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So we are doing calculations with the first law of thermodynamics: $$U = q + w$$

And we have been studying $\mathrm dU(T,V)$. So it is true that for some equations of state other than the ideal gas equation, $\mathrm dU(T)$ only, so therefore it follows that $U= C_V(T_2-T_1)$. So I am wondering is it even possible that $C_V$ is used for non-ideal gasses? And would it also be equal to $\frac 32R$? And if $C_V$ can be used, why? (Also is this true for $C_p$?)

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  • $\begingroup$ You can still use the heat capacities as they are defined by the equations: $C_V = (\partial U / \partial T)_V$ and $C_p = (\partial H / \partial T)_p$. The fact that $U = U(T)$ for an ideal gas is just a special case, so in general, the heat capacities of a real gas will not be the same as for the ideal gas. $\endgroup$ – orthocresol Oct 12 '15 at 22:39
  • $\begingroup$ By the way: it's $\Delta U = q + w$ with the triangle. $\endgroup$ – orthocresol Oct 12 '15 at 22:53
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We usually don't use the real gas $C_V(T,p)$ directly in calculating the change in U in going from thermodynamics equilibrium state A to thermodynamic equilibrium state B, if both these states are beyond the ideal gas region. Starting from state A, what we do first is use the real-gas p-v-T equation to determine the change in U in going from the initial pressure $p_A$ to a low pressure (or large specific volume) in the ideal gas region $p^*$ at constant temperature $T_A$. Then, our second step is to hold the pressure constant at pressure $p^*$ and raise the temperature from $T_A$ to $T_B$, using the ideal gas heat capacity to determine the change in internal energy for this step (since we are now in the ideal gas region). Finally, we hold the temperature constant at $T_B$ and raise the pressure from $p^*$ to $p_B$ using the real-gas p-v-T equation again to determine the change in U. The sum of the changes in U for these three steps adds up to the change in U in going from $(T_A,p_A)$ to $(T_B,p_B)$.

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