# $\Delta U = n C_v \Delta T$ non-constant volume [duplicate]

In the formula $\Delta U = C_v\Delta T$, we get the constant volume heat capacity by assuming constant volume conditions. However, I've understood that a constant volume is not required when using the formula.

I tried understanding the solution through several other posts but it still doesn't make sense to me. Unless my base knowledge is wrong (I have a hard time understanding thermodynamics), why are we allowed to use $\Delta U = C_v\Delta T$ when volume changes?

• For solids and liquids, $c_v$ is almost equal to $c_p$. It is easy to measure $c_p$. – user137289 Feb 4 '18 at 22:40
• " I've understood that a constant volume is not required when using the formula." please provide a source – pentane Feb 5 '18 at 0:03
• physics.stackexchange.com/questions/336945/… Maybe I don't understand what they're saying. – Dylan Patel Feb 5 '18 at 1:13
• Note that you can make your equations pretty by wrapping them in dollar -signs. Furthermore, aren't you missing an $n$ in the formulas written in the text? – Steeven Feb 5 '18 at 6:56
• I've just offered an answer to a later question, titled "Internal energy of an ideal gas". The question was almost the same as yours. – Philip Wood Feb 5 '18 at 15:12

For an ideal gas, the equilibrium value per unit mass does not depend on two intensive properties; it depends only on temperature, but not specific volume. So the change in internal energy between two thermodynamic equilibrium states depends only on the temperatures at these two states. If we change from state A to state B by first changing the temperature at constant volume, the difference in internal energy per unit mass will be $C_v(T_B-T_A)$. If we then change the specific volume from $V_A$ to $V_B$, there will be no change in internal energy. Since the internal energy is a function of state, the change in internal energy in going from $(T_A,V_A)$ to $(T_B,V_B)$ will be just $C_v(T_B-T_A)$, irrespective of whether the process is at constant volume.