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Is the formula $∆U = nC_v ∆T$ true for all processes? That is to say is this true for an isobaric, isochoric, adiabatic and isothermal process as well as any arbitrary process?

If not for which processes is it true?

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    $\begingroup$ It is true only for an ideal gas, for any process. $\endgroup$
    – Deep
    Commented Apr 13, 2017 at 6:01
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    $\begingroup$ In general: $$dU =C_{V}dT +\left[T\left(\frac{\partial P}{\partial T}\right)_{V} - P\right]dV$$ In case of an ideal gas the term proportional to dV is zero. $\endgroup$ Commented Apr 13, 2017 at 7:22

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Internal energy is a state function, i.e: for a given ideal gas it depends only on the thermodynamic parameters. Now consider an isobaric process from $A(P,V_1,T_1)$ to $B(P,V_2,T_2)$. Clearly heat procduced $dQ = nC_p.dT$, and work done $dW = p.dV$. For an ideal geas $PV = n.R.T$. So $pdV + Vdp = nR.dT$. But $dP$ is zero for an isobaric process, hence $dW = nRdT$. We know: $dQ = dU + dW$ => $dU = nR.(C_p - R)dT = nR.C_v.dT$. Now consider any other path from $A$ to $B$. The change is internal energy must be same since its a state function and depends only on the final and initial configuration of the system. Thus for all processes between any 2 given states, the internal enengy change is given by $dU = n.C_v.dT$

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The most common, yes, most common application of the formula $\Delta U = nC_v(T)\Delta T$ is actually not for an ideal gas but as a finite model of an ideal heat source/sink. i.e., thermostat being attached to the thermodynamic system under consideration. Considered that way the heat capacity $nC_v(T)$, or just $C(T)$, being dependent only on the temperature assures that its entropy change is exactly $\int\frac{q}{T}=\int\frac{C(T)}{T}dT$. The larger the heat capacity $C(T)$ is relative to that of the system under consideration to which the thermostat is attached the closer the thermostat will have nearly constant temperature.

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