Why does the negative sign arise in this thermodynamic relation?












2












$begingroup$


I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?






thermodynamics differentiation







share|cite|improve this question









New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
    $endgroup$
    – Andrew Steane
    7 hours ago
















2












$begingroup$


I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?






thermodynamics differentiation







share|cite|improve this question









New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
    $endgroup$
    – Andrew Steane
    7 hours ago














2












2








2


0



$begingroup$


I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?






thermodynamics differentiation







share|cite|improve this question









New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?






thermodynamics differentiation




thermodynamics differentiation






share|cite|improve this question









New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









ACuriousMind♦

72.5k18126318




72.5k18126318






New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









Srijan GhoshSrijan Ghosh

133




133




New contributor




Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    $begingroup$
    Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
    $endgroup$
    – Andrew Steane
    7 hours ago














  • 2




    $begingroup$
    Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
    $endgroup$
    – Andrew Steane
    7 hours ago








2




2




$begingroup$
Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
$endgroup$
– Andrew Steane
7 hours ago




$begingroup$
Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
$endgroup$
– Andrew Steane
7 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
$$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
holds.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
    $$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "151"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






      Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465965%2fwhy-does-the-negative-sign-arise-in-this-thermodynamic-relation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
      $$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
      holds.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
        $$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
        holds.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
          $$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
          holds.






          share|cite|improve this answer









          $endgroup$



          This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
          $$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
          holds.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          ACuriousMind♦ACuriousMind

          72.5k18126318




          72.5k18126318























              3












              $begingroup$

              This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
              $$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
                $$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
                  $$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$






                  share|cite|improve this answer









                  $endgroup$



                  This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
                  $$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  Chester MillerChester Miller

                  15.3k2824




                  15.3k2824






















                      Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.













                      Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.












                      Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.
















                      Thanks for contributing an answer to Physics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465965%2fwhy-does-the-negative-sign-arise-in-this-thermodynamic-relation%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      How to make a Squid Proxy server?

                      Image
                      1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari
                      Read more

                      Is this a new Fibonacci Identity?

                      Image
                      6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w
                      Read more

                      19世紀

                      Image
                      この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料
                      Read more