High Entropy alloys are considered as a breakthrough in the present generation of materials science but what if all metals(in solid state) in periodic table would have been mixed, would that still be a "High entropy alloy"?
In principle you are right. The entropy of mixing increases with the number of components by:
ΔS= R (X1lnX1 + X2lnX2)
ΔS= R (X1lnX1 + X2lnX2 + X3lnX3)
ΔS= R (X1lnX1 + X2lnX2 + X3lnX3 + X4lnX4)
......
R - gas constant, Xi - mol fraction.
The maximum entropy for the system is reached when X1=X2=X3... With increasing number of components the entropy of mixing is growing, however the value ΔS added is decreasing with increasing number of components. On the other side with increasing number of components the probability that components with high binding forces are added is also increasing. This means that the system will rather choose to form compounds to minimize the Gibbs free energy ΔG=ΔH-TΔS and the small amount ΔS is not able to compensate the negative ΔH. This is the reason that 5 to 6 components can be found with small binding enthalpies and the entropy is responsible that solid solutions are formed. Higher numbers of components should create again (e.g. intermetallic) phases that cause brittleness.
There are books published on High Entropy Alloys that give a good overview about these basic principles.
I guess the attached article will answer your question:
Gludovatz, B., Hohenwarter, A., Catoor, D., Chang, E.H., George, E.P. and Ritchie, R.O., 2014. A fracture-resistant high-entropy alloy for cryogenic applications. Science, 345(6201), pp.1153-1158.
During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
The so-called entropy was such a concept that was derived by mistake in history.
It is well known that calculus has a definition,
any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.
Based on the definition of calculus, we know:
to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.
As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,
∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.
1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.
2) On the other hand, In fact, Q=f(P, V, T), then
∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).
We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.
that is, the so-called "entropy " doesn't exist at all.
Mainak Saha An ideal solution is a valid where R is the gas constant (= 8.314 J · (mol · K)−1). For an equimolar alloy (where all elements have the same concentration) with n different components that means that there is a configurational entropy of S = R ln(n), which equals to 1.61R when n = 5.
It is suggested that obtaining the complete solid solution phase (it is either FCC, BCC or FCC+BCC) without any inter-metallic by addition of number of elements can be termed as High entropy alloy. To confirm the solid solution formation there are three major criteria 1. atomic size difference, 2. enthalpy of mixing and 3. entropy of mixing. The range of the above mentioned criteria can be easily find through literature.
Shufeng Zhang I agree that entropy doesn't exist but his question is basically related to physical metallurgy and formation of high entropy alloy in real. Thank you.