{"id":1534,"date":"2021-03-26T23:30:00","date_gmt":"2021-03-27T03:30:00","guid":{"rendered":"http:\/\/underthehood.blogwyrm.com\/?p=1534"},"modified":"2022-07-28T06:28:23","modified_gmt":"2022-07-28T10:28:23","slug":"maxwells-relations-in-action","status":"publish","type":"post","link":"https:\/\/underthehood.blogwyrm.com\/?p=1534","title":{"rendered":"Maxwell&#8217;s Relations in Action"},"content":{"rendered":"\n<p>Last month\u2019s column introduced the notion of the thermodynamic square as a mnemonic for organizing certain second-order partial derivatives amongst the various thermodynamic potentials: the internal energy $$U$$, the Gibbs and Helmholtz free energies $$G$$ and $$F$$, and the enthalpy $$H$$.\u00a0 As previously alluded to, physicist primarily use these relations (called the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Maxwell_relations\">Maxwell Relations<\/a>) to eliminate the terms difficult or impossible to measure experimentally in favor of parameters that are easily measured in the lab.<\/p>\n\n\n\n<p>A practical example of the application of the Maxwell relations is the simplification of the \u2018first\u2019 and \u2018second $$T dS$$ equations\u2019 as listed in Exercise 7.3-1 on page 189 of Herbert B. Callen\u2019s book <em>Thermodynamics and an Introduction to Thermostatistics<\/em>, 2<sup>nd<\/sup> edition.&nbsp; The relevant physical properties in these equations are (Sec. 3.9 of Callen):<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>the number of particles $$N$$,<\/li><li>the differential heat $$dQ = T dS$$,<\/li><li>the heat capacity at constant volume $$c_V = \\frac{T}{N} \\left( \\frac{\\partial S }{\\partial T} \\right)_V= \\frac{1}{N} \\left(\\frac{\\partial Q }{\\partial T}\\right)_{V}$$,<\/li><li>the heat capacity at constant pressure $$c_P = \\frac{T}{N} \\left( \\frac{\\partial S }{\\partial T} \\right)_P = \\frac{1}{N}\\left(\\frac{\\partial Q}{\\partial T}\\right)_{P}$$,<\/li><li>the coefficient of thermal expansion $$\\alpha = \\frac{1}{V} \\left(\\frac{\\partial V}{\\partial T}\\right)_P $$, and<\/li><li>the isothermal compressibility $$\\kappa_T = &#8211; \\frac{1}{V} \\left( \\frac{\\partial V}{\\partial P} \\right)_T$$.<\/li><\/ul>\n\n\n\n<h1 class=\"wp-block-heading\">First $$TdS$$ Relation<\/h1>\n\n\n\n<p>The first relation we want to verify is<\/p>\n\n\n\n<p>\\[ T dS = N c_V dT + \\frac{T \\alpha}{\\kappa_T} dV \\; .\\]<\/p>\n\n\n\n<p>From the form of this equation, assume that the quantity in question is the entropy as a function of the temperature and volume $$S = S(T,V)$$.&nbsp; Taking the first differential gives<\/p>\n\n\n\n<p>\\[ dS = \\left(\\frac{\\partial S}{\\partial T}\\right)_{V} dT + \\left(\\frac{\\partial S}{\\partial V}\\right)_{T} dV \\; .\\]<\/p>\n\n\n\n<p>The first term is relatively easy to deal with in terms of the heat capacity at constant volume $$c_V$$:<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial S}{\\partial T}\\right)_{V} = \\frac{N c_V}{T} \\; .\\]<\/p>\n\n\n\n<p>The second term requires a bit more work.&nbsp; First use the Maxwell relation associated with the Helmholtz free energy $$F$$<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/first_tdS.png\" alt=\"\" class=\"wp-image-1532\" width=\"417\" height=\"502\" srcset=\"https:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/first_tdS.png 639w, https:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/first_tdS-249x300.png 249w\" sizes=\"auto, (max-width: 417px) 100vw, 417px\" \/><\/figure><\/div>\n\n\n\n<p>to get<\/p>\n\n\n\n<p>\\[ &#8211; &nbsp;\\left(\\frac{\\partial S}{\\partial V}\\right)_{T} = &#8211; \\left(\\frac{\\partial P}{\\partial T}\\right)_{V} \\; .\\]<\/p>\n\n\n\n<p>Next use the second classical partial derivative identity<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial P}{\\partial T}\\right)_{V} \\left(\\frac{\\partial T}{\\partial V}\\right)_{P} \\left(\\frac{\\partial V}{\\partial P}\\right)_{T} = &#8211; 1 \\; ,\\]<\/p>\n\n\n\n<p>and solve for<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial P}{\\partial T}\\right)_{V} = \\frac{-1}{\\left(\\frac{\\partial V}{\\partial P}\\right)_{T} \\left(\\frac{\\partial T}{\\partial V}\\right)_{P} } \\; . \\]<\/p>\n\n\n\n<p>Use another of the classical partial derivative identities to move the $$\\left(\\frac{\\partial T}{\\partial V}\\right)_{P}$$ to the numerator to get<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial P}{\\partial V}\\right)_{T} = &#8211; \\left(\\frac{\\partial V}{\\partial T}\\right)_{P} \/ \\left(\\frac{\\partial V}{\\partial P}\\right)_{T} \\; .