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In Internal energy of a gas ($U$), we have to include ALL types of energies possible like $KE$, bond energies, nuclear interaction energies, etc..,

However we have derived that $U = (f/2)nRT$ from only the $KE$ in the equipartion theorem ignoring all other types of energies?

Even if we have an ideal gas, there should still be some nuclear binding energy?

Or from this.. we know that for all proceses $\mathrm{d}U = nC_v\mathrm{d}T$ integrating both sides from $0$ to $T$, we should get $$U(\text{at }T) - U(\text{at } 0K) = nC_v(T-0)$$ So at a particular $T$ $$U(T) = nC_vT + U(0) \quad[C_v = fR/2]$$ This gives $$U = (f/2)nRT + U(0)$$ This $U(\text{at }0 K)$ should include all those types of energies which are not taken in account of motion like all that bond energy, nuclear binding energy.

So why do we not account for all that and say $U = (f/2)nRT$ only?

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    $\begingroup$ Here we do consider more about energy change. Energy change will be used to compute other parameters such as work. For most processes, potential energy or bond energy do not change and thus are not included. But that is not always the case. Sometimes, we do consider potential energy. And in mixture or chemical reaction, we also consider other energy when the changes are important. $\endgroup$ – user115350 Jul 20 '16 at 15:02
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    $\begingroup$ The transition energies for nuclear processes are typically of the order of MeV. Room temperature is around 1/40 eV, so the contribution to the heat capacity will be essentially 0 until you reach very high temperatures. This means that the nuclear binding energy will simply be a constant shift to the internal energy, which makes no difference. $\endgroup$ – By Symmetry Jul 20 '16 at 15:03
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If you consider a very diluted atomic gas with $N$ atoms (say hydrogen atoms), it won't be equivalent to an ideal gas with $3N$ degrees of freedom for all temperature ranges. Instead, at very low temperature but very low density, you will expect these atoms to form molecules e.g. $H_2$ molecules and the internal energy will be $5N k_B T/4$ as there are $5N/2$ distinct degrees of freedom. As you increase the temperature, you will reach a monoatomic gas regime where the internal energy will read approximately $3N k_B T/2$ as there are about $3N$ degrees of freedom. If you increase the temperature even further, still in the very dilute regime, the gas will be fully ionised and if we talk about hydrogen, you will end up with an internal energy that is $3N k_B T$ as there are now $2N$ particles moving around. If you continue to increase the temperature, you will start probing nuclear energies and you will start exciting energy levels of the proton itself (that's quite huge energies in that case but for other types of atoms that can happen at energies in the range of gamma rays energies). So, to summarise, the ideal gas law with a certain number of "active" degrees of freedom depend on the temperature range you are probing. Thus, in substance, all the details you are talking about do matter if you want to study an unbounded range of temperatures. If however, you are interested in temperatures that stick to a quasi-invariant regime (where the number of degrees of freedom is approximately fixed), then you can "forget" about those details.

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