# Change in entropy of the system

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Thread starter 6 years ago
#1
A box is divided into two compartments of equal volume. N ideal gas molecules are confined to one half of the box and the other half is empty. When divider is removed what is the change in entropy of the system? Explain your reasoning.

So far this is what I've done
Ssubscript(AB)=ksubscript(b)*ln(2m*2m)
=ksubscript(b)*ln((2m)^2)
=2ksubscript(b)*(ln(m)+ln(2))

Once the divider is removed the number of micro states of a particle doubles to 2m
m=microstate(s)
Thanks for the help in advance.
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Thread starter 6 years ago
#2
Am I doing this wrong or right?
How do I explain my reasoning?
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6 years ago
#3
You need to think in terms of 'PositionalEntropy' vs 'Change in Entropy'. Positional Entropy is the disorder of a systemat a specific temperature. Example, the values posted in most Thermodynamicstables are 'Positional Entropys' at 25oC. However, for a 'Changein Entropy' (∆So) would be the sum of Positional Entropies ofProducts minus the sum of Positional Entropies of reactants.
∆So= ∑Sofinal - ∑Soinitisl (FollowsHess’s Law)
For your problem, Let’s assign Soinitial of the particles in the box before removing the partition to be So(1). This is also the positional entropy before removing the partition. Now, assign So(2)to be the positional entropy after removing the partition and the particles have reached equilibrium. The issue is to determine So(2) in termsof So(1). Since the overall volume doubles the particles will assume positions in space that minimize interactive effects. In general, at equilibrium the positional entropy doubles to 2So(1) as the volume doubles; that is, the positional entropy So(2) = 2So(1),Now use the equation ∆So = ∑Sofinal - ∑Soinitislto calculate The Change in Entropy (∆So).

∆So = ∑Sofinal- ∑Soinitisl = So(2)positionalEntropy after removing partition - So(1)positionalEntropy before removing partition = 2So(1) – So(1)= So(1). That is, the Change in entropy is equal to the valueof positional entropy before removing the partition. ∆So = So(1).

Numerically,assume ...
So(1) = 10 J/K;
So(2) = 20 J/K.
Note:The higher number value of positional entropy is an indication of 'moredisorder' in the system, and the lower number means 'less disorder' in the system.
∆So = So(2) - So(1) = (20 - 10)J/K = +10 J/K <=> the change in entropy is equal to the value of positional entropy, So(1) and suggests the system has 2x more disorder after removing the partition. That is, the new positional Entropy is (10 + 10) J/K = 20 J/K
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