{"id":7171,"date":"2011-11-15T17:51:46","date_gmt":"2011-11-15T17:51:46","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7171"},"modified":"2022-01-15T21:26:02","modified_gmt":"2022-01-15T21:26:02","slug":"volume-e-pressao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7171","title":{"rendered":"Volume e press\u00e3o"},"content":{"rendered":"<p><ul id='GTTabs_ul_7171' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7171' class='GTTabs_curr'><a  id=\"7171_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7171' ><a  id=\"7171_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_7171'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Sabemos que, se exercermos press\u00e3o sobre o \u00eambolo de uma seringa tapando o orif\u00edcio com o dedo de modo a n\u00e3o deixar sair o ar, o volume diminui \u00e0 medida que a press\u00e3o aumenta.<\/p>\n<p>\u00c0 temperatura de 0 \u00baC registaram-se os seguintes valores:<\/p>\n<table class=\" aligncenter\" style=\"width: 60%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">Press\u00e3o $p$ (em atmosferas)<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">224<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">89,6<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">44,8<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">22,4<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">11,2<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">5,6<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">2,24<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">0,448<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">Volume $v$ (em litros)<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">0,1<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">0,25<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">0,5<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">1<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">2<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">4<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">10<\/td>\n<td style=\"text-align: center; border: #8b008b 1px solid;\">50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>Que acontece ao volume se a press\u00e3o aumenta? E se a press\u00e3o diminui?<\/li>\n<li>O volume, \u00e0 temperatura de 0 \u00baC, \u00e9 inversamente proporcional \u00e0 press\u00e3o sobre ele exercida? Porqu\u00ea?<\/li>\n<li>Escreve uma express\u00e3o que permita obter $v$ em fun\u00e7\u00e3o de $p$.<\/li>\n<li>Determina:<br \/>\n(i) o volume de ar correspondente a uma press\u00e3o de 50 atmosferas;<br \/>\n(ii) a press\u00e3o correspondente a um volume de g\u00e1s de 2,5 litros.<\/li>\n<li>Representa graficamente $v$ em fun\u00e7\u00e3o de $p$.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7171' onClick='GTTabs_show(1,7171)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7171'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Se a press\u00e3o aumenta, o volume diminui; se a press\u00e3o diminui, o volume aumenta.<br \/>\n\u00ad<\/li>\n<li>O produto dos valores correspondentes das duas grandezas \u00e9 constante, logo, \u00e0 temperatura de 0 \u00baC, as duas grandezas s\u00e3o inversamente proporcionais:$$224\\times 0,1=89,6\\times 0,25=44,8\\times 0,5=22,4\\times 1=11,2\\times 2=5,6\\times 4=2,24\\times 10=0,448\\times 50=22,4$$<\/li>\n<li>Como $v\\times p=22,4$ (al\u00ednea anterior), ent\u00e3o: $$v=\\frac{22,4}{p}$$<\/li>\n<li>(i) Pretende-se $v=?$, sabendo que $p=50$.\n<p>Substituindo, o valor conhecido na express\u00e3o anterior, temos: $$v=\\frac{22,4}{50}\\Leftrightarrow v=0,448$$<br \/>\nPara uma press\u00e3o de 50 atmosferas, \u00e9 de $0,448$ litros o volume de ar correspondente.<\/p>\n<p>\u00ad<\/p>\n<p>(ii) Pretende-se $p=?$, sabendo que $v=2,5$.<\/p>\n<p>Substituindo, temos: \\[2,5=\\frac{22,4}{p}\\Leftrightarrow p=\\frac{22,4}{2,5}\\Leftrightarrow p=8,96\\]<br \/>\nPara um volume de g\u00e1s de 2,5 litros, a press\u00e3o correspondente \u00e9 8,96 atmosferas.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Como os valores da tabela variam entre valores pequenos e valores relativamente elevados, o gr\u00e1fico \u00e9 de dif\u00edcil representa\u00e7\u00e3o e leitura. 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