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Is the differential, dp, exact or not? [on hold]


How to make approximation of rotational partition function of diatomic linear molecules?How to derive the pressure dependency for the Gibbs free energy?Why is dU an exact differential and dq an inexact differential?Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature?How to prove that entropy is a state function?Gas expansion through adiabatic turbineDetermining pressure of sublimation of iodineHow to properly rearrange the first law of thermodynamics?Entropy and specific heat capacityExpression for fugacity coefficient derived from a pressure-explicite EOS













6












$begingroup$



Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV$$ (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.










share|improve this question











$endgroup$



put on hold as off-topic by Mithoron, A.K., Todd Minehardt, Melanie Shebel, Jon Custer 6 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – Mithoron, Todd Minehardt, Melanie Shebel, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    I'm voting to close this question as off-topic because it is a mathmatics question.
    $endgroup$
    – A.K.
    yesterday






  • 10




    $begingroup$
    Not all chemistry is done in a vial.
    $endgroup$
    – Charlie Crown
    yesterday















6












$begingroup$



Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV$$ (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.










share|improve this question











$endgroup$



put on hold as off-topic by Mithoron, A.K., Todd Minehardt, Melanie Shebel, Jon Custer 6 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – Mithoron, Todd Minehardt, Melanie Shebel, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    I'm voting to close this question as off-topic because it is a mathmatics question.
    $endgroup$
    – A.K.
    yesterday






  • 10




    $begingroup$
    Not all chemistry is done in a vial.
    $endgroup$
    – Charlie Crown
    yesterday













6












6








6


2



$begingroup$



Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV$$ (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.










share|improve this question











$endgroup$





Consider the differential $$mathrm dp=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV$$ (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















share|improve this question













share|improve this question




share|improve this question








edited yesterday









Charlie Crown

447115




447115










asked yesterday









NicciNicci

602




602




put on hold as off-topic by Mithoron, A.K., Todd Minehardt, Melanie Shebel, Jon Custer 6 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – Mithoron, Todd Minehardt, Melanie Shebel, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by Mithoron, A.K., Todd Minehardt, Melanie Shebel, Jon Custer 6 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – Mithoron, Todd Minehardt, Melanie Shebel, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    I'm voting to close this question as off-topic because it is a mathmatics question.
    $endgroup$
    – A.K.
    yesterday






  • 10




    $begingroup$
    Not all chemistry is done in a vial.
    $endgroup$
    – Charlie Crown
    yesterday
















  • $begingroup$
    I'm voting to close this question as off-topic because it is a mathmatics question.
    $endgroup$
    – A.K.
    yesterday






  • 10




    $begingroup$
    Not all chemistry is done in a vial.
    $endgroup$
    – Charlie Crown
    yesterday















$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
yesterday




$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
yesterday




10




10




$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
yesterday




$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
yesterday










2 Answers
2






active

oldest

votes


















13












$begingroup$

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



$$mathrmdp = A(V),mathrmdT + B(T,V),mathrmdV$$



where



$$A(V) = fracRV$$



and



$$ B(T,V) = left( frac2aV^3 - fracRTV^2 right) $$



A differential is exact if



$$left( fracpartial Apartial V right)_T = left( fracpartial Bpartial T right)_V $$



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 $mathrmdp$ 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



$$left( fracpartial Apartial V right)_T = -fracRV^2$$



$$left( fracpartial Bpartial T right)_V = -fracRV^2$$



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






share|improve this answer











$endgroup$












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






  • 1




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






  • 2




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











  • $begingroup$
    @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
    $endgroup$
    – Mithoron
    yesterday






  • 2




    $begingroup$
    No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
    $endgroup$
    – Charlie Crown
    yesterday


















7












$begingroup$

For a given function, $F(x,y,z,...)$, it's differential $textdF$ is given by:
$$
textdF = left(fracpartial Fpartial xright)_y,z text d x +left(fracpartial Fpartial yright)_x,z text d y + left(fracpartial Fpartial zright)_x,y text d z +; ...
$$



To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.



For the case given:



$$
mathrm dp(T,V)=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV
$$



If $mathrm dp$ is an exact differential, that would mean that:



$$
left(fracpartial ppartial Tright)_V = frac RV textand left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right)
$$



There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.



