An oxygen o2 molecule is adsorbed on a patch of surface (see sketch at right). this patch is known to contain 484 adsorption sites. the o2 molecule has enough energy to move from site to site, so it could be on any one of them. suppose additional surface becomes exposed, so that 729 adsorption sites are now available for the molecule. calculate the change in entropy. round your answer to 3 significant digits, and be sure it has the correct unit symbol.

We have Boltzmann's equation S = k ln W
Boltzmann's constant k = 1.381 x 10^-23 J/K
W = Number of absorption sites
At W = 484, Entropy S1 = 1.381 x 10^-23 ln 484 = 8.537 x 10^-23 J/K
At W = 729, Entropy S2 = 1.381 x 10^-23 ln 729 = 9.103 x 10^-23 J/K
Change of Entropy = S2 - S1 = 0.566 x 10^-23 J/K

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Tringa0 Tringa0 · Answer 2 · 2019-08-14T10:39:06+02:00

Answer:

The change in entropy is [tex]5.65\times 10^{-24} J/K[/tex].

Explanation:

The entropy can be determined from Boltzmann equation of entropy:

[tex]S=K_b\ln w[/tex]

S = Entropy of the system

[tex]K_b=1.38\times 10^{-23} J/K[/tex]

w = