## Question 9.21

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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### Question 9.21

Can someone please explain question 21:
Nanotechnologies have found ways to create and manipulate structures containing only a few molecules. However, orienting the molecule in specific ways to assemble such a structure can be difficult. Calculate the entropy of a solid nanostructure made of 64 molecules in which the molecules (a) are all aligned in the same direction; (b) lie in any one of the four orientations with the same energy.

Lisa Tang 1C
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### Re: Question 9.21

This question wants you to use your knowledge about the possible states/number of ways the molecules can be arranged to calculate the overall entropy. This can be done using the equation S=k*lnW, where k is Boltmann's constant: 1.381 x 10-23J/K and W is the number of ways that the molecules can be arranged but still have the same total energy. So, in part a, all the 64 molecules are arranged in the same direction. This means that each of the 64 molecules take up one orientation. W in this case is equal to 164 (which equals 1). So, when you solve for the entropy, S=k*ln(1)=0. For part b, there are four orientations that give rise to the same total energy. W is then equal to 464. Solving the equation, you get S=k*ln(464)=1.22x10-21J/K.

AtreyiMitra2L
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### Re: Question 9.21

If it is in the same direction, you know there is only one possible orienation. Therefore, regardless of how many molecules, the entropy using the boltzman equation will still be found to be 0. This occurs because the W in this equation is found by (# of possible orientations) ^ (# of molecules). In the second part, you use the same logic. The number of possible orientations would be 4.

204932558
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### Re: Question 9.21

For the first part, the degeneracy is 1^64, so S = ln(1^64) = 0. For the second part, S = ln(4^64).