\\left[ \\left(10^{50}\\right)! \\right] \\approx \\log(e)\\left[50\\ln(10)-1 \\right]10^{50}<\/span>\n<\/p>\n<p>Finalmente, ao aplicar a exponencial de base 10, obtemos:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!  \\approx  10^{\\log(e)\\left[50\\ln(10)-1 \\right]10^{50}} = 10^{49,5657 \\cdot 10^{50}}<\/span>\n<\/p>\n<p>O expoente sobre o 10 representa a ordem de magnitude, fornecendo uma estimativa da quantidade de d\u00edgitos que o n\u00famero <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span> possui.<\/p>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Caso 2: Grandes Combinat\u00f3rias<\/h3>\n<p>Uma casa m\u00e9dia possui aproximadamente 12 interruptores de luz, que podem estar ligados ou desligados. Em m\u00e9dia, cada casa abriga 4 pessoas. Se uma cidade tem 5 milh\u00f5es de habitantes, de quantas formas poss\u00edveis podem estar ligados metade dos interruptores da cidade?<\/p>\n<h5>Solu\u00e7\u00e3o<\/h5>\n<p>O n\u00famero <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> de interruptores totais na cidade \u00e9:<\/p>\n<p style=\"text-align:center;\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\nn &amp;=&amp;\\dfrac{\\text{habitantes na cidade}}{\\text{pessoas por casa}} \\times \\text{interruptores por casa} \\\\ \\\\\n\n&amp;=&amp; \\dfrac{5\\cdot 10^6}{4}\\cdot 12 = 15\\cdot 10^6\n\n\\end{array}<\/span>\n<\/p>\n<p>O macroestado formado por todos os microestados em que metade dos interruptores est\u00e3o ligados coincide com o macroestado que possui o maior n\u00famero de configura\u00e7\u00f5es poss\u00edveis. Denotando esse n\u00famero m\u00e1ximo como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max}<\/span><\/span>, podemos obter as seguintes estimativas conforme cada m\u00e9todo:<\/p>\n<ul>\n<li><strong>Estimativa ordin\u00e1ria:<\/strong> <span dir=\"ltr\"><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><\/span><\/li>\n<li><strong>Estimativa simplificada:<\/strong> <span dir=\"ltr\"><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><\/span><\/li>\n<\/ul>\n<p>Ainda que entre ambas as aproxima\u00e7\u00f5es haja uma diferen\u00e7a pr\u00f3xima a 4 ordens de magnitude (o que pode parecer consider\u00e1vel), na verdade isso n\u00e3o \u00e9 significativo em compara\u00e7\u00e3o com mais de 6,5 milh\u00f5es de ordens de magnitude.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problemas de Combinat\u00f3rias em Termodin\u00e2mica Quantas maneiras existem de organizar um sistema f\u00edsico composto por milh\u00f5es de elementos? Nesta aula, abordaremos como a matem\u00e1tica permite responder a perguntas como essa no contexto da termodin\u00e2mica, desde a distribui\u00e7\u00e3o de quantas de energia em sistemas at\u00f4micos at\u00e9 o c\u00e1lculo de configura\u00e7\u00f5es poss\u00edveis em sistemas de grande escala. [&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":0,"footnotes":""},"categories":[637,921],"tags":[],"class_list":["post-31190","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fisica-pt","category-termodinamica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Problemas de Combinat\u00f3rias em Termodin\u00e2mica - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Nesta aula, revisaremos uma aplica\u00e7\u00e3o de combinat\u00f3ria e a f\u00f3rmula de Stirling em problemas de termodin\u00e2mica com grandes n\u00fameros.\" \/>\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\/pt\/problemas-de-combinatorias-em-termodinamica\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Problemas de Combinat\u00f3rias em Termodin\u00e2mica\" \/>\n<meta property=\"og:description\" content=\"Nesta aula, revisaremos uma aplica\u00e7\u00e3o de combinat\u00f3ria e a f\u00f3rmula de Stirling em problemas de termodin\u00e2mica com grandes n\u00fameros.\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/\" \/>\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=\"2021-03-28T13:00:51+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-01-12T13:03:44+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=\"Problemas de Combinat\u00f3rias em Termodin\u00e2mica\" \/>\n<meta name=\"twitter:description\" content=\"Nesta aula, revisaremos uma aplica\u00e7\u00e3o de combinat\u00f3ria e a f\u00f3rmula de Stirling em problemas de termodin\u00e2mica com grandes n\u00fameros.\" \/>\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=\"7 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Problemas de Combinat\u00f3rias em Termodin\u00e2mica\",\"datePublished\":\"2021-03-28T13:00:51+00:00\",\"dateModified\":\"2025-01-12T13:03:44+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/\"},\"wordCount\":1813,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/12\/PROBLEMASDECOMBINATORIA.jpg\",\"articleSection\":[\"F\u00edsica\",\"Termodin\u00e2mica\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/\",\"url\":\"http:\/\/toposuranos.com\/material\/pt\/problemas-de-combinatorias-em-termodinamica\/\",\"name\":\"Problemas de Combinat\u00f3rias em Termodin\u00e2mica - 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