Increase Entropy in Sealed, Thermally Insulated Container

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In a sealed, thermally insulated container divided by an impermeable wall, the left half contains n moles of an ideal gas at temperature T. When the wall is removed, the gas expands to fill the entire volume, leading to a change in entropy. The relevant equation for entropy change is ΔS(T,P) = (3/2)Nk ln(T2/T1) + Nk ln(V2/V1), where T2 equals T1 due to thermal insulation, making the first term zero. The second term, which accounts for the increase in volume, is positive, indicating that the entropy of the system increases. This discussion clarifies the relationship between entropy, temperature, and volume in the context of ideal gas behavior.
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A sealed and thermally insulated container of total volume V is divided into two equal volumes by an impermeable wall. The left half of the container is initially occupied by n moles of an ideal
gas at temperature T. Which of the following gives the change in entropy of the system when
the wall is suddenly removed and the gas expands to fill the entire volume?

thermo is a very weak subject for me... any help is appreciated

I know that entropy is s=k*g where g is the number of accessible states, but I'm not sure where to go from here
 
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\Delta S(T,P)=\frac{3}{2}Nk ln( \frac{T_2}{T_1} ) + Nk ln( \frac{V_2}{V_1} )

This is the state equation for the entropy in terms of temperature and volume. Since the container is thermally insulated, T2=T1 and the first term drops out since ln(1)=0. Since V2 > V1, the second term is positive and the entropy of the system will INCREASE.
 
where did you get this equation from? Is this a standard equation?
 
Look under the entropy section here:http://en.wikipedia.org/wiki/Ideal_gas"
 
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