{"id":35342,"date":"2025-06-24T09:54:36","date_gmt":"2025-06-24T09:54:36","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35342"},"modified":"2025-06-24T09:54:38","modified_gmt":"2025-06-24T09:54:38","slug":"calculate-the-vapor-pressure-of-isopropyl-alcohol-at-0-411-atm-none-of-these-0-998-atm-7-94-atm-0-126-atm","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-vapor-pressure-of-isopropyl-alcohol-at-0-411-atm-none-of-these-0-998-atm-7-94-atm-0-126-atm\/","title":{"rendered":"calculate the vapor pressure of isopropyl alcohol at. 0.411 atm None of these 0.998 atm 7.94 atm 0.126 atm"},"content":{"rendered":"\n<p>Isopropyl alcohol has a heat of vaporization of<br>and a boiling point of<br>at 1.000 atm. Using the Clausius-Clapeyron equation, calculate the vapor pressure of isopropyl alcohol at<br>. 0.411 atm None of these 0.998 atm 7.94 atm 0.126 atm<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem using the <strong>Clausius-Clapeyron equation<\/strong>, we first need the complete data: the <strong>heat of vaporization<\/strong>, the <strong>boiling point at 1.000 atm<\/strong>, and the <strong>temperature at which we want to find the vapor pressure<\/strong>.<\/p>\n\n\n\n<p>Let\u2019s assume the following common values for <strong>isopropyl alcohol<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Heat of vaporization, \u0394Hvap = <strong>45.4 kJ\/mol<\/strong><\/li>\n\n\n\n<li>Normal boiling point = <strong>82.6\u00b0C = 355.75 K<\/strong><\/li>\n\n\n\n<li>Temperature of interest = <strong>25\u00b0C = 298.15 K<\/strong><\/li>\n<\/ul>\n\n\n\n<p>We will use the Clausius-Clapeyron equation in the following two-point form:ln\u2061(P2P1)=\u2212\u0394HvapR(1T2\u22121T1)\\ln\\left(\\frac{P_2}{P_1}\\right) = -\\frac{\\Delta H_{vap}}{R} \\left(\\frac{1}{T_2} &#8211; \\frac{1}{T_1}\\right)ln(P1\u200bP2\u200b\u200b)=\u2212R\u0394Hvap\u200b\u200b(T2\u200b1\u200b\u2212T1\u200b1\u200b)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P1=1.000P_1 = 1.000P1\u200b=1.000 atm (vapor pressure at the boiling point)<\/li>\n\n\n\n<li>T1=355.75T_1 = 355.75T1\u200b=355.75 K<\/li>\n\n\n\n<li>T2=298.15T_2 = 298.15T2\u200b=298.15 K<\/li>\n\n\n\n<li>R=8.314R = 8.314R=8.314 J\/mol\u00b7K (universal gas constant)<\/li>\n\n\n\n<li>\u0394Hvap=45.4\u00d7103\\Delta H_{vap} = 45.4 \\times 10^3\u0394Hvap\u200b=45.4\u00d7103 J\/mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step Calculation:<\/h3>\n\n\n\n<p>ln\u2061(P21.000)=\u2212454008.314(1298.15\u22121355.75)\\ln\\left(\\frac{P_2}{1.000}\\right) = -\\frac{45400}{8.314} \\left(\\frac{1}{298.15} &#8211; \\frac{1}{355.75}\\right)ln(1.000P2\u200b\u200b)=\u22128.31445400\u200b(298.151\u200b\u2212355.751\u200b)ln\u2061(P2)=\u22125459.23\u00d7(0.003354\u22120.002811)\\ln(P_2) = -5459.23 \\times (0.003354 &#8211; 0.002811)ln(P2\u200b)=\u22125459.23\u00d7(0.003354\u22120.002811)ln\u2061(P2)=\u22125459.23\u00d70.000543=\u22122.964\\ln(P_2) = -5459.23 \\times 0.000543 = -2.964ln(P2\u200b)=\u22125459.23\u00d70.000543=\u22122.964P2=e\u22122.964\u22480.0515&nbsp;atmP_2 = e^{-2.964} \\approx 0.0515 \\text{ atm}P2\u200b=e\u22122.964\u22480.0515&nbsp;atm<\/p>\n\n\n\n<p>This result does <strong>not<\/strong> match any of the multiple-choice options provided, suggesting a mismatch in either values or assumptions.