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Is the differential, dp, exact or not?

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5

$begingroup$

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

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

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

thermodynamics

share|improve this question

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

$endgroup$

add a comment | 

5

$begingroup$

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

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

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

thermodynamics

share|improve this question

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

$endgroup$

add a comment | 

$begingroup$

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

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

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

thermodynamics

share|improve this question

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

$endgroup$

share|improve this question

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

share|improve this question

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

edited 50 mins ago

Charlie Crown

348113

edited 50 mins ago

Charlie Crown

348113

asked 4 hours ago

NicciNicci

492

asked 4 hours ago

NicciNicci

492

1 Answer
1

active

oldest

votes

6

$begingroup$

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

Let me rewrite the differential as

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

where

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

and

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

A differential is exact if

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

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

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

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

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

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

share|improve this answer

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I will figure it out one day, it can't be harder than thermo ;)
  $endgroup$
  – Charlie Crown
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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6

$begingroup$

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

Let me rewrite the differential as

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

where

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

and

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

A differential is exact if

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

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

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

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

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

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

share|improve this answer

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I will figure it out one day, it can't be harder than thermo ;)
  $endgroup$
  – Charlie Crown
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

6

$begingroup$

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

Let me rewrite the differential as

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

where

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

and

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

A differential is exact if

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

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

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

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

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

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

share|improve this answer

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I will figure it out one day, it can't be harder than thermo ;)
  $endgroup$
  – Charlie Crown
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

Let me rewrite the differential as

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

where

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

and

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

A differential is exact if

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

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

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

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

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

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

share|improve this answer

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

$endgroup$

share|improve this answer

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

edited 32 mins ago

orthocresol♦

39.4k7114241

edited 32 mins ago

orthocresol♦

39.4k7114241

answered 4 hours ago

Charlie CrownCharlie Crown

348113

answered 4 hours ago

Charlie CrownCharlie Crown

348113