Integration



Taking indefinite integrals of the suspected derivatives:



$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
to p(T,V) = fracRT-2aV + c
$$



Differentiation



It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:



$$
fracpartial partial T _V left(fracpartial p partial V right)_T = fracpartial partial V _T left( fracpartial p partial T right)_V
$$



Assuming:
$$ left(fracpartial ppartial Tright)_V = frac RV \
to fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$



Assuming



$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
to fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$



Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.






share|improve this answer











$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    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



    $$mathrmdp = A(V),mathrmdT + B(T,V),mathrmdV$$



    where



    $$A(V) = fracRV$$



    and



    $$ B(T,V) = left( frac2aV^3 - fracRTV^2 right) $$



    A differential is exact if



    $$left( fracpartial Apartial V right)_T = left( fracpartial Bpartial T right)_V $$



    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 $mathrmdp$ 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



    $$left( fracpartial Apartial V right)_T = -fracRV^2$$



    $$left( fracpartial Bpartial T right)_V = -fracRV^2$$



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






    share|improve this answer











    $endgroup$












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






    • 1




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






    • 2




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











    • $begingroup$
      @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
      $endgroup$
      – Mithoron
      yesterday






    • 2




      $begingroup$
      No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
      $endgroup$
      – Charlie Crown
      yesterday















    13












    $begingroup$

    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



    $$mathrmdp = A(V),mathrmdT + B(T,V),mathrmdV$$



    where



    $$A(V) = fracRV$$



    and



    $$ B(T,V) = left( frac2aV^3 - fracRTV^2 right) $$



    A differential is exact if



    $$left( fracpartial Apartial V right)_T = left( fracpartial Bpartial T right)_V $$



    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 $mathrmdp$ 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



    $$left( fracpartial Apartial V right)_T = -fracRV^2$$



    $$left( fracpartial Bpartial T right)_V = -fracRV^2$$



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






    share|improve this answer











    $endgroup$












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






    • 1




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






    • 2




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











    • $begingroup$
      @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
      $endgroup$
      – Mithoron
      yesterday






    • 2




      $begingroup$
      No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
      $endgroup$
      – Charlie Crown
      yesterday













    13












    13








    13





    $begingroup$

    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



    $$mathrmdp = A(V),mathrmdT + B(T,V),mathrmdV$$



    where



    $$A(V) = fracRV$$



    and



    $$ B(T,V) = left( frac2aV^3 - fracRTV^2 right) $$



    A differential is exact if



    $$left( fracpartial Apartial V right)_T = left( fracpartial Bpartial T right)_V $$



    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 $mathrmdp$ 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



    $$left( fracpartial Apartial V right)_T = -fracRV^2$$



    $$left( fracpartial Bpartial T right)_V = -fracRV^2$$



    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



    $$mathrmdp = A(V),mathrmdT + B(T,V),mathrmdV$$



    where



    $$A(V) = fracRV$$



    and



    $$ B(T,V) = left( frac2aV^3 - fracRTV^2 right) $$



    A differential is exact if



    $$left( fracpartial Apartial V right)_T = left( fracpartial Bpartial T right)_V $$



    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 $mathrmdp$ 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



    $$left( fracpartial Apartial V right)_T = -fracRV^2$$



    $$left( fracpartial Bpartial T right)_V = -fracRV^2$$



    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 13 hours ago









    Loong

    33.7k884175




    33.7k884175










    answered yesterday









    Charlie CrownCharlie Crown

    447115




    447115











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






    • 1




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






    • 2




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











    • $begingroup$
      @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
      $endgroup$
      – Mithoron
      yesterday






    • 2




      $begingroup$
      No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
      $endgroup$
      – Charlie Crown
      yesterday
















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






    • 1




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






    • 2




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











    • $begingroup$
      @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
      $endgroup$
      – Mithoron
      yesterday






    • 2




      $begingroup$
      No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
      $endgroup$
      – Charlie Crown
      yesterday















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




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




    1




    1




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




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




    2




    2




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





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













    $begingroup$
    @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
    $endgroup$
    – Mithoron
    yesterday




    $begingroup$
    @orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.
    $endgroup$
    – Mithoron
    yesterday




    2




    2




    $begingroup$
    No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
    $endgroup$
    – Charlie Crown
    yesterday




    $begingroup$
    No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.
    $endgroup$
    – Charlie Crown
    yesterday











    7












    $begingroup$

    For a given function, $F(x,y,z,...)$, it's differential $textdF$ is given by:
    $$
    textdF = left(fracpartial Fpartial xright)_y,z text d x +left(fracpartial Fpartial yright)_x,z text d y + left(fracpartial Fpartial zright)_x,y text d z +; ...
    $$



    To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.



