Problem — Internal energy of helium, monatomic and diatomic

Problem

A rigid balloon contains $2$ mol of helium (monatomic) at $T = 250$ K. Calculate the internal energy of the gas and state by how much it would increase if the same gas were diatomic at the same temperature.

Solution

For a monatomic gas ($f=3$) the internal energy is $E_\text{int} = \tfrac{3}{2}nRT$: $E_\text{int}^\text{mono} = \tfrac{3}{2}\cdot 2\cdot 8{,}31\cdot 250 \approx 6{,}23\cdot 10^{3}\ \text{J}$ If the same gas were diatomic ($f=5$) the internal energy would be $E_\text{int} = \tfrac{5}{2}nRT$: $E_\text{int}^\text{bi} = \tfrac{5}{2}\cdot 2\cdot 8{,}31\cdot 250 \approx 1{,}04\cdot 10^{4}\ \text{J}$ The increase (from the rotational degrees of freedom that are added) is the difference: $\Delta E_\text{int} = E_\text{int}^\text{bi} - E_\text{int}^\text{mono} = nRT \approx 4{,}16\cdot 10^{3}\ \text{J}$ $\ev{E_\text{int}^\text{mono} \approx 6{,}23\cdot 10^{3}\ \text{J}\ ;\quad E_\text{int}^\text{bi} \approx 1{,}04\cdot 10^{4}\ \text{J}}$

Connections

Topics: Teoria cinetica dei gas Concepts: Energia interna · Temperatura Skills: Analisi di casi limite e fantafisica