\\]<\/p>\n\n\n\n<p>Multiply the numerator and denominator by $$1\/V$$ and use the definitions of $$\\alpha$$ and $$\\kappa_T$$ to get<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial P}{\\partial V}\\right)_{T} = \\frac{\\alpha}{\\kappa_T} \\; . \\]<\/p>\n\n\n\n<p>At this point, the first differential stands as<\/p>\n\n\n\n<p>\\[ dS = \\frac{N c_V}{T} dT + \\frac{\\alpha}{\\kappa_T} dV \\; . \\]<\/p>\n\n\n\n<p>Multiplying each side by $$T$$ gets us to the final form of the first $$T dS$$ equation<\/p>\n\n\n\n<p>\\[ T dS = N c_V dT + \\left(\\frac{T \\alpha}{\\kappa_T} \\right) dV \\; .\\]<\/p>\n\n\n\n<h1 class=\"wp-block-heading\">Second $$TdS$$ Relation<\/h1>\n\n\n\n<p>The second relation is<\/p>\n\n\n\n<p>\\[ T dS = N c_P dT \u2013 T V \\alpha dP \\; . \\]<\/p>\n\n\n\n<p>From the form of this equation, assume that the quantity in question is the entropy as a function of the temperature and pressure $$S = S(T,P)$$.&nbsp; As in the first $$T dS$$ relation, taking the first differential gives<\/p>\n\n\n\n<p>\\[ dS = \\left(\\frac{\\partial S}{\\partial T}\\right)_{P} dT + \\left(\\frac{\\partial S}{\\partial P}\\right)_{T} dV \\; .\\]<\/p>\n\n\n\n<p>The first term, as in the case above, is also relatively easy to deal with in terms of the heat capacity, this time at constant volume, $$c_P$$:<\/p>\n\n\n\n<p>\\[ \\left(\\frac{\\partial S}{\\partial T}\\right)_{P} = \\frac{N c_P}{T} \\; .\\]<\/p>\n\n\n\n<p>The second term only requires a Maxwell relation in terms of the Gibbs free energy<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/second_TdS.png\" alt=\"\" class=\"wp-image-1533\" width=\"349\" height=\"421\" srcset=\"https:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/second_TdS.png 626w, https:\/\/underthehood.blogwyrm.com\/wp-content\/uploads\/2021\/02\/second_TdS-248x300.png 248w\" sizes=\"auto, (max-width: 349px) 100vw, 349px\" \/><\/figure><\/div>\n\n\n\n<p>\\[ -\\left(\\frac{\\partial S}{\\partial P}\\right)_{T} = \\left(\\frac{\\partial V}{\\partial T}\\right)_{P} \\; .\\]<\/p>\n\n\n\n<p>The first differential becomes<\/p>\n\n\n\n<p>\\[ dS = \\frac{N c_P}{T} dT \u2013 &nbsp;\\left(\\frac{\\partial V}{\\partial T}\\right)_{P} dP \\; .\\]<\/p>\n\n\n\n<p>Multiplying the second term by $$V\/V$$ and simplifying gives<\/p>\n\n\n\n<p>\\[ dS = \\frac{N c_P}{T} dT \u2013 &nbsp;V \\alpha &nbsp;dP \\; ,\\]<\/p>\n\n\n\n<p>which becomes the desired relation when multiplying both sides by $$T$$<\/p>\n\n\n\n<p>\\[ T dS = N c_P dT \u2013 T V \\alpha dP \\; .\\]<\/p>\n\n\n\n<p>To summarize, these two relations show how to express the heat $$T dS$$, expressed in terms of the entropy (which cannot be measured) in terms of parameters, such as temperature, pressure, and heat capacity, that are easily measured in the lab.\u00a0\u00a0 All that is needed is the machinery of partial derivatives.\u00a0 This observation is the reason that so many textbooks on Thermodynamics have specific sections devoted to these approaches.<\/p>\n\n\n\n \n","protected":false},"excerpt":{"rendered":"<p>Last month\u2019s column introduced the notion of the thermodynamic square as a mnemonic for organizing certain second-order partial derivatives amongst the various thermodynamic potentials: the internal energy $$U$$, the Gibbs&#8230; <a class=\"read-more-button\" href=\"https:\/\/underthehood.blogwyrm.com\/?p=1534\">Read more &gt;<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1534","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1534","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1534"}],"version-history":[{"count":16,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1534\/revisions"}],"predecessor-version":[{"id":1766,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1534\/revisions\/1766"}],"wp:attachment":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1534"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1534"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1534"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}