Free vs isothermal expansion

Reviewing isothermal expansion

Consider the isothermal expansion of an ideal gas, doubling the initial volume from $V_i=V_0$ to $V_f=2V_0$. Write an expression for the work, $\Delta W$, in terms of $V_0$, $T_0$, and $n$.
Temperature is constant for an isothermal process. The final volume, $V_f=2V_0$ is twice the initial volume $V_0$. $$\Delta W = nRT_0\ln\left(\frac{V_f}{V_i}\right)=nRT\ln 2$$ For 1 kilomole at $T_0=300K$,this works out to
$nRT\ln 2= 1\text{ kMole }*8.314\times 10^3\text{ J/(kMole*K)}*300\text{ K }*0.693$
= $1.73 \times 10^6$ Joules.
But for an ideal gas, the internal energy only dependss on temperature. Using the first law, what is the heat transfer $\Delta Q$? Does the system absorb heat or do the surroundings absorb heat?
First law: $\Delta U = \Delta Q - \Delta W$.
1. This is an ideal gas.
2. The internal energy of an ideal gas only depends on temperature.
3. This is an isothermal process in which $\Delta T=0$
$\Rightarrow \Delta U=0$, and therefore $\Delta W = $ $\Delta Q=1.73 \times 10^6$ Joules . The convention is that positive $\Delta Q$ means heat has been absorbed by the sytem, from its surroundings.
What are the final volume and final temperature of the system?
Initial volume and temperature were $V_0$ and $T_0$. Final volume and temperature are $2V_0$ and $T_0$

Free expansion

Now, consider the "free expansion" of an ideal gas from $V_0$ to $2V_0$. Imagine that we have a rigid, thermally insulated container , the system, with a total volume $2V_0$. Initially, there is a flimsy partition separating the two sides of the container in half, and all the gas occupies one half with a volume of $V_0$ and temperature $T_0$:

A remote controlled dart (of vanishingly small volume) from the vacuum side is launched, puncturing the partition, and the gas explodes out, eventually filling the whole volume, $2V_0$.

Did this system give off or absorb heat from its surroundings?
No because the walls of the container are insulated, allowing no heat to flow either in or out of the system. $\Delta Q = 0$.
Did this system perform work on its surroundings or have work performed on it?
No because the walls of the system are rigid meaning $dV=0$, and since $\Delta W=\int P\,dV$, it's not possible that any work was done by the surroundings or by the system. $\Delta W = 0$.
Using the first law, calculate the change to the gas' internal energy, and therefore, calculate the final temperature of the system?
According to the first law, $\Delta U=\Delta Q - \Delta W$. But we just argued above that $\Delta Q=0$ and $\Delta W=0$. And $\Rightarrow\Delta U=0$. Since this is an ideal gas, and the internal energy only depends on temperature, $\Delta U=0\ \Rightarrow \Delta T = 0$ So, the final temperature is $T_0$, and the final volume of the gas is $2V_0$.

And now, a calculation of the entropy change, if possible.

Which process is reversible? Which one irreversible?
The isothermal expansion is reversible. Running the "movie" of the free expansion backwards would mean that all the gas molecules in the right half of the $2V_0$ container would simultaneously approach the left side through a hole in the partition. _Unlikely_ So the free expansion is irreversible.
If you can, calculate the change in entropy using the reversible process.
For a reversible process, we can find the change in entropy by integrating an expression involving the reversible heat flow, $dQ_r$. This is a reversible process, so... $$\begineq \Delta S&=\int\frac{dQ_r}{T}\\ &=\frac1T\int dQ_r\\ &=\frac{\Delta Q_r}{T} \endeq $$ The second line is possible, because we are considering a reversible isothermal process. Plugging in our numbers for 1 kiloMole of an ideal gas at 300 K:
$\Delta S$ $ =1.73 \times 10^6\text{ Joules }/300\text{ K}=$ 5,767 Joules/K.
Is it possible to know the change in entropy of the irreversible process?
Yes! Since $dS$ is an exact differential, the integral of $\Delta S=\int dS$ is path independent. The initial conditions $V_0,T_0$, and the final conditions $2V_0,T_0$ were the same for both the reversible isothermal expansion, and for the irreversible free expansion of the gas. So the change in entropy for the free expansion is the same, +5,767 Joules/K, as the isothermal expansion.

Spakovsky's lecture notes have a thorough discussion of these two expansions. He points out that the net change of entropy of the universe (system + surroundings) is zero for the isothermal expansion, but positive for the free expansion.