{"id":213762,"date":"2025-05-12T10:06:12","date_gmt":"2025-05-12T10:06:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213762"},"modified":"2025-05-12T10:06:14","modified_gmt":"2025-05-12T10:06:14","slug":"consider-an-ideal-gas-enclosed-in-a-1-00-l-container-at-an-internal-pressure-of-24-0-atm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/12\/consider-an-ideal-gas-enclosed-in-a-1-00-l-container-at-an-internal-pressure-of-24-0-atm\/","title":{"rendered":"Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 24.0 atm"},"content":{"rendered":"\n<p>Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 24.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 24.0L W = -55931. Now calculate the work done if this process is carried out in two steps. 1. First, let the gas expand against a constant external pressure of 1.50 atm to a volume of 16.0L. 2. From the end point of step 1, let the gas expand to 24.0 L against a constant external pressure of 1.00 atm. W = -55931. Now calculate the work done if this process is carried out in two steps. 1. First, let the gas expand against a constant external pressure of 1.50 atm to a volume of 16.01. 2. From the end point of step 1, let the gas expand to 24.0L against a constant external pressure of 1.00 atm. -57146.9<\/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, we calculate the <strong>work (w)<\/strong> done during each step of the gas expansion using the formula for work done by a gas against constant external pressure: w=\u2212Pext\u0394Vw = -P_{\\text{ext}} \\Delta V<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>ww is work (in L\u00b7atm, can be converted to joules)<\/li>\n\n\n\n<li>PextP_{\\text{ext}} is the external pressure (in atm)<\/li>\n\n\n\n<li>\u0394V=Vf\u2212Vi\\Delta V = V_f &#8211; V_i is the change in volume (in L)<\/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-by-Step Work Calculation<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 1: Expansion from 1.00 L to 16.0 L against 1.50 atm<\/strong><\/h4>\n\n\n\n<p>\u0394V1=16.0\u2009L\u22121.00\u2009L=15.0\u2009L\\Delta V_1 = 16.0\\,\\text{L} &#8211; 1.00\\,\\text{L} = 15.0\\,\\text{L} w1=\u22121.50\u2009atm\u00d715.0\u2009L=\u221222.5\u2009L\\cdotpatmw_1 = -1.50\\,\\text{atm} \\times 15.0\\,\\text{L} = -22.5\\,\\text{L\u00b7atm}<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 2: Expansion from 16.0 L to 24.0 L against 1.00 atm<\/strong><\/h4>\n\n\n\n<p>\u0394V2=24.0\u2009L\u221216.0\u2009L=8.0\u2009L\\Delta V_2 = 24.0\\,\\text{L} &#8211; 16.0\\,\\text{L} = 8.0\\,\\text{L} w2=\u22121.00\u2009atm\u00d78.0\u2009L=\u22128.0\u2009L\\cdotpatmw_2 = -1.00\\,\\text{atm} \\times 8.0\\,\\text{L} = -8.0\\,\\text{L\u00b7atm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Total Work (Two-Step Process)<\/strong><\/h3>\n\n\n\n<p>wtotal=w1+w2=\u221222.5\u2009L\\cdotpatm+(\u22128.0\u2009L\\cdotpatm)=\u221230.5\u2009L\\cdotpatmw_{\\text{total}} = w_1 + w_2 = -22.5\\,\\text{L\u00b7atm} + (-8.0\\,\\text{L\u00b7atm}) = -30.5\\,\\text{L\u00b7atm}<\/p>\n\n\n\n<p>Convert to joules (1 L\u00b7atm = 101.325 J): w=\u221230.5\u00d7101.325=\u22123090.56\u2009Jw = -30.5 \\times 101.325 = \\boxed{-3090.56\\,\\text{J}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Correct Answer Explanation (300+ words)<\/strong><\/h3>\n\n\n\n<p>The work done by an expanding gas depends heavily on the external pressure it expands against. In this problem, an ideal gas initially confined to a volume of 1.00 L at high pressure expands in a two-step process. In each step, the gas expands against a <strong>constant external pressure<\/strong>, which allows us to use the equation w=\u2212Pext\u0394Vw = -P_{\\text{ext}} \\Delta V.<\/p>\n\n\n\n<p>In the <strong>first step<\/strong>, the gas expands from 1.00 L to 16.0 L against an external pressure of 1.50 atm. The change in volume is 15.0 L. Multiplying this by the external pressure (and applying a negative sign, as expansion does work by the gas), the gas performs \u201322.5 L\u00b7atm of work.<\/p>\n\n\n\n<p>In the <strong>second step<\/strong>, the gas continues to expand from 16.0 L to 24.0 L, but now against a lower pressure of 1.00 atm. This change in volume (8.0 L) leads to an additional \u20138.0 L\u00b7atm of work.<\/p>\n\n\n\n<p>Adding the two contributions gives a total work of \u201330.5 L\u00b7atm. To express this in SI units (joules), we use the conversion factor 1 L\u00b7atm = 101.325 J. Thus, the total work is: \u221230.5\u00d7101.325=\u22123090.56\u2009J-30.5 \\times 101.325 = -3090.56 \\,\\text{J}<\/p>\n\n\n\n<p>This is <strong>less negative<\/strong> than the single-step expansion against 1.00 atm from 1.00 L to 24.0 L, which gives: w=\u22121.00\u00d7(24.0\u22121.00)=\u221223.0\u2009L\\cdotpatm=\u22122320.48\u2009Jw = -1.00 \\times (24.0 &#8211; 1.00) = -23.0\\,\\text{L\u00b7atm} = -2320.48\\,\\text{J}<\/p>\n\n\n\n<p>So why is the <strong>total work in the two-step process greater in magnitude<\/strong> than the single-step? Because in step 1, the gas expands partway against <strong>higher pressure<\/strong>, meaning <strong>more work is done<\/strong> than if the entire expansion occurred against the lower pressure.<\/p>\n\n\n\n<p>This illustrates how <strong>work depends on the external pressure path<\/strong>, not just the initial and final states\u2014a key point in thermodynamics. Would you like a graph of this pressure-volume path to visualize the difference?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 24.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 24.0L W = -55931. Now calculate the work done if this process is carried out in two [&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-213762","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213762","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=213762"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213762\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213762"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213762"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213762"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}