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In thermodynamics we have an equation for the change in internal energy:

$$\mathrm dU = nC_V\,\mathrm dT\tag{1}$$

$$\int{\mathrm dU} = \int{nC_V\,\mathrm dT}\tag{2}$$

Since $C_V$ is independent of temperature, for an ideal gas

$$ΔU = nC_VΔT\tag{3}$$

Why is $C_V$ independent of temperature for an ideal gas? On the other hand, in case of solids and liquids, $C_V$ may or may not depend on temperature. Why?

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    $\begingroup$ Ideal gases are made of moving points, without any volume, and without any internal degree of freedom like vibration and rotation. So $C_v$ is only due to the three degree of freedom called translations, and this number does not depend on temperature. $\endgroup$
    – Maurice
    Commented Apr 16, 2021 at 10:03
  • $\begingroup$ In engineering, we consider the heat capacity of a real gas in the ideal gas limit to be a function of temperature, and we call this the ideal gas heat capacity. $\endgroup$ Commented Apr 16, 2021 at 11:42

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The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.

The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, vibrational and rotational energy contribution depends on ratio of mean energy per degree of freedom $E=\frac{1}{2} \cdot k_\mathrm{B} \cdot T$ versus the energy of quantization steps.

The capacity raises as rotation energy contribution kicks in, and later vibration energy as well, and finally electronic energy, as their quantization steps grows in this order.

There is a similar temperature dependent effect for liquids and solids, where quantized vibration is essentional, especially for solids.


The classical model of ideal gas does not take into account quantum effects. By warming up of an ideal gas, the mechanical energy gets evenly distributed between energy of linear motion, bond vibration and molecular rotation, proportionally to absolute temperature.

Quantum effects, taking part in behaviour of real gases, liquids and solids affects energy distribution. Vibration, rotation or electron energy are allowed by laws of quantum mechanics to have just some energy levels.

Imagine you throw small stones upwards to a steep hill. Stronger you throw it, the higher it would stay. This is analogous to an ideal gas.

Now you try to throw such stones to open windows of a tall building. If you make weak throwing, it will not flow even to the nearest window and will fall back. Stronger and stronger throwing ( = higher temperature ) would eventually cause some stones would reach at least the 2nd floor. Later even 3rd one etc. This is like quantization of possible energies for real gases and condensed phases.

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  • $\begingroup$ Can you please explain in simple language? I am just a Grade 12 student and I do not have higher knowledge about science. $\endgroup$
    – user281837
    Commented Apr 16, 2021 at 12:16
  • $\begingroup$ Well, asking about temperature dependency of heat capacity does not match your statement much. :-) I will try... $\endgroup$
    – Poutnik
    Commented Apr 16, 2021 at 12:50
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An ideal gas is defined as one of non-interacting point particles and its internal energy depends on temperature alone, thus the slope of a plot of $U$ vs $T$ is a straight line with slope $C_V$.

A real gas (or liquid or solid) is made up of molecules and these have internal energy levels due to a molecule's ability to rotate and vibrate. At low temperatures these energy levels are mostly not excited (or occupied) and so the internal energy is low, however, as the temperature is increased more levels become excited and the internal energy increases. These levels are not equally spaced from one another (due to the nature of the molecule) and so the internal energy initially rises slowly with temperature then more rapidly, i.e. initially the slope of $U$ vs $T$ is small but gradually increases until at v high temperature the slope becomes effectively constant. Thus as the heat capacity is the slope of $U$ vs $T$ it is initially small and increases to a constant value at high temperature.

(At very, very low temperatures all molecules are in their lowest energy levels and a slight increase in temperature may not be enough to excite any other levels so the heat capacity remains at zero (from zero K up to a few degrees K) then increases as more levels become excited as the temperature increases)

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