{"id":20727,"date":"2017-03-29T21:57:31","date_gmt":"2017-03-29T21:57:31","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=20727"},"modified":"2017-03-29T22:03:32","modified_gmt":"2017-03-29T22:03:32","slug":"chemistry-grahams-law-of-effusion","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/chemistry-grahams-law-of-effusion\/","title":{"rendered":"Chemistry:  Graham&#8217;s Law of Effusion"},"content":{"rendered":"<h1>Tutoring chemistry, you may mention kinetic energy of particles, diffusion, and effusion.  The tutor gives a brief explanation.<\/h1>\n<p>In my <a href=\"?p=19657\">post from Jan 18, 2017,<\/a> I define <em>effuse<\/em>:  it means to escape from a container through a porous boundary.<\/p>\n<p>Graham&#8217;s Law of Effusion compares the rate at which two different gases will effuse, based on their comparative molecular masses.  If e<sub>A<\/sub> is the effusion rate of gas A, and e<sub>B<\/sub> that of gas B, then<\/p>\n<p>e<sub>A<\/sub>\/e<sub>B<\/sub> = (MM<sub>B<\/sub>\/MM<sub>A<\/sub>)<sup>1\/2<\/sup><\/p>\n<p>where MM<sub>A<\/sub> is molecular mass of gas A, and MM<sub>B<\/sub>, that of gas B.<\/p>\n<p>The reasoning behind the formula is that effusion depends on molecular motion, which is quantified by kinetic energy, KE:<\/p>\n<p>KE = 0.5MMv<sup>2<\/sup><\/p>\n<p>where, once again, MM means molecular mass, while v means velocity.<\/p>\n<p>Two gases of the same temperature have the same kinetic energy:<\/p>\n<p>0.5MM<sub>A<\/sub>v<sub>A<\/sub><sup>2<\/sup>=0.5MM<sub>B<\/sub>v<sub>B<\/sub><sup>2<\/sup><\/p>\n<p>Dividing both sides by 0.5MM<sub>A<\/sub>v<sub>B<\/sub><sup>2<\/sup> gives<\/p>\n<p>v<sub>A<\/sub><sup>2<\/sup>\/v<sub>B<\/sub><sup>2<\/sup> = MM<sub>B<\/sub>\/MM<sub>A<\/sub><\/p>\n<p>square rooting both sides gives<\/p>\n<p>v<sub>A<\/sub>\/v<sub>B<\/sub> = (MM<sub>B<\/sub>\/MM<sub>A<\/sub>)<sup>1\/2<\/sup><\/p>\n<p>Since effusion is motion through pores, the ratio of velocities is the ratio of effusion:<\/p>\n<p>e<sub>A<\/sub>\/e<sub>B<\/sub>=v<sub>A<\/sub>\/v<sub>B<\/sub> = (MM<sub>B<\/sub>\/MM<sub>A<\/sub>)<sup>1\/2<\/sup><\/p>\n<p>Example: Compare the effusion rate of methane, CH<sub>4<\/sub>, with that of propane, C<sub>3<\/sub>H<sub>8<\/sub>.<\/p>\n<p>Solution:  the ratio of effusion should be<\/p>\n<p>e<sub>A<\/sub>\/e<sub>B<\/sub>=(MM<sub>B<\/sub>\/MM<sub>A<\/sub>)<sup>1\/2<\/sup><\/p>\n<p>Note that methane, CH<sub>4<\/sub>, has MM=16, while propane, C<sub>3<\/sub>H<sub>8<\/sub>, has MM=44.  Therefore, <\/p>\n<p>e<sub>methane<\/sub>\/e<sub>propane<\/sub>=(44\/16)<sup>1\/2<\/sup>=1.66<\/p>\n<p>Methane should escape from a porous container 1.66 times the rate that propane escapes.<\/p>\n<p>Note that, in a more general sense, this is a law of <em>diffusion<\/em>.  Therefore, if propane and methane are released at one end of a room, methane should reach the other end 1.66 times as quickly as propane.<\/p>\n<p>HTH:)<\/p>\n<p>Source:<\/p>\n<p>Mortimer, Charles E.  <u>Chemistry<\/u>, sixth ed.  Belmont:  Wadsworth, 1986.<\/p>\n<p>White, J. Edmund.  <u>Physical Chemistry<\/u>, College Outline Series.  New York:  Harcourt Brace Jovanovich, 1987.<\/p>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Tutoring chemistry, you may mention kinetic energy of particles, diffusion, and effusion. The tutor gives a brief explanation. In my post from Jan 18, 2017, I define effuse: it means to escape from a container through a porous boundary. Graham&#8217;s &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/chemistry-grahams-law-of-effusion\/\"> <span class=\"screen-reader-text\">Chemistry:  Graham&#8217;s Law of Effusion<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[2204,2205,2206],"class_list":["post-20727","post","type-post","status-publish","format-standard","hentry","category-chemistry","tag-grahams-law-of-effusion","tag-kinetic-energy-vs-molar-mass","tag-kinetic-energy-vs-molecular-mass"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/20727","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=20727"}],"version-history":[{"count":33,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/20727\/revisions"}],"predecessor-version":[{"id":20760,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/20727\/revisions\/20760"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=20727"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=20727"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=20727"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}