Question

Can irreversible isothermal work be used to predict spontaneity of processes for thermodynamic systems in the same way as mechanical systems (which follows the principle: Mechanical systems with spontaneously lower their potential energy V)?

Please provide and argument for your answer. This is supposed to be a debatable question.

Answer 1

For a Thermodynamics process we tell about spontaneity of a system by using change in gibbs's free energy (G) which is calculated using change in enthalpy of system (H) and chane in Entropy of system (S) according to the following equation:

G_system = H_system - T S_system

on the basis of sign of G we tell about the spontaneity of thermodynamic system

If G is positive than process is Non-spontaneous

If G is negative than process is Spontaneuos

Now for irreversible isothermal process:

Hence Entropy is a state function that's why S_{reversible process} = S_{irreversible process}

And Entropy is given by S = Q_system / Temp.

according to first law of Thermodynamic U = Q + W

And for an isotehrmal process U = 0 (Temp. is constant)

so Q_system = - W_system

So in case of reversible and irreversible work is different (Because work is a path function)

But S is a path function so it comes out to be equal in both reversible and irreversible process and it is qual to

S_system = n R ln(V_f/V_i)

where n = no. of moles of gas

R = gas constant

V_f = final volume of th system

V_i = intial volume of the system.

so for an irreversible procss we don't use irreversible work to calculate Entropy change of the process (S) and also in that way we don't use irreversible isothermal work to calculate the spontaneity of the process.