{"id":219850,"date":"2025-05-27T05:38:37","date_gmt":"2025-05-27T05:38:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219850"},"modified":"2025-05-27T05:38:39","modified_gmt":"2025-05-27T05:38:39","slug":"what-is-the-heat-of-vaporization-of-ch4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/what-is-the-heat-of-vaporization-of-ch4\/","title":{"rendered":"What is the heat of vaporization of CH4"},"content":{"rendered":"\n

Methane (CH4) has a boiling point of -164 C at 1 atm and a vapor pressure of 42.8 atm at -100 C. What is the heat of vaporization of CH4?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the heat of vaporization<\/strong> (\u0394Hvap) of methane (CH\u2084), we use the Clausius-Clapeyron equation<\/strong>, which relates the vapor pressure and temperature to the heat of vaporization: ln\u2061(P2P1)=\u2212\u0394HvapR(1T2\u22121T1)\\ln\\left(\\frac{P_2}{P_1}\\right) = \\frac{-\\Delta H_{\\text{vap}}}{R} \\left(\\frac{1}{T_2} – \\frac{1}{T_1}\\right)<\/p>\n\n\n\n

Given:<\/h3>\n\n\n\n
    \n
  • P1=1.00\u00a0atmP_1 = 1.00\\ \\text{atm} at T1=\u2212164\u2218C=109\u00a0KT_1 = -164^\\circ C = 109\\ \\text{K}<\/li>\n\n\n\n
  • P2=42.8\u00a0atmP_2 = 42.8\\ \\text{atm} at T2=\u2212100\u2218C=173\u00a0KT_2 = -100^\\circ C = 173\\ \\text{K}<\/li>\n\n\n\n
  • R=8.314\u00a0J\/mol\\cdotpKR = 8.314\\ \\text{J\/mol\u00b7K}<\/li>\n<\/ul>\n\n\n\n

    Step-by-step calculation:<\/h3>\n\n\n\n
      \n
    1. Convert temperatures to Kelvin<\/strong>: T1=\u2212164+273=109\u00a0K,T2=\u2212100+273=173\u00a0KT_1 = -164 + 273 = 109\\ \\text{K},\\quad T_2 = -100 + 273 = 173\\ \\text{K}<\/li>\n\n\n\n
    2. Plug into the Clausius-Clapeyron equation:<\/li>\n<\/ol>\n\n\n\n

      ln\u2061(42.81.00)=\u2212\u0394Hvap8.314(1173\u22121109)\\ln\\left(\\frac{42.8}{1.00}\\right) = \\frac{-\\Delta H_{\\text{vap}}}{8.314} \\left(\\frac{1}{173} – \\frac{1}{109}\\right) ln\u2061(42.8)\u22483.755\\ln(42.8) \\approx 3.755 (1173\u22121109)=0.00578\u22120.00917=\u22120.00339 K\u22121\\left(\\frac{1}{173} – \\frac{1}{109}\\right) = 0.00578 – 0.00917 = -0.00339\\ \\text{K}^{-1} 3.755=\u2212\u0394Hvap8.314\u00d7(\u22120.00339)3.755 = \\frac{-\\Delta H_{\\text{vap}}}{8.314} \\times (-0.00339) \u0394Hvap=3.755\u00d78.3140.00339\u224831.230.00339\u22489211 J\/mol\u22489.21 kJ\/mol\\Delta H_{\\text{vap}} = \\frac{3.755 \\times 8.314}{0.00339} \\approx \\frac{31.23}{0.00339} \\approx 9211\\ \\text{J\/mol} \\approx \\boxed{9.21\\ \\text{kJ\/mol}}<\/p>\n\n\n\n

      \n\n\n\n

      Explanation<\/h3>\n\n\n\n

      The heat of vaporization<\/strong> is the energy required to convert one mole of a substance from liquid to gas at constant pressure. Methane (CH\u2084), being a small nonpolar molecule, has relatively weak intermolecular forces (London dispersion forces), resulting in a low boiling point of -164 \u00b0C.<\/p>\n\n\n\n

      To determine its heat of vaporization, we used the Clausius-Clapeyron equation<\/strong>, which describes how vapor pressure varies with temperature for phase changes. The equation is derived from thermodynamic principles and assumes that \u0394Hvap remains constant over the temperature range considered \u2014 a reasonable approximation here.<\/p>\n\n\n\n

      In this case, we are given two vapor pressures at two different temperatures: 1 atm at -164 \u00b0C and 42.8 atm at -100 \u00b0C. After converting temperatures to Kelvin and plugging values into the Clausius-Clapeyron equation, we solved for \u0394Hvap using natural logarithms and algebra.<\/p>\n\n\n\n

      The result is approximately 9.21 kJ\/mol<\/strong>, indicating the amount of energy needed to vaporize one mole of liquid methane at its boiling point. This relatively low value aligns with methane\u2019s weak intermolecular forces and low molecular weight.<\/p>\n\n\n\n

      Understanding \u0394Hvap is essential in areas like natural gas processing, cryogenics, and climate science. Methane is a potent greenhouse gas, and its evaporation behavior influences atmospheric concentrations and energy balance. Accurately determining its thermodynamic properties helps in designing storage systems and understanding its environmental impact.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Methane (CH4) has a boiling point of -164 C at 1 atm and a vapor pressure of 42.8 atm at -100 C. What is the heat of vaporization of CH4? The Correct Answer and Explanation is: To calculate the heat of vaporization (\u0394Hvap) of methane (CH\u2084), we use the Clausius-Clapeyron equation, which relates the vapor […]<\/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-219850","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219850","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=219850"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219850\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219850"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219850"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}