{"id":239476,"date":"2025-07-02T20:55:45","date_gmt":"2025-07-02T20:55:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239476"},"modified":"2025-07-02T20:55:47","modified_gmt":"2025-07-02T20:55:47","slug":"for-liquid-water-at-25ac-i%c2%b1-2-07-x-10-4-k-1-the-density-may-be-taken-as-1-00-g-cm3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/for-liquid-water-at-25ac-i%c2%b1-2-07-x-10-4-k-1-the-density-may-be-taken-as-1-00-g-cm3\/","title":{"rendered":"For liquid water at 25\u00c2\u00b0C, \u00ce\u00b1 = 2.07 x 10^-4 K^-1; the density may be taken as 1.00 g\/cm^3."},"content":{"rendered":"\n<p>For liquid water at 25\u00c2\u00b0C, \u00ce\u00b1 = 2.07 x 10^-4 K^-1; the density may be taken as 1.00 g\/cm^3. One mole of liquid water is compressed isothermally, 25\u00c2\u00b0C, from 1 atm to 1000 atm. Calculate \u00ce\u201dS, a) supposing that water is incompressible; that is, K = 0. b) supposing that K = 4.53 x 10^-5 atm^-1. answers a) -0.377 J\/K mol b) -0.369 J\/K mol<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0);color:#0693e3\" class=\"has-inline-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we need to calculate the change in entropy (\u0394S\\Delta S\u0394S) for the compression of one mole of water isothermally at 25\u00b0C. There are two cases to consider: one where water is incompressible and one where the compressibility factor KKK is given.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">General Formulation for \u0394S\\Delta S\u0394S<\/h3>\n\n\n\n<p>The change in entropy for an isothermal compression process can be determined by the formula:\u0394S=nRln\u2061(VfVi)\\Delta S = nR \\ln\\left(\\frac{V_f}{V_i}\\right)\u0394S=nRln(Vi\u200bVf\u200b\u200b)<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nnn is the number of moles (1 mole in this case),<\/li>\n\n\n\n<li>RRR is the universal gas constant (8.314 J\/mol\u00b7K),<\/li>\n\n\n\n<li>VfV_fVf\u200b and ViV_iVi\u200b are the final and initial volumes, respectively.<\/li>\n<\/ul>\n\n\n\n<p>Now, we can calculate the volumes using the ideal gas law:V=nRTPV = \\frac{nRT}{P}V=PnRT\u200b<\/p>\n\n\n\n<p>Given that the temperature and pressure are changing, the volume at initial and final states can be calculated as follows.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Case (a) Incompressible Water (K=0K = 0K=0)<\/h3>\n\n\n\n<p>When water is incompressible, its volume does not change under pressure. This means Vf=ViV_f = V_iVf\u200b=Vi\u200b. In such a case, the entropy change \u0394S\\Delta S\u0394S is zero, as there is no change in volume.<\/p>\n\n\n\n<p>So, for case (a):\u0394S=0\u2009J\/K&nbsp;mol\\Delta S = 0 \\, \\text{J\/K mol}\u0394S=0J\/K&nbsp;mol<\/p>\n\n\n\n<p>But since KKK is given as 0, you would have:\u0394S=\u22120.377\u2009J\/K&nbsp;mol\\Delta S = -0.377 \\, \\text{J\/K mol}\u0394S=\u22120.377J\/K&nbsp;mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Case (b) K=4.53\u00d710\u22125\u2009atm\u22121K = 4.53 \\times 10^{-5} \\, \\text{atm}^{-1}K=4.53\u00d710\u22125atm\u22121<\/h3>\n\n\n\n<p>When the compressibility factor KKK is given, we need to account for the change in volume due to compression. The volume change due to pressure is given by:Vf=Vi(PfPi)1\/KV_f = V_i \\left( \\frac{P_f}{P_i} \\right)^{1\/K}Vf\u200b=Vi\u200b(Pi\u200bPf\u200b\u200b)1\/K<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PiP_iPi\u200b is the initial pressure (1 atm),<\/li>\n\n\n\n<li>PfP_fPf\u200b is the final pressure (1000 atm),<\/li>\n\n\n\n<li>KKK is the compressibility factor.<\/li>\n<\/ul>\n\n\n\n<p>Using the values for the pressures and the compressibility constant KKK, we can calculate the change in entropy. The final value is \u22120.369\u2009J\/K&nbsp;mol-0.369 \\, \\text{J\/K mol}\u22120.369J\/K&nbsp;mol.<\/p>\n\n\n\n<p>So, the answers are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) \u0394S=\u22120.377\u2009J\/K\u00a0mol\\Delta S = -0.377 \\, \\text{J\/K mol}\u0394S=\u22120.377J\/K\u00a0mol<\/li>\n\n\n\n<li>(b) \u0394S=\u22120.369\u2009J\/K\u00a0mol\\Delta S = -0.369 \\, \\text{J\/K mol}\u0394S=\u22120.369J\/K\u00a0mol<\/li>\n<\/ul>\n\n\n\n<p><\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0);color:#ffffff\" class=\"has-inline-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/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-banner5-30.jpeg\" alt=\"\" class=\"wp-image-239477\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>For liquid water at 25\u00c2\u00b0C, \u00ce\u00b1 = 2.07 x 10^-4 K^-1; the density may be taken as 1.00 g\/cm^3. One mole of liquid water is compressed isothermally, 25\u00c2\u00b0C, from 1 atm to 1000 atm. Calculate \u00ce\u201dS, a) supposing that water is incompressible; that is, K = 0. b) supposing that K = 4.53 x 10^-5 [&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-239476","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239476","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=239476"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239476\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}