Is the differential, dp, exact or not?












3












$begingroup$



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



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




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






thermodynamics







share|improve this question











$endgroup$












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
















3












$begingroup$



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



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




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






thermodynamics







share|improve this question











$endgroup$












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














3












3








3





$begingroup$



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



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




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






thermodynamics







share|improve this question











$endgroup$





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



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




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






thermodynamics




thermodynamics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









Charlie Crown

368113




368113










asked 6 hours ago









NicciNicci

452




452












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


















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
















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




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










1 Answer
1






active

oldest

votes


















8












$begingroup$

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



Let me rewrite the differential as



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



where



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



and



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



A differential is exact if



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



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



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



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



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



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






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    I will figure it out one day, it can't be harder than thermo ;)
    $endgroup$
    – Charlie Crown
    5 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
    5 hours ago












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






  • 1




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






  • 2




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













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









8












$begingroup$

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



Let me rewrite the differential as



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



where



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



and



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



A differential is exact if



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



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



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



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



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



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






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    I will figure it out one day, it can't be harder than thermo ;)
    $endgroup$
    – Charlie Crown
    5 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
    5 hours ago












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






  • 1




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






  • 2




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


















8












$begingroup$

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



Let me rewrite the differential as



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



where



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



and



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



A differential is exact if



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



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



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



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



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



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






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    I will figure it out one day, it can't be harder than thermo ;)
    $endgroup$
    – Charlie Crown
    5 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
    5 hours ago












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






  • 1




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






  • 2




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
















8












8








8





$begingroup$

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



Let me rewrite the differential as



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



where



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



and



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



A differential is exact if



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



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



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



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



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



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






share|improve this answer











$endgroup$



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



Let me rewrite the differential as



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



where



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



and



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



A differential is exact if



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



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



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



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



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



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







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago









orthocresol♦

39.4k7114241




39.4k7114241










answered 6 hours ago









Charlie CrownCharlie Crown

368113




368113








  • 1




    $begingroup$
    I will figure it out one day, it can't be harder than thermo ;)
    $endgroup$
    – Charlie Crown
    5 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
    5 hours ago












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






  • 1




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






  • 2




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
















  • 1




    $begingroup$
    I will figure it out one day, it can't be harder than thermo ;)
    $endgroup$
    – Charlie Crown
    5 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
    5 hours ago












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






  • 1




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






  • 2




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










1




1




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




$begingroup$
I will figure it out one day, it can't be harder than thermo ;)
$endgroup$
– Charlie Crown
5 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
5 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
5 hours ago














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




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




1




1




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




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




2




2




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






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




















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國家象徵 事物 國號 國旗 國徽 國歌 國旗歌 國慶日 國家格言 國家代表色 國籍標誌 國家代碼 國家航空公司 國技 國樂 國璽 國石 （ 日语 ： 国石 ） 國菜 人物或拟人形象 國家元首 国父 国家化身 非人生物 國獸 （ 英语 ： List of national animals ） 國鳥 國花 國樹 查 论 编 本條目 論述以部分區域為主 ，未必具有普世通用的觀點。 請 協助補充內容 以避免偏頗，或討論本文的問題。   「 国玺 」重定向至此。關於法國法兰西岛大区上塞纳省的一个市镇，詳見「 索城 」。 太平天國玉璽 璽 是帝王用以署名的印章，武则天以后则多以 宝 名之。「璽」字的由來是秦始皇以和氏璧（一說是用藍田玉）製造玉璽，璽為印記之意。古時不少國家的君主均有專用的印章為印記，其中又以秦始皇的 傳國璽 最为知名。現代也以「璽」稱呼國家使用的印章，稱為 國璽 。 目录 1 中國 2 日本 3 參見 4 参考文献 5 外部链接 中國 秦、兩漢、新、魏、西晉、前趙、東晉、宋、南齊、梁、陳、隋、唐、後梁及後唐各朝沿用秦始皇之傳國玉璽，上面刻有「受命於天，既壽永昌」，以和氏璧製造，其中在王莽篡漢時，曾向當時的太皇太后王政君索取傳國玉璽，王政君非常憤怒，便將玉璽取出摔砸於地，為此傳國璽還碎了一角，雖然日後王莽以黃金補上缺角，卻仍看的出摔碎的痕跡。傳國玉璽在後唐末帝李從珂自焚之時失蹤。 在宋朝以前，皇帝通常也同时沿用秦朝 乘舆七玺 的制度，即在仅用于礼仪场合的传国玺外，另作皇帝日常办公用玺六方（皇帝之玺、天子之玺、皇帝行玺、天子行玺、皇帝信玺、天子信玺），重大礼仪时与传国玺一道作为礼器同列于御座御舆之前，象征皇权至上。惟唐初失传国玺，唐太宗另作一方宝玺代替，后复得传国玺，则自作宝玺易名“皇帝神玺”，传国玺称“受命玺”，与皇帝天子六玺合为乘舆八玺。五季纷扰，玺制混乱，至宋，皇帝就只用三方宝玺，一方为礼仪用玺受命宝（傳國璽此時已失踪，該玺為自作），另两方为日常办公所用的“书诏之宝”与“天下合同之宝”。宋徽宗时虽复活乘舆八玺之制，但为了强调自己的...
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