According to my Equation sheet, $$\delta U=\left(\frac{\delta U}{\delta T}\right)_V\,\delta T+\left(\frac{\delta U}{\delta V}\right)_T\,\delta V=C_V\,\delta T-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$$ Where $\delta U$ is change in internal energy.

So one of the homework solutions said that for a system with a constant Temperature, $\delta T=0$ $\Delta U=0$. And I remember one of the TAs saying that $\Delta U=C_V\,\delta T$ and since $\delta T=0$ the value goes to zero, which makes sense. But she didn't even include the second term: $-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$.

I was trying to make sense of the derivative but I'm really having trouble with the second term. The small letter $U$ outside the parenthesis is supposed to mean constant $U$.

Why am I allowed to ignore the second term?

The equation for enthalpy is very similar, and I have the same problem:

$$\delta H=\left(\frac{\delta H}{\delta T}\right)_p\,\delta T+\left(\frac {\delta H}{\delta H}\right)_T\,\delta H=C_p\,\delta T-C_p\left(\frac{\delta T}{\delta p}\right)_H\,\delta p$$

  • 2
    $\begingroup$ $\Delta T = 0$ only implies $\Delta U = 0$ under certain conditions (the same conditions apply to the differentials $\mathrm{d}T$ and $\mathrm{d}U$). The single most important condition is that the system must consist of only an ideal gas, in which case the partial derivative $(\partial T/\partial V)_U$ is equal to zero. I wrote quite a bit about this particular topic here. $\endgroup$ Commented Oct 18, 2016 at 21:45
  • $\begingroup$ Thanks for the link. it is very helpful. I will learn to derive the thermo equations! The problem did specify that the gas was ideal. $\endgroup$
    – Jess L
    Commented Oct 18, 2016 at 21:52

1 Answer 1


The second term in each of your two equations can be expressed solely in terms of the equation of state for the solid, liquid, or gas:$$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$ In the case of an ideal gas, the right hand sides of both these equations are equal to zero.


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