{"id":246707,"date":"2025-07-07T10:56:25","date_gmt":"2025-07-07T10:56:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246707"},"modified":"2025-07-07T10:56:27","modified_gmt":"2025-07-07T10:56:27","slug":"io-10-0-l-of-a-gas-is-found-t0-exert-97-0-kpa-at-25-0-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/io-10-0-l-of-a-gas-is-found-t0-exert-97-0-kpa-at-25-0-c\/","title":{"rendered":"IO: 10.0 L of a gas is found t0 exert 97.0 kPa at 25.0 %C"},"content":{"rendered":"\n<p>IO: 10.0 L of a gas is found t0 exert 97.0 kPa at 25.0 %C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?&#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 problem, we can use <strong>Gay-Lussac&#8217;s Law<\/strong>, which relates the pressure and temperature of a gas when the volume and amount of gas remain constant. The formula is:P1\/T1=P2\/T2P_1\/T_1 = P_2\/T_2P1\u200b\/T1\u200b=P2\u200b\/T2\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P1P_1P1\u200b is the initial pressure (97.0 kPa),<\/li>\n\n\n\n<li>T1T_1T1\u200b is the initial temperature in Kelvin,<\/li>\n\n\n\n<li>P2P_2P2\u200b is the final pressure (standard pressure = 101.3 kPa),<\/li>\n\n\n\n<li>T2T_2T2\u200b is the final temperature in Kelvin, which we need to find.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert the initial temperature to Kelvin<\/h3>\n\n\n\n<p>The initial temperature is given as 25.0\u00b0C. To convert to Kelvin, add 273.15:T1=25.0\u00b0C+273.15=298.15\u2009KT_1 = 25.0\u00b0C + 273.15 = 298.15 \\, \\text{K}T1\u200b=25.0\u00b0C+273.15=298.15K<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Set up the equation using Gay-Lussac\u2019s Law<\/h3>\n\n\n\n<p>Using the formula P1\/T1=P2\/T2P_1\/T_1 = P_2\/T_2P1\u200b\/T1\u200b=P2\u200b\/T2\u200b, we substitute the known values:97.0\u2009kPa298.15\u2009K=101.3\u2009kPaT2\\frac{97.0 \\, \\text{kPa}}{298.15 \\, \\text{K}} = \\frac{101.3 \\, \\text{kPa}}{T_2}298.15K97.0kPa\u200b=T2\u200b101.3kPa\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Solve for T2T_2T2\u200b<\/h3>\n\n\n\n<p>Rearranging the equation to solve for T2T_2T2\u200b:T2=101.3\u2009kPa\u00d7298.15\u2009K97.0\u2009kPaT_2 = \\frac{101.3 \\, \\text{kPa} \\times 298.15 \\, \\text{K}}{97.0 \\, \\text{kPa}}T2\u200b=97.0kPa101.3kPa\u00d7298.15K\u200bT2=30272.39597.0\u2248312.7\u2009KT_2 = \\frac{30272.395}{97.0} \\approx 312.7 \\, \\text{K}T2\u200b=97.030272.395\u200b\u2248312.7K<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Convert the final temperature back to Celsius<\/h3>\n\n\n\n<p>To convert the final temperature back to Celsius:T2=312.7\u2009K\u2212273.15=39.5\u00b0CT_2 = 312.7 \\, \\text{K} &#8211; 273.15 = 39.5\u00b0CT2\u200b=312.7K\u2212273.15=39.5\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The required temperature to change the pressure to standard pressure is <strong>39.5\u00b0C<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>This problem illustrates how the pressure of a gas is directly proportional to its temperature when the volume is constant. By using Gay-Lussac\u2019s Law, we can predict that increasing the temperature of the gas will increase its pressure, and conversely, decreasing the temperature will lower the pressure. In this case, to reach standard pressure (101.3 kPa), the temperature needs to increase from 25\u00b0C to 39.5\u00b0C. This demonstrates the relationship between temperature and pressure in gases, assuming no volume change.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-922.jpeg\" alt=\"\" class=\"wp-image-246708\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>IO: 10.0 L of a gas is found t0 exert 97.0 kPa at 25.0 %C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?&#8217; The Correct Answer and Explanation is: To solve this problem, we can use Gay-Lussac&#8217;s Law, which relates the pressure and temperature of a gas when [&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-246707","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246707","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=246707"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246707\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246707"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246707"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}