Partial derivatives and thermodynamics

[voila]

Hi all.

Suppose I have the ideal gas law $$P=\frac{RT}{v}$$If I'm asked about the partial derivative of P with respect to molar energy $u$, I may think "derivative of P keeping other quantities (whatever those are) constant", so from the formula above I get $$\frac{\partial P}{\partial u}=0$$However, one of the consequences of the ideal gas law is that energy is only dependant on T as in $u=cRT$. If I reintroduce this in the ideal gas law, I get $$P=\frac{u}{cv}$$ What if I know want to take the partial derivative of P with respect to u? Then $$\frac{\partial P}{\partial u}=\frac{1}{cv}\neq 0$$which doesn't match the result above.

What am I doing wrong? I've tried to show my general problem using this particular example, but you may guess this kind of confusion causes problems everywhere in thermodynamics.

 

[Charles Link]

When you write $P=P(T, V)$, you can readily take $(\frac{\partial{P}}{\partial{T}})_V$ and $(\frac{\partial{P}}{\partial{V}})_T$, but if you want to find $(\frac{\partial{P}}{\partial{x}})_y$, you need to use the chain rule, and also consider both parameters $T$ , and $V$ to see if they might have an $x$ dependence. In full detail (this will take a minute because it's not easy to do in Latex) $(\frac{\partial{P}(T,V)}{\partial{x}})_y=(\frac{\partial{P}}{\partial{T}})_V (\frac{\partial{T}}{\partial{x}})_y +(\frac{\partial{P}}{\partial{V}})_T (\frac{\partial{V}}{\partial{x}})_y$. They teach this in detail in an advanced calculus class, but if you study this example, it might help clear up the difficulty. $\\$ In your example above, I presume they are writing $P=P(u, V)$ and finding $(\frac{\partial{P}}{\partial{u}})_V$.

 

[voila]

When you write $P=P(T, V)$, you can readily take $(\frac{\partial{P}}{\partial{T}})_V$ and $(\frac{\partial{P}}{\partial{V}})_T$, but if you want to find $(\frac{\partial{P}}{\partial{x}})_y$, you need to use the chain rule, and also consider both parameters $T$ , and $V$ to see if they might have an $x$ dependence. In full detail (this will take a minute because it's not easy to do in Latex) $(\frac{\partial{P}(T,V)}{\partial{x}})_y=(\frac{\partial{P}}{\partial{T}})_V (\frac{\partial{T}}{\partial{x}})_y +(\frac{\partial{P}}{\partial{V}})_T (\frac{\partial{V}}{\partial{x}})_y$. They teach this in detail in an advanced calculus class, but if you study this example, it might help clear up the difficulty. $\\$ In your example above, I presume they are writing $P=P(u, V)$ and finding $(\frac{\partial{P}}{\partial{u}})_V$.

The problem I seem to have is the following. Say I want the variation of P with respect to u keeping the v and T constants. Starting from the ideal gas law: $$\left(\frac{\partial P}{\partial u}\right)_{T,v}=0$$ since T and v re constants.

However, if I start from $P=\frac{u}{cv}$, I get $$\left(\frac{\partial P}{\partial u}\right)_{T,v}=\frac{1}{cv}\left(\frac{\partial u }{\partial u}\right)_{T,v}=\frac{1}{cv}$$ which is the same I had above.

 

[voila]

It looks like you are confusing partial and total derivatives.
It is true that $\frac{\partial}{\partial u} \left( \frac{RT}{v} \right) =0$ and $\frac{\partial}{\partial u} \left( \frac{u}{cv} \right) =\frac{1}{cv}$
It is also true that $\frac{dP}{du}=\frac{\partial P}{\partial T}\frac{\partial T}{\partial u}=\frac{1}{cR}\frac{\partial P}{\partial T}=\frac{1}{cR}\frac{R}{v}=\frac{1}{cv}$

Then how is it that $\left(\frac{\partial P}{\partial u}\right)_{T,v}$ can give a different result depending on how I pick it?

 

[Charles Link]

@voila In the case above, you have two degrees of freedom in this problem. You can write $P=P(T,V)$ and you can also write $P(u, V)$. You cannot take a derivative w.r.t. $u$ when you are keeping both $T$ and $V$ constant. That completely prevents you from varying $u$, because $u=u(T,V)$.

 

[kuruman]

Then how is it that $\left(\frac{\partial P}{\partial u}\right)_{T,v}$ can give a different result depending on how I pick it?

Sorry, I deleted my post because you already are in good hands. @Charles Link will see you through.

 

[voila]

@voila In the case above, you have two degrees of freedom in this problem. You can write $P=P(T,V)$ and you can also write $P(u, V)$. You cannot take a derivative when you are keeping both $T$ and $V$ constant. That completely prevents you from varying $u$, because $u=u(T,V)$.

I know keeping T, v constants forbid me from changing the energy, but still I applied the definition correctly, right? The thing is, I know you are right: "the variation of pressure with respect to energy, keeping v and T constants" makes no sense because it is physically impossible. However, it kind of seems like an ad hoc argument. Not only that, but what about in other cases where I may be doing something inappropriate like this but not quite as obvious? Seems like I can run into trouble even if I just stick to the definition.

 

[voila]

@Charles Link Let me ask it to the point: is $\left(\frac{\partial u}{\partial u}\right)_T=0$ (assuming $u=cRT$)??

I think that could kind of make sense, though I had never considered it. Maybe knowing that I may avoid future problems (but I can't promise it!).

 

[Charles Link]

I know keeping T, v constants forbid me from changing the energy, but still I applied the definition correctly, right? The thing is, I know you are right: "the variation of pressure with respect to energy, keeping v and T constants" makes no sense because it is physically impossible. However, it kind of seems like an ad hoc argument. Not only that, but what about in other cases where I may be doing something inappropriate like this but not quite as obvious? Seems like I can run into trouble even if I just stick to the definition.

One suggestion would be to study the application of the partial derivative in post 2 in detail, and I think you will find with practice, that any ambiguities go away. One thing I could add to post 2 is that it is assumed $x=x(T,V)$ and $y=y(T,V)$, $x$ and $y$ are alternative parameters that can be used to specify the system, (in this case $x=u$ and $y=V$). If you are taking a course in thermodynamics, you will likely see a lot of these partial derivatives, but once you get some practice with them, they should start to become more commonplace.

 

[Charles Link]

@Charles Link Let me ask it to the point: is $\left(\frac{\partial u}{\partial u}\right)_T=0$ (assuming $u=cRT$)??

I think that could kind of make sense, though I had never considered it. Maybe knowing that I may avoid future problems (but I can't promise it!).

A good puzzle=I don't know that this one has an answer, because the derivative in that form and holding T constant is invalid. I don't know that such an expression could ever arise. $\\$ Editing: $(\frac{\partial{u}}{\partial{u}})_V=1$, and $(\frac{\partial{u}}{\partial{V}})_u =(\frac{\partial{u}}{\partial{V}})_T=0$. For the case of $(\frac{\partial{u}}{\partial{u}})_T=(\frac{\partial{u}}{\partial{u}})_u$, I think is an impossible expression that is simply invalid. (Holding $T$ constant for the case you have is the same as holding $u$ constant. We have in general $u=u(T,V)$, but in this case $u=cRT$ with no $V$ dependence.).