Internal energy change in isothermal process vs. adiabatic process with $\gamma$ approaching 1

Question

This might be a somewhat academic question (since there's no gas with $\gamma = 1$), but I am still interested in this.

Let's consider ideal gas and polytropic process with index $n$. We know, that when $n \to 1$ we get $\mathrm{d} E = 0$, i.e. the energy is constant.

On the other hand, for adiabatic process we have the following $$ \mathrm{d} E = - p \mathrm{d} V = - p_0 \frac{V_0^\gamma}{V^\gamma} \mathrm{d} V $$

One integration later... $$ E = - p_0 V_0^\gamma \int \limits_{V_0}^V \frac{\mathrm{d} V}{V^\gamma} = \frac{p_0 V_0^\gamma}{\gamma - 1} \left( \frac{1}{V^{\gamma-1}} - \frac{1}{V_0^{\gamma-1}} \right) = \frac{p V}{\gamma-1} \left( 1 - \left( \frac{V}{V_0} \right)^{\gamma-1} \right) $$ In the limit where $\gamma \to 1$ we get $$ E = p V \log (V_0 / V) $$

Isn't this a bit of a contradiction? If $n = 1$ (isothermal process) we get $E = \text{const}.$, however, if $n = \gamma$ and we let $\gamma \to 1$, then $E = p V \log (V_0 / V)$ which is not a constant. Note that $\gamma$ doesn't have to be strictly 1, we get a curve that's close to $p V \log V$ even when $\gamma = 1.1$.

Answer 1

See the mistake here you are doing is, taking $\gamma$ as just a coefficient. The thing is that $\gamma$ is a special value and is different for different gases, you can't just say I'll set it to 1, hence there is absolutely no controversy here. Because if you are setting $\gamma$ as ( which is theoretically impossible due to Mayer's relation) you are saying your process is a reversible isothermal, as well as adiabatic process.

Answer 2

There is no contradiction. Your final equation is in fact the work done for an ideal gas isothermal process, not the change in internal energy. When you made $\gamma=n=1$, $\gamma$ is no longer the ratio of specific heats because you changed the process from adiabatic to isothermal.

Hope this helps.