Yes quite! The ideal gas law (and in general any state equation) holds only on equillibrium while the 1st law (and all the rest) hold in general. So, your anslysis mr James Hoyland is inacurrate.
Mr Steven, the post includes the word "expansion" so volume changes by assumption.
And mr or ms PhysC, the first two cases are correct. About the third, expansion would occur by pulling the piston thus removing energy from the system and causing decrease on temperature
In free expansion there is no work done as there is no external external pressure.
That's certainly true, in fact free expansion is an irreversible process in which a gas expands into an insulated evacuated chamber, you can think of it like ann container with a piston and the gas is left to expand in vacuum.
Hence, it is evident that $P_{ext}=0$ during the expansion, so the $W=0$. Now for a ideal gas this process occurs quickly so there is not temperature rise as well, so $dT=0$, so as per first law of thermodynamics $Q=0=W$ and since internal energy is only a function of temperature, so $dU=0$ as well.
So that was for free expansion.
Now for isothermal expansion:
Here if we see and characterise the states after and before isothermal expansion we can see: $$T_1=T_2$$ but other quantities differ,like the external pressure is constant, not necessarily zero.
Hence work done can be given by:
$$W_{1 \rightarrow 2}=- \int_{1}^{2}pdV$$ and as $$p=\frac{nRT}{V}$$
Work done can be given as,
$$W_{1 \rightarrow 2}=- \int_{1}^{2}\frac{nRT}{V}dV$$
$$W_{1 \rightarrow 2}=-nRT ln\frac{V_2}{V_1}$$
And hence, work done differs in case of isothermal expansion as compared to that of free expansion.
Best Answer
Remember what internal energy $U$ is. It is the sum:
But, doing work $W$ on an object could e.g. be displacing it sideways a distance $x$ - that is, putting it somewhere else. This would require some force $F$, and:
$$W=F \cdot x$$
In this new position no changes are done in chemical composition, no changes in potential energy, no kinetic energy (it lies still on the table), and no temperature change (since you want it to be a isothermal process). So this is an example of work done with no changes in internal energy.
For an isothermal process temperature has to be constant. After you have displaced (pushed) the box and done work on it, friction might stop it. This would generate heat. For an isothermal process this heat $Q$ has to be removen right away (maybe you use a cooling system or liquid). Now, conservation of energy from thermodynamics means that:
$$\Delta U = Q-W$$
And since - as discussed above - internal energy is unchanged, $\Delta U=0$ and $W=Q$.