Which thermodynamic property does the PdV for a free expansion define if work is zero? Considering there is a finite change in PdV.
For a closed system (no mass exchange, only heat exchange is possible), the chagne in the internal energy is the obtained by subtracting the expansion work from heat flow in to the system. So, if there is no free expansion ,i.e. no work is done on the system or by the system. The change in internal energy is equal to the heat supplied.
dQ = dU
For a open system (where there is mass and heat exchange), enthalpy is the term that corresponds to free expansion. Enthalpy is given by, H = U + (P.V)
I don't think Pdv on its own represents any meaningful quantity in free expansion, as you rightly pointed out that, work done in free expansion is zero, since there is no external pressure to work against. Thus pressure of the gas remains same*(see edit), even though the volume changes. Now, consider the set-up, where you have a adiabatic container, having a removable partition, with one side containing ideal gas(for simplicity) and the other side being vacuum. If we now, remove the partition, and allow the gas to fill the container and come into equilibrium, then at this stage entropy has increased. This increase in entropy is given by, Integral of Pdv/T, between the limits Vi & Vf. Here Temperature T, will also remain constant, since the container is adiabatic and W=0.See the link for mathematical form
The increase in entropy takes place because of the finite pressure difference that exists between the two partitions, which is an irreversibility and hence leads to entropy change. * Edit: The pressure doesn't remain constant in free expansion, it changes. Consider the same set up i have described above, since temperature is constant and amount of substance remains unchanged, therefore according to boyle's law, as volume increases pressure decreases.
During a gas free expansion in an adiabatic chamber, experimentally (according to Joule's experiment) no change in temperature is observed; then, if the gas can be considered ideal, its internal energy $U$ does not change during the transformation.
Let us now state the First Principle of Thermodynamics in its differential form: $$dU=dQ+dW$$ We can make the following assumptions:
- we have just said that $U$ does not change, then $dU=0$;
- remember that the chamber is adiabatic, then $dQ=0$.
Eventually, what we obtain is
so no work is performed within the system.
Please notice that I have not used the definition $dW=PdV$.
That is because free expansion is an irreversible transformation (to perform the reverse process, i.e. compression, a non-null work is required) and elementary work cannot be expressed as $PdV$, since this definition is valid only for reversible work.