Does Work become state function in an Isothermal Process and what are other processes in which it happens? In a reversible isothermal process and for an ideal gas we know from the definition of Helmholtz free Energy
$dF= -SdT -PdV$.
And as temperature doesn't change for an isothermal process, $dT$ must be zero. So dF can be written negative of change in Helmholtz free Energy. Since $F$ is a state function and $dF$ a perfect differential, work also should be.
Also, does work become state function for adiabatic processes also? Please throw light on it.
 A: The fact that work can equal the change in a state function, as in the case of an adiabatic process where work equals the change in internal energy, does not mean that work is a state function.
A state function is a system property. Work (and heat) is never a state function because work is not a property of a system. Work is the transfer of energy to or from a system. It is not the energy of system itself, which is its internal energy.
Hope this helps.
A: The equation dF=TdS-PdV applies only to two closely neighboring (i.e., differentially separated) thermodynamic equilibrium states, where P is the pressure calculated from the (equilibrium) equation of state (e.g., the ideal gas law) for the fluid.  In an irreversible process, even if the boundary of the system is held at a constant temperature, this does not mean that the temperature interior to the system is uniform spatially.  This spatial non-uniformity will also apply to adiabatic irreversible processes.  So the entire fluid is isothermal only for a reversible path.    In addition, in an irreversible expansion or compression, the force per unit area at the interface where work is being done (e.g., the inside face of a piston) is not equal to the pressure calculated from the equation of state.  This force also includes viscous stresses resulting from rapid deformation of the fluid.  Therefore, the equation for dF cannot be applied to this, and it is not a perfect differential all along an irreversible path.  In addition, from this it follows that, for the irreversible path, the work is not equal to the change in F.
A: Here is an additional analysis that is consistent with my previous answer:
For a process in a closed system, the first law of thermodynamics tells us that $$\Delta U=Q-W$$Now, if we define an isothermal process (either reversible or irreversible) as one in which the temperature of the system in its initial and final thermodynamic equilibrium states is T, and that, during the process, all heat transfer takes place (at the interface) between the system and surroundings at temperature T, then from the Clausius Inequality, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the entropy generated during the process as a result of irreversibility (a positive definite quantity).  If we combine these two equations, we have $$\Delta U=T\Delta S-T\sigma-W$$or$$W=-\Delta F-T\sigma$$From this it follows that for any isothermal process path, the work done by the system on the surroundings is less than (irreversible) or equal to (reversible) the decrease in the Helmholtz free energy.
Here is a specific example:  If we have one mole of an ideal gas at $P_1$ and $V_1$ and we suddenly drop the external pressure on the gas to $P_2$, and then let it equilibrate, what is the change in F and how much work is done on the surroundings.  Well, the change in F is just $$\Delta F=-\int_{V_1}^{V_2}{\frac{RT}{V}dV}=-RT\ln{\left(\frac{V_2}{V_1}\right)}=-RT\ln{\left(\frac{P_1}{P_2}\right)}$$The work done on the surroundings is $$W=P_2(V_2-V_1)=P_2V_2\left(1-\frac{V_1}{V_2}\right)=RT\left(1-\frac{P_2}{P_1}\right)$$Mathematically, the decrease in F, given by $RT\ln{\left(\frac{P_1}{P_2}\right)}$ is always greater than the work W, given by $RT\left(1-\frac{P_2}{P_1}\right)$, irrespective of the pressure ratio (even for compression).
