Changes in internal energy at

  1. At constant temperature, $$d U = \left({\partial U \over \partial V}\right)_TdV$$

  2. At constant volume, $$d U = \left({\partial U \over \partial T}\right)_VdT$$

  3. When both temperature and volume is changing, $$d U = \left({\partial U \over \partial V}\right)_TdV +\left({\partial U \over \partial T}\right)_VdT$$

I want to know how do we get these equations.

I think I can speculate a bit about how do we get the first and second equation.

For the first equation, since the temperature is kept constant, we can plot a U-V graph.

Tell me if I am wrong here but for that graph the slope at a given point is $\displaystyle \left({\partial U \over \partial V}\right)_T$.

Now since the change is a infinitesimal change, the slope of tangent at that point is given by $\displaystyle \left({\partial U \over \partial V}\right)_T$.

So $\displaystyle {\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$, since the change is infinitesimal, $\Delta U = dU, \Delta V = dV$ and substituting this we will get out equation 1. Same can be said for equation 2.

I have no idea how do we get the equation 3, I think it is a clever use of product rule or something like that.

  • Is my understanding for the equations 1 and 2 correct ?
  • How do we get equation 3 from 1 and 2 ? because simply adding 1 and 2 does only give the RHS of 3 not the LHS.

1 Answer 1


One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general

$$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$

There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.

$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i \overset{\mathrm{d}n_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V.$$

A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, e.g., $S = S(T,V)$. From the equation

$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V$$

you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.

  • $\begingroup$ So my way of thinking of how first and second equation is incorrect ? $\endgroup$
    – user31607
    Feb 11, 2017 at 19:11
  • $\begingroup$ Also can please you explain a bit more about how did you get the second equation. I did not understand that. $\endgroup$
    – user31607
    Feb 11, 2017 at 19:12
  • $\begingroup$ @A---B You are right that if temperature is constant, then you are essentially in a sliver where $U$ varies as $V$ is changed. And yes, the derivative $\left({\partial U \over \partial V}\right)_T$ will thus be the "slope" in that particular direction. Using $${\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$$ is not a rigorous way of thinking about it though. I would focus on the general case (the total or exact differential) and derive equations $1$ and $2$ from there. $\endgroup$ Feb 11, 2017 at 19:23
  • 1
    $\begingroup$ @A---B Oh, just a random variable. Later on I took $x_1 = T, x_2 = V, x_3 = n_1$ and so on. $\endgroup$ Feb 11, 2017 at 19:39
  • 1
    $\begingroup$ @A---B Glad I could help! $\endgroup$ Feb 11, 2017 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy