Internal energy is purely kinetic

Why does the internal energy of an ideal gas coincide with kinetic energy alone? A minimal model of a “gas” helps to see this: two balls connected by a spring falling under gravity. The spring force is internal to the system, the weight is external. The system’s total energy then splits into several contributions:

$E_{\text{tot}} = E_{\text{cin,CM}} + \underbrace{E_{\text{cin,int}} + E_{\text{pot,molla}}}_{E_{\text{interna}}} + E_{\text{pot,g,CM}}$

The internal energy is the part that does not depend on the overall motion (of the centre of mass) nor on the system’s position in the gravitational field: it is the system’s “own” energy as seen from within, the sum of the internal kinetic energy and the potential energy of the interactions between its parts.

Now comes the key point: in an ideal gas the particles do not interact, so $E_{\text{pot,molla}} = 0$ (there is no “spring” between the molecules). It follows that the internal energy is purely kinetic:

Key consequence

$E_{\text{interna}} = E_{\text{cin,int}} \quad(\text{gas ideale})$ With no interactions, the internal energy of an ideal gas is only thermal agitation: it is this identity that, in the next section, allows the internal energy to be linked directly to the temperature.

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Topics: Teoria cinetica dei gas Concepts: Energia interna · Energia cinetica Skills: Interpretazione micro-macro

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