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Is the differential, dp, exact or not?
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
$begingroup$
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.
thermodynamics
$endgroup$
add a comment |
$begingroup$
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.
thermodynamics
$endgroup$
$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
2
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago
add a comment |
$begingroup$
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.
thermodynamics
$endgroup$
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
thermodynamics
edited 6 hours ago
Charlie Crown
368114
368114
asked 11 hours ago
NicciNicci
552
552
$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
2
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago
add a comment |
$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
2
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago
$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
2
2
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools likepandoc
and wrappers such asbookdown
to exist:)
$endgroup$
– andselisk
9 hours ago
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
$begingroup$
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
5 hours ago
|
show 4 more comments
$begingroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
implies p(T,V) = fracRT-2aV + c
$$
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:
$$ left(fracpartial ppartial Tright)_V = frac RV \
implies fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$
Assuming
$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
implies fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$
Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools likepandoc
and wrappers such asbookdown
to exist:)
$endgroup$
– andselisk
9 hours ago
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
$begingroup$
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
5 hours ago
|
show 4 more comments
$begingroup$
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.
$endgroup$
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools likepandoc
and wrappers such asbookdown
to exist:)
$endgroup$
– andselisk
9 hours ago
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
$begingroup$
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
5 hours ago
|
show 4 more comments
$begingroup$
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.
$endgroup$
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.
edited 6 hours ago
orthocresol♦
39.4k7114241
39.4k7114241
answered 10 hours ago
Charlie CrownCharlie Crown
368114
368114
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools likepandoc
and wrappers such asbookdown
to exist:)
$endgroup$
– andselisk
9 hours ago
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
$begingroup$
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
5 hours ago
|
show 4 more comments
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
No worries, many find $mathrmLaTeX$ and mathJax syntax cumbersome and frustrating. There is a good reason for tools likepandoc
and wrappers such asbookdown
to exist:)
$endgroup$
– andselisk
9 hours ago
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
$begingroup$
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
5 hours ago
1
1
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
9 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
9 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
9 hours ago
1
1
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
$begingroup$
In my experience people mess up the differentiation and then think they did everything wrong
$endgroup$
– Charlie Crown
9 hours ago
2
2
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
$begingroup$
Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/…
$endgroup$
– orthocresol♦
6 hours ago
1
1
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
|
show 4 more comments
$begingroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
implies p(T,V) = fracRT-2aV + c
$$
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:
$$ left(fracpartial ppartial Tright)_V = frac RV \
implies fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$
Assuming
$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
implies fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$
Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.
$endgroup$
add a comment |
$begingroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
implies p(T,V) = fracRT-2aV + c
$$
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:
$$ left(fracpartial ppartial Tright)_V = frac RV \
implies fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$
Assuming
$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
implies fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$
Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.
$endgroup$
add a comment |
$begingroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
implies p(T,V) = fracRT-2aV + c
$$
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:
$$ left(fracpartial ppartial Tright)_V = frac RV \
implies fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$
Assuming
$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
implies fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$
Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.
$endgroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
implies p(T,V) = fracRT-2aV + c
$$
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:
$$ left(fracpartial ppartial Tright)_V = frac RV \
implies fracpartial partial V _T left( fracpartial p partial T right)_V = -fracRV^2
$$
Assuming
$$ left(fracpartial ppartial Vright)_T = left(frac2aV^2-fracRTV^2right) \
implies fracpartial partial T _V left( fracpartial p partial V right)_T = -fracRV^2
$$
Both terms of the differential $mathrm dp$ imply the same mixed second derivative, hence $mathrm dp$ is an exact differential.
answered 4 hours ago
user213305user213305
946416
946416
add a comment |
add a comment |
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$begingroup$
I'm voting to close this question as off-topic because it is a mathmatics question.
$endgroup$
– A.K.
4 hours ago
2
$begingroup$
Not all chemistry is done in a vial.
$endgroup$
– Charlie Crown
1 hour ago