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The specific heat of a real gas, unlike an ideal gas, depends on temperature. How can we physically understand this? Thanks.

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  • $\begingroup$ It depends on how one defines an ideal gas. We engineers include temperature dependence of specific heat in our definition of ideal gases. Physicists, on the other hand do not. Engineers regard ideal gases as the limiting behavior of real gases at low specific volume. $\endgroup$ – Chet Miller Sep 25 '17 at 13:21
  • $\begingroup$ A physicist with a good knowledge of thermodynamics should know that the thermodynamic ideal gas definition does not require that the specific heat capacity is constant. Thus engineers and physicists agree if the latter have done their homework. $\endgroup$ – Andrew Steane Nov 29 '18 at 22:15
  • $\begingroup$ Based on the answers so far, there seems to be disagreement about if the question is about non-ideal gases (i.e., those with interactions between particles) or about the "freezing out" of vibrational and rotational degrees of freedom in gases of non-interacting molecules. $\endgroup$ – Endulum Nov 30 '18 at 15:54
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The heat capacity (specific heat times the mass of the gas) is defined to be how much the internal energy of the gas changes due to changes in temperature, which can be done either at constant pressure $$ C_P=\left.\frac{\partial U}{\partial T}\right)_P $$ or at constant volume $$ C_V = \left.\frac{\partial U}{\partial T}\right)_V. $$ Notice that both $C_P$ and $C_V$ will be constant if the internal energy $U$ is a linear function of the temperature. This is of course the case for the ideal case, for which $$ U_{\mathrm{ideal}} = \frac{3}{2}N k_B T, $$ where $N$ is the number of particles (I've assumed monatomic here, but the linear dependence in $T$ is true of diatomic and polyatomic ideal gases as well). Microscopically, this form of the internal energy results from the fact that all of an ideal gas's energy is kinetic. Real gases, however, also have internal energy due to the potential energy of the interactions between particles. So, the total internal energy of a real gas is $$ U_{\mathrm{real}} = U_{\mathrm{ideal}} + U_{\mathrm{pot}}. $$ In textbooks, the potential part is often called the "excess" internal energy. The way it depends on temperature is different for every gas, since it depends on the details of their interactions. In general, though, it will not depend linearly on $T$. Then, the heat capacity (either $C_V$ or $C_P$) also has two parts: $$ C = \frac{\partial U}{\partial T} = \frac{3}{2}Nk_B + \frac{\partial U_{\mathrm{pot}}}{\partial T} $$ for a montaomic gas.

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  • $\begingroup$ In some cases $ U_{\mathrm{pot}}$ is not a function of T, and C is identical to that for ideal gases. $\endgroup$ – juanrga May 8 '19 at 10:38
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Ideal gas in real sense is not a physical reality. We treat it as monatomic gas. In monatomic gases only translational degree of freedom is effective,which is three. At any high temperature rotational and vibrational degree of freedom are not effective. Thus its specific heat is independent of temperature. On the other hand real gases may be monatomic,diatomic or in general polyatomic. In polyatomic gases vibrational degree of freedom becomes effective at higher temperature. For example in case of diatomic gases two vibrational degree of freedom becomes effective at higher temperature and their molar sp heat becomes 7/2 R from 5/2R.

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The tempurature dependence of heat capacitance was one of the historical failures of classical physics, which predicts constant heat capacitance (i.e. $\frac{\partial}{\partial T} \frac{n}{2}k_\text{B}T$). The temperature dependence of solids and gases was not explained until the advent of quantum mechancis, in which the vibrational states are quantized.

An ideal gas is a classical gas and will have constant heat capacitance.

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