Quantification of entropy mathematically when $T$ and $V$ both change 
$$ln\frac{W_f}{W_i}=N ln\frac{V_f}{V_i}=n N_a ln\frac{V_f}{V_i}$$
$$\Delta S=nRln\frac{V_f}{V_i}$$ 
$$ln\frac{V_f}{V_i}=\frac{1}{n N_a}ln\frac{W_f}{W_i}$$
$$\Delta S=\frac{R}{N_a}ln\frac{W_f}{W_i}=kln\frac{W_f}{W_i}=klnW_f-klnW_i$$
hence  $$S=klnW$$
For T change taken from Atkins physical chemistry:

Above micro entropy S=klnW and macro entropy $\Delta S=\frac{\Delta Q}{T}$ are united isolated for T and V change. Entropy being a state function my problem is how one can understand that this is the $quantification$ that works for paths in change of both V and T? How does T and V quantify against each other? Can this be justified or explained?
 A: It seems to me the first derivation is simply assuming that S = k ln W, because it looks like it was inserted along the way and then pulled out at the end as if it was a conclusion of the logic.  What's more, by considering only position and not momentum, it seems to me that derivation is assuming the particles are not changing kinetic energy.  So that's why it ends up looking like there's no explicit mention of heat input or dQ/T, because to change V but not change kinetic energy, there would need to be heat input that is implicit.  The second derivation shows more clearly where the connection between dQ/T and k ln W comes from, but it does it only for constant volume.  But a reversible change in V does not change the entropy, so as long as everything is done reversibly, it doesn't matter if V changes or not, and the second derivation is less general than it could be.
A: For stable pressure it is ΔS= nCplnTf/Ti= nCplnVf/Vi (Cp=molecular specific heat)
To quantify,we set initial values Ti=1=Vi and derive: S= nCplnT = nCplnV
