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
Why is it common in European airports not to display the gate for flights until around 45-90 minutes before departure, unlike other places?
Help! My Character is too much for her story!
If nine coins are tossed, what is the probability that the number of heads is even?
What do you call someone who likes to pick fights?
Why aren't there more Gauls like Obelix?
How can a demon take control of a human body during REM sleep?
Use Mercury as quenching liquid for swords?
I can't die. Who am I?
Can I negotiate a patent idea for a raise, under French law?
Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?
What sort of fish is this
How exactly does an Ethernet collision happen in the cable, since nodes use different circuits for Tx and Rx?
Movie: boy escapes the real world and goes to a fantasy world with big furry trolls
Should we avoid writing fiction about historical events without extensive research?
Did Amazon pay $0 in taxes last year?
Numerical value of Determinant far from what it is supposed to be
Locked Away- What am I?
Do similar matrices have same characteristic equations?
Why is there an extra space when I type "ls" on the Desktop?
How is it possible to drive VGA displays at such high pixel clock frequencies?
What can I do if someone tampers with my SSH public key?
Origin of the word “pushka”
What will happen if my luggage gets delayed?
What is this tube in a jet engine's air intake?
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
$begingroup$
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
$endgroup$
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
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
$endgroup$
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
$begingroup$
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
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 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
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
$begingroup$
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
add a comment |
$begingroup$
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
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.
$endgroup$
$begingroup$
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
|
show 2 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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
to p(T,V) = fracRT-2aV + c
$$
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.
$endgroup$
add a comment |
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.
$endgroup$
$begingroup$
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
|
show 2 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.
$endgroup$
$begingroup$
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
|
show 2 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 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
|
show 2 more comments
$begingroup$
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
|
show 2 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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
to p(T,V) = fracRT-2aV + c
$$
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.
$endgroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
to p(T,V) = fracRT-2aV + c
$$
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.
$endgroup$
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:
$$int frac RV mathrm d T = fracRTV + g(V) \
int left(frac2aV^2-fracRTV^2right) mathrm d V = frac-2a+RTV + h(T) \
to p(T,V) = fracRT-2aV + c
$$
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.
$endgroup$
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.
edited 12 hours ago
Tyberius
7,04832160
7,04832160
answered yesterday
user213305user213305
1,031416
1,031416
add a comment |
add a comment |
$begingroup$
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