Why does the negative sign arise in this thermodynamic relation?












1












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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







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  • 1




    $begingroup$
    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
    4 hours ago
















1












$begingroup$


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.







$endgroup$








  • 1




    $begingroup$
    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
    4 hours ago














1












1








1





$begingroup$


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









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









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edited 5 hours ago









ACuriousMind♦

72.5k18126318




72.5k18126318






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asked 5 hours ago









Srijan GhoshSrijan Ghosh

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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.








  • 1




    $begingroup$
    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
    4 hours ago














  • 1




    $begingroup$
    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
    4 hours ago








1




1




$begingroup$
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
4 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
4 hours ago










2 Answers
2






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$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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









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      2 Answers
      2






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      2 Answers
      2






      active

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      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 4 hours ago









          ACuriousMind♦ACuriousMind

          72.5k18126318




          72.5k18126318























              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 4 hours ago









                  Chester MillerChester Miller

                  15.3k2824




                  15.3k2824






















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