# Why is $\Delta U = mc_v \Delta T$ applicable in non-constant volume processes (ideal gas )

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 ?

• 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) 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.
\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,
$$\mathrm d U=nC_v\mathrm d T$$