Calculating Entropy

[clarissa_ram 2L]

In professor's lecture on 02/06, we ended the class with the example of 4CO molecules in a 2-state system. We then used the equation (S=(KB)lnW) to solve for entropy.

I was wondering why we didn't multiply by the 4 molecules that are present by doing: S= 4(KB)lnW?

[Aiden_Wong_3E]

It's because the number of particles affects entropy in an exponential way, not in a multiplier way. When you have 4CO, you would set W=2^4 and plug it into the entropy equation to solve for S. I think this is because probability is affected exponentially, not multiplication-ally as the number of choices made increases.

[hiranya sundar 2D]

So we could have said S = (Kb)lnW = Kb ln (2^4) = 4kb ln(2) and done that multiplication trick.
But that would mean what's in the logarithm isn't W anymore when we multiply it, it's just a step in the process.
The same thing applies to the mole thing we were talking about: if we have 2 states, our equation would be S = (Kb)lnW = 6.022 * 10^23 * (Kb) * ln(2), but W would actually be 2 ^ (6.022 * 10^23).

[Kushal Chatterjee 1J]

The W term already takes into account the number of molecules. In this case, we said that W = 2⁴ = 16 because we had 4 CO molecules, each of which could occupy two possible positions. Multiplying by 4 would not be correct because the number of arrangements is not a scalar multiplication issue, but rather exponential. When we calculate W, we already take the number of molecules into consideration, so multiplying by 4 is not needed and would be incorrect.

[Levon_Avedian_2H]

What does this equation " (S=(KB)lnW)" represent in calculating entropy?

[BeauBrown]

The 4 comes into play with 2^4, so multiplying the number by 4 would just throw the calculation off.

[Angie Tan 2L]

When calculating W, it is the (number of possible in the orientations)^(number of molecules), so 4 is already included in the calculations and is not needed to multiply. Additionally, 4 molecules does not have a direct relationship with the number of possible states that exists in the same energy. Therefore, we never need to multiply the number of molecules since it is calculated in W.