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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.

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    $\begingroup$ Can't you see that Chet Miller has proven that work can not be a state function? $\endgroup$
    – Bob D
    Commented Dec 31, 2020 at 16:28

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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.

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  • $\begingroup$ So, sir @Bob D, if somebody somebody asks me that does work become a state funcn. in an adiabatic or say a isothermal process, you mean clearcut answer should be no.. But these questions on work becoming state funcn. have been asked several times in Top level University entrance exams, in India and Outside also and official answer keys always ticks correct the option which says it does become State funcn. For your confirmation you can google this que., and check that it has been asked many times. $\endgroup$
    – user283695
    Commented Dec 31, 2020 at 3:05
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    $\begingroup$ @PARADA Any basic thermodynamics text book will tell you that work and heat are path functions and not state or point functions. Give me some links that state otherwise. $\endgroup$
    – Bob D
    Commented Dec 31, 2020 at 14:55
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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.

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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).

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    $\begingroup$ But the OP keeps insisting that work can be a state function. See comments below. You haven't addressed that. $\endgroup$
    – Bob D
    Commented Dec 31, 2020 at 15:00
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    $\begingroup$ @BobD To prove that the OP is wrong, all that needs to be done is provide an example of two different isothermal paths between the same two isothermal equilibrium end states that gives different values for the amount of work. This is done in the example I provided in which the expansion work for the reversible path is greater than for a specified irreversible work. $\endgroup$
    – Chet Miller
    Commented Dec 31, 2020 at 15:59
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    $\begingroup$ Yes you proved it in the example for a reversible vs irreversible isothermal process, but I feel you didn't give the knockout punch, which is to explicitly state that it proves work is not a state function. Moreover, to state that work is never a state function regardless of the process. $\endgroup$
    – Bob D
    Commented Dec 31, 2020 at 16:11
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    $\begingroup$ @BobD Well, this is just your judgment. What does the OP say? $\endgroup$
    – Chet Miller
    Commented Dec 31, 2020 at 16:20
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    $\begingroup$ Let's find out. $\endgroup$
    – Bob D
    Commented Dec 31, 2020 at 16:27

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