{"id":218209,"date":"2025-05-23T11:35:13","date_gmt":"2025-05-23T11:35:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218209"},"modified":"2025-05-23T11:35:16","modified_gmt":"2025-05-23T11:35:16","slug":"a-2-liter-dumas-bulb-contains-n-moles-of-nitrogen-at-0-5-atm-pressure-and-at-tk-on-addition-of-0-01-moles-of-oxygen","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/a-2-liter-dumas-bulb-contains-n-moles-of-nitrogen-at-0-5-atm-pressure-and-at-tk-on-addition-of-0-01-moles-of-oxygen\/","title":{"rendered":"A 2-liter Dumas bulb contains n moles of nitrogen at 0.5 atm pressure and at T\u00b0K. On addition of 0.01 moles of oxygen,"},"content":{"rendered":"\n<p>A 2-liter Dumas bulb contains n moles of nitrogen at 0.5 atm pressure and at T\u00b0K. On addition of 0.01 moles of oxygen, it is necessary to cool the bulb to a temperature of 10\u00b0C in order to maintain the same pressure. Calculate n and T<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we will use the <strong>Ideal Gas Law<\/strong>:<\/p>\n\n\n\n<p>$$<br>PV = nRT<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$P$ is the pressure (in atm),<\/li>\n\n\n\n<li>$V$ is the volume (in liters),<\/li>\n\n\n\n<li>$n$ is the number of moles of gas,<\/li>\n\n\n\n<li>$R$ is the ideal gas constant = 0.0821 L\u00b7atm\/mol\u00b7K,<\/li>\n\n\n\n<li>$T$ is the temperature in Kelvin (K).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Initial Condition (Nitrogen only)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume $V = 2$ L<\/li>\n\n\n\n<li>Pressure $P = 0.5$ atm<\/li>\n\n\n\n<li>Moles of nitrogen $= n$<\/li>\n\n\n\n<li>Temperature $= T$ K<\/li>\n<\/ul>\n\n\n\n<p>Using the Ideal Gas Law:<\/p>\n\n\n\n<p>$$<br>0.5 \\cdot 2 = n \\cdot 0.0821 \\cdot T \\quad \\text{(1)}<br>$$<\/p>\n\n\n\n<p>$$<br>1 = 0.0821nT \\quad \\text{(1)}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: After Adding Oxygen<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>0.01 moles of O\u2082 are added \u2192 Total moles = $n + 0.01$<\/li>\n\n\n\n<li>Volume $= 2$ L<\/li>\n\n\n\n<li>Pressure remains $0.5$ atm<\/li>\n\n\n\n<li>Final temperature $= 10^\\circ \\text{C} = 283 \\text{ K}$<\/li>\n<\/ul>\n\n\n\n<p>Again using the Ideal Gas Law:<\/p>\n\n\n\n<p>$$<br>0.5 \\cdot 2 = (n + 0.01) \\cdot 0.0821 \\cdot 283 \\quad \\text{(2)}<br>$$<\/p>\n\n\n\n<p>$$<br>1 = (n + 0.01) \\cdot 23.2343 \\quad \\text{(2)}<br>$$<\/p>\n\n\n\n<p>Solving equation (2):<\/p>\n\n\n\n<p>$$<br>n + 0.01 = \\frac{1}{23.2343} \\approx 0.043<br>$$<\/p>\n\n\n\n<p>$$<br>n = 0.043 &#8211; 0.01 = 0.033 \\text{ moles}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Find Initial Temperature $T$<\/strong><\/h3>\n\n\n\n<p>Plug $n = 0.033$ into equation (1):<\/p>\n\n\n\n<p>$$<br>1 = 0.0821 \\cdot 0.033 \\cdot T<br>$$<\/p>\n\n\n\n<p>$$<br>T = \\frac{1}{0.0821 \\cdot 0.033} \\approx 369.4 \\text{ K}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$n = 0.033 \\text{ moles}$<\/li>\n\n\n\n<li>$T = 369.4 \\text{ K}$<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><\/h3>\n\n\n\n<p>This problem revolves around the <strong>ideal gas law<\/strong>, which relates pressure, volume, temperature, and the amount of gas. Initially, a Dumas bulb contains nitrogen gas, and we are given its volume, pressure, and unknown temperature and moles. We apply the ideal gas law to set up an equation in terms of $n$ and $T$.<\/p>\n\n\n\n<p>Upon adding 0.01 moles of oxygen, the number of moles increases. To maintain the same pressure in the same volume, the system must adjust its temperature. We&#8217;re told that to keep pressure constant at 0.5 atm, the bulb must be cooled to 10\u00b0C (or 283 K). Using this new condition, we substitute the known values into the ideal gas law again\u2014this time with $n + 0.01$ moles and the known temperature.<\/p>\n\n\n\n<p>This second equation gives us the total number of moles after the oxygen addition. Subtracting the 0.01 moles of added oxygen gives us the original moles of nitrogen. With that value of $n$, we substitute back into the first equation to solve for the original temperature $T$.<\/p>\n\n\n\n<p>The result\u20140.033 moles of nitrogen at approximately 369.4 K\u2014shows that the addition of oxygen increases the total gas amount. To maintain constant pressure, the temperature must be reduced, demonstrating a key principle of gas behavior: <strong>if the amount of gas increases and volume and pressure are constant, temperature must decrease<\/strong>.<\/p>\n\n\n\n<p>This is a classic example of how interdependent gas variables are and how we can manipulate one to maintain equilibrium when another changes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-176.png\" alt=\"\" class=\"wp-image-218210\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A 2-liter Dumas bulb contains n moles of nitrogen at 0.5 atm pressure and at T\u00b0K. On addition of 0.01 moles of oxygen, it is necessary to cool the bulb to a temperature of 10\u00b0C in order to maintain the same pressure. Calculate n and T The Correct Answer and Explanation is: To solve this [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-218209","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218209","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=218209"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218209\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218209"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218209"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218209"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}