{"id":187225,"date":"2025-02-03T06:25:41","date_gmt":"2025-02-03T06:25:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187225"},"modified":"2025-02-03T06:25:42","modified_gmt":"2025-02-03T06:25:42","slug":"a-sample-of-neon-gas-ne-has-a-density-of-g-l-at-a-pressure-of-1-13-atm-and-a-temperature-of-57c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/03\/a-sample-of-neon-gas-ne-has-a-density-of-g-l-at-a-pressure-of-1-13-atm-and-a-temperature-of-57c\/","title":{"rendered":"A sample of neon gas, Ne, has a density of g\/L at a pressure of 1.13 atm and a temperature of 57\u00b0C"},"content":{"rendered":"\n<p>A sample of neon gas, Ne, has a density of g\/L at a pressure of 1.13 atm and a temperature of 57\u00b0C?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the density of neon gas ((\\text{Ne})) at a given pressure and temperature, we use the <strong>Ideal Gas Law<\/strong> in its density form:<\/p>\n\n\n\n<p>[<br>\\text{Density} (\\rho) = \\frac{PM}{RT}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( P ) = Pressure = <strong>1.13 atm<\/strong><\/li>\n\n\n\n<li>( M ) = Molar mass of neon = <strong>20.18 g\/mol<\/strong><\/li>\n\n\n\n<li>( R ) = Ideal gas constant = <strong>0.0821 L\u00b7atm\/(mol\u00b7K)<\/strong><\/li>\n\n\n\n<li>( T ) = Temperature in Kelvin = <strong>57\u00b0C + 273.15 = 330.15 K<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Plug in the values<\/h3>\n\n\n\n<p>[<br>\\rho = \\frac{(1.13 \\text{ atm}) (20.18 \\text{ g\/mol})}{(0.0821 \\text{ L\u00b7atm\/(mol\u00b7K)}) (330.15 \\text{ K})}<br>]<\/p>\n\n\n\n<p>[<br>\\rho = \\frac{22.8034}{27.107415}<br>]<\/p>\n\n\n\n<p>[<br>\\rho \\approx 0.841 \\text{ g\/L}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The Ideal Gas Law, ( PV = nRT ), describes the behavior of gases. To find density, we rearrange the equation in terms of mass per unit volume.<\/p>\n\n\n\n<p>Since ( n = \\frac{m}{M} ), replacing in ( PV = nRT ):<\/p>\n\n\n\n<p>[<br>P V = \\frac{m}{M} RT<br>]<\/p>\n\n\n\n<p>Dividing both sides by ( V ):<\/p>\n\n\n\n<p>[<br>P = \\frac{m}{V} \\frac{RT}{M}<br>]<\/p>\n\n\n\n<p>Since density ( \\rho = \\frac{m}{V} ), the equation simplifies to:<\/p>\n\n\n\n<p>[<br>\\rho = \\frac{PM}{RT}<br>]<\/p>\n\n\n\n<p>Using this equation, we calculated that neon gas at 1.13 atm and 57\u00b0C has a density of <strong>0.841 g\/L<\/strong>.<\/p>\n\n\n\n<p>This result aligns with the expected behavior of gases\u2014higher temperature reduces density because the gas expands, while higher pressure increases density by compressing the gas.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A sample of neon gas, Ne, has a density of g\/L at a pressure of 1.13 atm and a temperature of 57\u00b0C? The Correct Answer and Explanation is : To determine the density of neon gas ((\\text{Ne})) at a given pressure and temperature, we use the Ideal Gas Law in its density form: [\\text{Density} (\\rho) [&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-187225","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187225","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=187225"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187225\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187225"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187225"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187225"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}