    For the case given:



    $$
    mathrm dp(T,V)=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV
    $$



    If $mathrm dp$ is an exact differential, that would mean that:



    $$
    left(fracpartial ppartial Tright)_V = frac RV textand left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right)
    $$



    There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.



    Integration



    Taking indefinite integrals of the suspected derivatives:



    $$int frac RV mathrm d T = fracRTV + g(V) \
    int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
    to p(T,V) = fracRT-2aV + c
    $$



    Differentiation



    It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:



    $$
    fracpartial partial T _V left(fracpartial p partial V right)_T = fracpartial partial V _T left( fracpartial p partial T right)_V
    $$



    Assuming:
    $$ left(fracpartial ppartial Tright)_V = frac RV \
    to fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
    $$



    Assuming



    $$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
    to fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
    $$



    Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.






    share|improve this answer











    $endgroup$

















      7












      $begingroup$

      For a given function, $F(x,y,z,...)$, it's differential $textdF$ is given by:
      $$
      textdF = left(fracpartial Fpartial xright)_y,z text d x +left(fracpartial Fpartial yright)_x,z text d y + left(fracpartial Fpartial zright)_x,y text d z +; ...
      $$



      To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.



      For the case given:



      $$
      mathrm dp(T,V)=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV
      $$



      If $mathrm dp$ is an exact differential, that would mean that:



      $$
      left(fracpartial ppartial Tright)_V = frac RV textand left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right)
      $$



      There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.



      Integration



      Taking indefinite integrals of the suspected derivatives:



      $$int frac RV mathrm d T = fracRTV + g(V) \
      int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
      to p(T,V) = fracRT-2aV + c
      $$



      Differentiation



      It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:



      $$
      fracpartial partial T _V left(fracpartial p partial V right)_T = fracpartial partial V _T left( fracpartial p partial T right)_V
      $$



      Assuming:
      $$ left(fracpartial ppartial Tright)_V = frac RV \
      to fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
      $$



      Assuming



      $$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
      to fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
      $$



      Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.






      share|improve this answer











      $endgroup$















        7












        7








        7





        $begingroup$

        For a given function, $F(x,y,z,...)$, it's differential $textdF$ is given by:
        $$
        textdF = left(fracpartial Fpartial xright)_y,z text d x +left(fracpartial Fpartial yright)_x,z text d y + left(fracpartial Fpartial zright)_x,y text d z +; ...
        $$



        To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.



        For the case given:



        $$
        mathrm dp(T,V)=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV
        $$



        If $mathrm dp$ is an exact differential, that would mean that:



        $$
        left(fracpartial ppartial Tright)_V = frac RV textand left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right)
        $$



        There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.



        Integration



        Taking indefinite integrals of the suspected derivatives:



        $$int frac RV mathrm d T = fracRTV + g(V) \
        int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
        to p(T,V) = fracRT-2aV + c
        $$



        Differentiation



        It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:



        $$
        fracpartial partial T _V left(fracpartial p partial V right)_T = fracpartial partial V _T left( fracpartial p partial T right)_V
        $$



        Assuming:
        $$ left(fracpartial ppartial Tright)_V = frac RV \
        to fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
        $$



        Assuming



        $$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
        to fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
        $$



        Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.






        share|improve this answer











        $endgroup$



        For a given function, $F(x,y,z,...)$, it's differential $textdF$ is given by:
        $$
        textdF = left(fracpartial Fpartial xright)_y,z text d x +left(fracpartial Fpartial yright)_x,z text d y + left(fracpartial Fpartial zright)_x,y text d z +; ...
        $$



        To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.



        For the case given:



        $$
        mathrm dp(T,V)=frac RV,mathrm dT+left(frac2aV^2-fracRTV^2right),mathrm dV
        $$



        If $mathrm dp$ is an exact differential, that would mean that:



        $$
        left(fracpartial ppartial Tright)_V = frac RV textand left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right)
        $$



        There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.



        Integration



        Taking indefinite integrals of the suspected derivatives:



        $$int frac RV mathrm d T = fracRTV + g(V) \
        int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
        to p(T,V) = fracRT-2aV + c
        $$



        Differentiation



        It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:



        $$
        fracpartial partial T _V left(fracpartial p partial V right)_T = fracpartial partial V _T left( fracpartial p partial T right)_V
        $$



        Assuming:
        $$ left(fracpartial ppartial Tright)_V = frac RV \
        to fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
        $$



        Assuming



        $$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
        to fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
        $$



        Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 12 hours ago









        Tyberius

        7,04832160




        7,04832160










        answered yesterday









        user213305user213305

        1,031416




        1,031416













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