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.

  • $\begingroup$ I'm voting to close this question as off-topic because it is a mathmatics question. $\endgroup$
    – A.K.
    Mar 8 '19 at 23:42
  • 11
    $\begingroup$ Not all chemistry is done in a vial. $\endgroup$
    – B. Kelly
    Mar 9 '19 at 2:45

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

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


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


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

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.

  • $\begingroup$ I don't think you should basically have revealed the full solution. $\endgroup$ Mar 8 '19 at 18:35
  • 1
    $\begingroup$ In my experience people mess up the differentiation and then think they did everything wrong $\endgroup$
    – B. Kelly
    Mar 8 '19 at 18:38
  • 2
    $\begingroup$ Full solutions are fine! We encourage them: chemistry.meta.stackexchange.com/questions/4287/… $\endgroup$
    – orthocresol
    Mar 8 '19 at 21:37
  • $\begingroup$ @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
    Mar 8 '19 at 22:18
  • 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. $\endgroup$
    – B. Kelly
    Mar 8 '19 at 22:29

For a given function, $F(x,y,z,...)$, it's differential $\text{d}F$ is given by: $$ \text{d}F = \left(\frac{\partial F}{\partial x}\right)_{y,z} \text d x +\left(\frac{\partial F}{\partial y}\right)_{x,z} \text d y + \left(\frac{\partial F}{\partial z}\right)_{x,y} \text d z +\; ... $$

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:

$$ \mathrm dp(T,V)=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV $$

If $\mathrm dp$ is an exact differential, that would mean that:

$$ \left(\frac{\partial p}{\partial T}\right)_{V} = \frac RV\ \text{and } \left(\frac{\partial p}{\partial V}\right)_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) $$

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.


Taking indefinite integrals of the suspected derivatives:

$$\int \frac RV\ \mathrm d T = \frac{RT}{V} + g(V) \\ \int \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \mathrm d V = \frac{-2a+RT}{V} + h(T) \\ \to p(T,V) = \frac{RT-2a}{V} + c $$


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:

$$ \frac{\partial }{\partial T }_V \left(\frac{\partial p }{\partial V }\right)_T = \frac{\partial }{\partial V }_T \left( \frac{\partial p }{\partial T } \right)_V $$

Assuming: $$ \left(\frac{\partial p}{\partial T}\right)_{V} = \frac RV\ \\ \to \frac{\partial }{\partial V }_T \left( \frac{\partial p }{\partial T } \right)_V = -\frac{R}{V^2} $$


$$ \left(\frac{\partial p}{\partial V}\right)_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \\ \to \frac{\partial }{\partial T }_V \left( \frac{\partial p }{\partial V } \right)_T = -\frac{R}{V^2} $$

Both terms of the differential $\mathrm dp$ imply the same mixed second derivative, hence $\mathrm dp$ is an exact differential.


Not the answer you're looking for? Browse other questions tagged or ask your own question.