{"id":246710,"date":"2025-07-07T10:58:28","date_gmt":"2025-07-07T10:58:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246710"},"modified":"2025-07-07T10:58:30","modified_gmt":"2025-07-07T10:58:30","slug":"io-10-0-l-of-a-gas-is-found-t0-exert-97-0-kpa-at-25-0-c-2","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-2\/","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?<\/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 <strong>Gay-Lussac&#8217;s Law<\/strong>, which states that for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its temperature in Kelvin. The formula for this law 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 = initial pressure (97.0 kPa)<\/li>\n\n\n\n<li>T1T_1T1\u200b = initial temperature in Kelvin (which we will convert from Celsius)<\/li>\n\n\n\n<li>P2P_2P2\u200b = final pressure (standard pressure = 101.3 kPa)<\/li>\n\n\n\n<li>T2T_2T2\u200b = final temperature in Kelvin (this is what we are solving for)<\/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 temperature in Kelvin can be calculated from Celsius using the formula:T(K)=T(C)+273.15T(K) = T(C) + 273.15T(K)=T(C)+273.15<\/p>\n\n\n\n<p>Given that the initial temperature is 25.0\u00b0C:T1=25.0+273.15=298.15\u2009KT_1 = 25.0 + 273.15 = 298.15 \\, KT1\u200b=25.0+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>We are given the following:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P1=97.0\u2009kPaP_1 = 97.0 \\, kPaP1\u200b=97.0kPa<\/li>\n\n\n\n<li>T1=298.15\u2009KT_1 = 298.15 \\, KT1\u200b=298.15K<\/li>\n\n\n\n<li>P2=101.3\u2009kPaP_2 = 101.3 \\, kPaP2\u200b=101.3kPa<\/li>\n<\/ul>\n\n\n\n<p>We need to find T2T_2T2\u200b. Using the equation:P1\/T1=P2\/T2P_1 \/ T_1 = P_2 \/ T_2P1\u200b\/T1\u200b=P2\u200b\/T2\u200b<\/p>\n\n\n\n<p>Rearrange the formula to solve for T2T_2T2\u200b:T2=(P2\u00d7T1)\/P1T_2 = (P_2 \\times T_1) \/ P_1T2\u200b=(P2\u200b\u00d7T1\u200b)\/P1\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Plug in the values<\/h3>\n\n\n\n<p>T2=(101.3\u2009kPa)\u00d7(298.15\u2009K)97.0\u2009kPaT_2 = \\frac{(101.3 \\, kPa) \\times (298.15 \\, K)}{97.0 \\, kPa}T2\u200b=97.0kPa(101.3kPa)\u00d7(298.15K)\u200bT2=30239.09597.0=311.5\u2009KT_2 = \\frac{30239.095}{97.0} = 311.5 \\, KT2\u200b=97.030239.095\u200b=311.5K<\/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 T2T_2T2\u200b from Kelvin to Celsius, we subtract 273.15:T2=311.5\u2009K\u2212273.15=38.35\u00b0CT_2 = 311.5 \\, K &#8211; 273.15 = 38.35\u00b0CT2\u200b=311.5K\u2212273.15=38.35\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>38.35\u00b0C<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>We applied Gay-Lussac&#8217;s Law to find the relationship between temperature and pressure. The key is understanding that when pressure increases, temperature must increase as well, assuming the volume remains constant. The conversion between Kelvin and Celsius is essential to properly apply the law and determine the change in temperature that would result in the standard pressure condition.<\/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-banner5-923.jpeg\" alt=\"\" class=\"wp-image-246711\"\/><\/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? The Correct Answer and Explanation is: To solve this problem, we will use Gay-Lussac&#8217;s Law, which states that for a given amount of gas at [&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-246710","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246710","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=246710"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246710\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246710"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246710"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}