Difference between Isothermal and Free Expansion I just don't get the exact difference between isothermal expansion and free expansion: in free expansion there is no work done since there is no external external pressure. As to the case of isothermal expansion there is no pressure applied, then how is work said to be done in that case?
 A: 
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.
A: There is no concept of temperature here because the initial condition of free expansion is assumed if the system is properly insulated that means, there is no heat interaction between system and surrounding is permitted, however in my personal opinion the temp of system will try to change during this rapid process, but as the system is insulated, no heat transfer will take place and in the end there will be no change in internal energy of the system.
I hope some of you may correct me.
Thanks
