Is the differential, dp, exact or not?












3












$begingroup$



Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac{2a}{V^2}-frac{RT}{V^2}right),mathrm dV$$ (where $a$ is a constant value)



(a) Determine whether the above differential, i.e. $mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!




I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.






thermodynamics







share|improve this question











$endgroup$

















    3












    $begingroup$



    Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac{2a}{V^2}-frac{RT}{V^2}right),mathrm dV$$ (where $a$ is a constant value)



    (a) Determine whether the above differential, i.e. $mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!




    I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.






    thermodynamics







    share|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac{2a}{V^2}-frac{RT}{V^2}right),mathrm dV$$ (where $a$ is a constant value)



      (a) Determine whether the above differential, i.e. $mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!




      I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.






      thermodynamics







      share|improve this question











      $endgroup$





      Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac{2a}{V^2}-frac{RT}{V^2}right),mathrm dV$$ (where $a$ is a constant value)



      (a) Determine whether the above differential, i.e. $mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!




      I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.






      thermodynamics




      thermodynamics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      Charlie Crown

      368113




      368113










      asked 5 hours ago









      NicciNicci

      452




      452






















          1 Answer
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          8












          $begingroup$

          Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.



          Let me rewrite the differential as



          $$mathrm{d}p = A(V)mathrm{d}T + B(T,V),mathrm{d}V$$



          where



          $$A(V) = frac{R}{V}$$



          and



          $$ B(T,V) = left( frac{2a}{V^3} - frac{RT}{V^2} right) $$



          A differential is exact if



          $$left( frac{partial A}{partial V} right)_T = left( frac{partial B}{partial T} right)_V $$



          Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $mathrm{d}p$ is exact!



          Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to



          $$left( frac{partial A}{partial V} right)_T = -frac{R}{V^2}$$



          $$left( frac{partial B}{partial T} right)_V = -frac{R}{V^2}$$



          If the two partial derivatives are the same, the differential is exact. I will let you be the judge.






          share|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I will figure it out one day, it can't be harder than thermo ;)
            $endgroup$
            – Charlie Crown
            4 hours ago










          • $begingroup$
            No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
            $endgroup$
            – andselisk
            4 hours ago












          • $begingroup$
            I don't think you should basically have revealed the full solution.
            $endgroup$
            – Chester Miller
            4 hours ago






          • 1




            $begingroup$
            In my experience people mess up the differentiation and then think they did everything wrong
            $endgroup$
            – Charlie Crown
            4 hours ago






          • 2




            $begingroup$
            Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
            $endgroup$
            – orthocresol♦
            1 hour ago













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          1 Answer
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          active

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          active

          oldest

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          8












          $begingroup$

          Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.



          Let me rewrite the differential as



          $$mathrm{d}p = A(V)mathrm{d}T + B(T,V),mathrm{d}V$$



          where



          $$A(V) = frac{R}{V}$$



          and



          $$ B(T,V) = left( frac{2a}{V^3} - frac{RT}{V^2} right) $$



          A differential is exact if



          $$left( frac{partial A}{partial V} right)_T = left( frac{partial B}{partial T} right)_V $$



          Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $mathrm{d}p$ is exact!



          Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to



          $$left( frac{partial A}{partial V} right)_T = -frac{R}{V^2}$$



          $$left( frac{partial B}{partial T} right)_V = -frac{R}{V^2}$$



          If the two partial derivatives are the same, the differential is exact. I will let you be the judge.






          share|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I will figure it out one day, it can't be harder than thermo ;)
            $endgroup$
            – Charlie Crown
            4 hours ago










          • $begingroup$
            No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
            $endgroup$
            – andselisk
            4 hours ago












          • $begingroup$
            I don't think you should basically have revealed the full solution.
            $endgroup$
            – Chester Miller
            4 hours ago






          • 1




            $begingroup$
            In my experience people mess up the differentiation and then think they did everything wrong
            $endgroup$
            – Charlie Crown
            4 hours ago






          • 2




            $begingroup$
            Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
            $endgroup$
            – orthocresol♦
            1 hour ago


















          8












          $begingroup$

          Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.



          Let me rewrite the differential as



          $$mathrm{d}p = A(V)mathrm{d}T + B(T,V),mathrm{d}V$$



          where



          $$A(V) = frac{R}{V}$$



          and



          $$ B(T,V) = left( frac{2a}{V^3} - frac{RT}{V^2} right) $$



          A differential is exact if



          $$left( frac{partial A}{partial V} right)_T = left( frac{partial B}{partial T} right)_V $$



          Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $mathrm{d}p$ is exact!



          Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to



          $$left( frac{partial A}{partial V} right)_T = -frac{R}{V^2}$$



          $$left( frac{partial B}{partial T} right)_V = -frac{R}{V^2}$$



          If the two partial derivatives are the same, the differential is exact. I will let you be the judge.






          share|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I will figure it out one day, it can't be harder than thermo ;)
            $endgroup$
            – Charlie Crown
            4 hours ago










          • $begingroup$
            No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
            $endgroup$
            – andselisk
            4 hours ago












          • $begingroup$
            I don't think you should basically have revealed the full solution.
            $endgroup$
            – Chester Miller
            4 hours ago






          • 1




            $begingroup$
            In my experience people mess up the differentiation and then think they did everything wrong
            $endgroup$
            – Charlie Crown
            4 hours ago






          • 2




            $begingroup$
            Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
            $endgroup$
            – orthocresol♦
            1 hour ago
















          8












          8








          8





          $begingroup$

          Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.



          Let me rewrite the differential as



          $$mathrm{d}p = A(V)mathrm{d}T + B(T,V),mathrm{d}V$$



          where



          $$A(V) = frac{R}{V}$$



          and



          $$ B(T,V) = left( frac{2a}{V^3} - frac{RT}{V^2} right) $$



          A differential is exact if



          $$left( frac{partial A}{partial V} right)_T = left( frac{partial B}{partial T} right)_V $$



          Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $mathrm{d}p$ is exact!



          Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to



          $$left( frac{partial A}{partial V} right)_T = -frac{R}{V^2}$$



          $$left( frac{partial B}{partial T} right)_V = -frac{R}{V^2}$$



          If the two partial derivatives are the same, the differential is exact. I will let you be the judge.






          share|improve this answer











          $endgroup$



          Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.



          Let me rewrite the differential as



          $$mathrm{d}p = A(V)mathrm{d}T + B(T,V),mathrm{d}V$$



          where



          $$A(V) = frac{R}{V}$$



          and



          $$ B(T,V) = left( frac{2a}{V^3} - frac{RT}{V^2} right) $$



          A differential is exact if



          $$left( frac{partial A}{partial V} right)_T = left( frac{partial B}{partial T} right)_V $$



          Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $mathrm{d}p$ is exact!



          Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to



          $$left( frac{partial A}{partial V} right)_T = -frac{R}{V^2}$$



          $$left( frac{partial B}{partial T} right)_V = -frac{R}{V^2}$$



          If the two partial derivatives are the same, the differential is exact. I will let you be the judge.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago









          orthocresol♦

          39.4k7114241




          39.4k7114241










          answered 5 hours ago









          Charlie CrownCharlie Crown

          368113




          368113








          • 1




            $begingroup$
            I will figure it out one day, it can't be harder than thermo ;)
            $endgroup$
            – Charlie Crown
            4 hours ago










          • $begingroup$
            No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
            $endgroup$
            – andselisk
            4 hours ago












          • $begingroup$
            I don't think you should basically have revealed the full solution.
            $endgroup$
            – Chester Miller
            4 hours ago






          • 1




            $begingroup$
            In my experience people mess up the differentiation and then think they did everything wrong
            $endgroup$
            – Charlie Crown
            4 hours ago






          • 2




            $begingroup$
            Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
            $endgroup$
            – orthocresol♦
            1 hour ago
















          • 1




            $begingroup$
            I will figure it out one day, it can't be harder than thermo ;)
            $endgroup$
            – Charlie Crown
            4 hours ago










          • $begingroup$
            No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
            $endgroup$
            – andselisk
            4 hours ago












          • $begingroup$
            I don't think you should basically have revealed the full solution.
            $endgroup$
            – Chester Miller
            4 hours ago






          • 1




            $begingroup$
            In my experience people mess up the differentiation and then think they did everything wrong
            $endgroup$
            – Charlie Crown
            4 hours ago






          • 2




            $begingroup$
            Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
            $endgroup$
            – orthocresol♦
            1 hour ago










          1




          1




          $begingroup$
          I will figure it out one day, it can't be harder than thermo ;)
          $endgroup$
          – Charlie Crown
          4 hours ago




          $begingroup$
          I will figure it out one day, it can't be harder than thermo ;)
          $endgroup$
          – Charlie Crown
          4 hours ago












          $begingroup$
          No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
          $endgroup$
          – andselisk
          4 hours ago






          $begingroup$
          No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools like pandoc and wrappers such as bookdown to exist:)
          $endgroup$
          – andselisk
          4 hours ago














          $begingroup$
          I don't think you should basically have revealed the full solution.
          $endgroup$
          – Chester Miller
          4 hours ago




          $begingroup$
          I don't think you should basically have revealed the full solution.
          $endgroup$
          – Chester Miller
          4 hours ago




          1




          1




          $begingroup$
          In my experience people mess up the differentiation and then think they did everything wrong
          $endgroup$
          – Charlie Crown
          4 hours ago




          $begingroup$
          In my experience people mess up the differentiation and then think they did everything wrong
          $endgroup$
          – Charlie Crown
          4 hours ago




          2




          2




          $begingroup$
          Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
          $endgroup$
          – orthocresol♦
          1 hour ago






          $begingroup$
          Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
          $endgroup$
          – orthocresol♦
          1 hour ago




















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