I can completely accept that the internal energy of an ideal gas is the function of the temperature only, namely $$U = \frac{f}{2} n R T$$

and that we can define 2 quantities ($c_v$ and $ c_p $) that can give the change in the temperature due to heat added when volume or pressure is held constant( $ dQ=mc_{\square} dT$), but why is $$ dU =m c_v dT$$ valid for a general state-change when volume is not constant ?

  • 1
    $\begingroup$ You answered your own question. If U is a linear function of temperature, then any change in U is proportional to the change in temperature. (Note: m is proportional to n) $\endgroup$
    – R.W. Bird
    Jun 8 '20 at 15:03

Any two states on the $P-V$ diagram can be connected by a combination of an isothermal (constant temperature) path and an isochoric (constant volume) path. Since, the internal energy of an ideal gas is a state function, so it doesn't matter which path you take from one point to another as long as the starting and the ending points are the same.

Thus, once you set up a path, which is a combination of isothermal and isochoric paths, you can directly add up the internal energy change in both the paths to get the total internal energy change. Thus

\begin{align} \Delta U_{\text{total}}&=\Delta U_{\text{isochoric}} + \Delta U_{\text{isothermal}}\\ &=nC_v \Delta T + 0\\ \Delta U_{\text{total}}&=nC_v \Delta T\tag{1} \end{align}

The $\Delta U_{\text{isothermal}}$ term evaluates to $0$ because $\displaystyle \left(\frac{\partial U}{\partial V} \right)_T=0$ for an ideal gas. Now since the equation $(1)$ is true for any general process, we can also convert it into the differential form,

$$\mathrm d U=nC_v\mathrm d T$$


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