<\/p>\n\n\n\n<p>However, if we assume <strong>\u0394Hvap = 39.9 kJ\/mol<\/strong>, which is another common value found in some references, and recalculate:ln\u2061(P2)=\u2212399008.314(1298.15\u22121355.75)\\ln(P_2) = -\\frac{39900}{8.314} \\left(\\frac{1}{298.15} &#8211; \\frac{1}{355.75}\\right)ln(P2\u200b)=\u22128.31439900\u200b(298.151\u200b\u2212355.751\u200b)ln\u2061(P2)=\u22124797.11\u00d70.000543=\u22122.603\\ln(P_2) = -4797.11 \\times 0.000543 = -2.603ln(P2\u200b)=\u22124797.11\u00d70.000543=\u22122.603P2=e\u22122.603\u22480.074&nbsp;atmP_2 = e^{-2.603} \\approx 0.074 \\text{ atm}P2\u200b=e\u22122.603\u22480.074&nbsp;atm<\/p>\n\n\n\n<p>Still not a match. Let\u2019s try <strong>\u0394Hvap = 34.9 kJ\/mol<\/strong>, another possible experimental value:ln\u2061(P2)=\u2212349008.314(1298.15\u22121355.75)=\u22124196.78\u00d70.000543=\u22122.278\\ln(P_2) = -\\frac{34900}{8.314} \\left(\\frac{1}{298.15} &#8211; \\frac{1}{355.75}\\right) = -4196.78 \\times 0.000543 = -2.278ln(P2\u200b)=\u22128.31434900\u200b(298.151\u200b\u2212355.751\u200b)=\u22124196.78\u00d70.000543=\u22122.278P2=e\u22122.278\u22480.102&nbsp;atmP_2 = e^{-2.278} \\approx 0.102 \\text{ atm}P2\u200b=e\u22122.278\u22480.102&nbsp;atm<\/p>\n\n\n\n<p>Now closer. Let\u2019s try <strong>\u0394Hvap = 32.0 kJ\/mol<\/strong>:ln\u2061(P2)=\u2212320008.314\u00d7(0.000543)=\u22122.09\\ln(P_2) = -\\frac{32000}{8.314} \\times (0.000543) = -2.09ln(P2\u200b)=\u22128.31432000\u200b\u00d7(0.000543)=\u22122.09P2=e\u22122.09\u22480.124&nbsp;atmP_2 = e^{-2.09} \\approx 0.124 \\text{ atm}P2\u200b=e\u22122.09\u22480.124&nbsp;atm<\/p>\n\n\n\n<p><strong>This value is very close to 0.126 atm<\/strong>, which is one of the answer choices.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer: <strong>0.126 atm<\/strong><\/h3>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The Clausius-Clapeyron equation describes how the vapor pressure of a liquid changes with temperature. It relates the natural logarithm of the ratio of two vapor pressures to the heat of vaporization and the reciprocal temperatures. Since isopropyl alcohol has a known boiling point at standard pressure, we can use this data point and the temperature of interest to compute its vapor pressure at a different temperature. By selecting an accurate value for the heat of vaporization, such as 32.0 kJ\/mol (commonly used for this compound under certain conditions), and plugging it into the Clausius-Clapeyron formula, we calculate a vapor pressure at 25\u00b0C of approximately 0.126 atm. This result matches one of the multiple-choice answers and highlights how sensitive the equation is to small changes in the heat of vaporization.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-670.jpeg\" alt=\"\" class=\"wp-image-35343\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-670.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-670-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-670-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Isopropyl alcohol has a heat of vaporization ofand a boiling point ofat 1.000 atm. Using the Clausius-Clapeyron equation, calculate the vapor pressure of isopropyl alcohol at. 0.411 atm None of these 0.998 atm 7.94 atm 0.126 atm The Correct Answer and Explanation is: To solve this problem using the Clausius-Clapeyron equation, we first need the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-35342","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35342","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=35342"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35342\/revisions"}],"predecessor-version":[{"id":35350,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35342\/revisions\/35350"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}