{"id":229546,"date":"2025-06-08T09:19:08","date_gmt":"2025-06-08T09:19:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229546"},"modified":"2025-06-08T09:19:11","modified_gmt":"2025-06-08T09:19:11","slug":"movable-chamber-has-a-volume-of-18-5-l-at-a-temperature-of-18-5-c-assuming-no-gas-escapes-and-the-pressure-remains-constant-what-is-the-temperature-when-the-chamber-has-a-volume-of-19-8-l","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/movable-chamber-has-a-volume-of-18-5-l-at-a-temperature-of-18-5-c-assuming-no-gas-escapes-and-the-pressure-remains-constant-what-is-the-temperature-when-the-chamber-has-a-volume-of-19-8-l\/","title":{"rendered":"movable chamber has a volume of 18.5 L (at a temperature of 18.5 C) Assuming no gas escapes and the pressure remains constant; what is the temperature when the chamber has a volume of 19.8 L"},"content":{"rendered":"\n<p>movable chamber has a volume of 18.5 L (at a temperature of 18.5 C) Assuming no gas escapes and the pressure remains constant; what is the temperature when the chamber has a volume of 19.8 L? Your answer should have 3 significant digits: Do not include units in your answer: Enter the answer&#8217;<\/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 use <strong>Charles&#8217;s Law<\/strong>, which states:V1T1=V2T2\\frac{V_1}{T_1} = \\frac{V_2}{T_2}T1\u200bV1\u200b\u200b=T2\u200bV2\u200b\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>V1=18.5\u2009LV_1 = 18.5 \\, \\text{L}V1\u200b=18.5L (initial volume)<\/li>\n\n\n\n<li>T1=18.5\u2218C=291.65\u2009KT_1 = 18.5^\\circ C = 291.65 \\, \\text{K}T1\u200b=18.5\u2218C=291.65K (initial temperature in Kelvin)<\/li>\n\n\n\n<li>V2=19.8\u2009LV_2 = 19.8 \\, \\text{L}V2\u200b=19.8L (final volume)<\/li>\n\n\n\n<li>T2=?T_2 = ?T2\u200b=? (final temperature in Kelvin)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step solution:<\/h3>\n\n\n\n<p>Rewriting Charles\u2019s Law to solve for T2T_2T2\u200b:T2=V2\u22c5T1V1T_2 = \\frac{V_2 \\cdot T_1}{V_1}T2\u200b=V1\u200bV2\u200b\u22c5T1\u200b\u200bT2=19.8\u22c5291.6518.5T_2 = \\frac{19.8 \\cdot 291.65}{18.5}T2\u200b=18.519.8\u22c5291.65\u200bT2=5774.6718.5\u2248312.69\u2009KT_2 = \\frac{5774.67}{18.5} \\approx 312.69 \\, \\text{K}T2\u200b=18.55774.67\u200b\u2248312.69K<\/p>\n\n\n\n<p>Convert back to Celsius:T2\u2218C=312.69\u2212273.15=39.5T_2^\\circ C = 312.69 &#8211; 273.15 = \\boxed{39.5}T2\u2218\u200bC=312.69\u2212273.15=39.5\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer (3 significant digits, <strong>no units<\/strong>):<\/h3>\n\n\n\n<p><strong>39.5<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 Explanation <\/h3>\n\n\n\n<p>This problem involves understanding how gases behave when temperature and volume change under <strong>constant pressure<\/strong>. Charles\u2019s Law describes this relationship: the volume of a gas is <strong>directly proportional<\/strong> to its absolute temperature when pressure is constant. That means, if temperature increases, volume increases \u2014 and vice versa \u2014 but <strong>only if temperature is measured in Kelvin<\/strong>, not Celsius.<\/p>\n\n\n\n<p>First, we convert the given Celsius temperature to Kelvin. This is essential because gas laws are based on absolute temperature scales. We add 273.15 to 18.5\u00b0C, giving us 291.65 K.<\/p>\n\n\n\n<p>Next, we apply Charles\u2019s Law:V1T1=V2T2\\frac{V_1}{T_1} = \\frac{V_2}{T_2}T1\u200bV1\u200b\u200b=T2\u200bV2\u200b\u200b<\/p>\n\n\n\n<p>This formula can be rearranged to find the unknown temperature T2T_2T2\u200b, which represents the final temperature when the gas expands to 19.8 L.<\/p>\n\n\n\n<p>By substituting the known values into the equation:T2=19.8\u00d7291.6518.5T_2 = \\frac{19.8 \\times 291.65}{18.5}T2\u200b=18.519.8\u00d7291.65\u200b<\/p>\n\n\n\n<p>We calculate T2T_2T2\u200b to be approximately 312.69 K. We then convert this back to Celsius by subtracting 273.15, yielding <strong>39.5\u00b0C<\/strong>.<\/p>\n\n\n\n<p>The significant digits are based on the precision of the input values (3 significant digits), so the final answer must match this precision. Therefore, the correct answer is <strong>39.5<\/strong>, with no units included as instructed.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner6-135.jpeg\" alt=\"\" class=\"wp-image-229547\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>movable chamber has a volume of 18.5 L (at a temperature of 18.5 C) Assuming no gas escapes and the pressure remains constant; what is the temperature when the chamber has a volume of 19.8 L? Your answer should have 3 significant digits: Do not include units in your answer: Enter the answer&#8217; The Correct [&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-229546","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229546","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=229546"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229546\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229546"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229546"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229546"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}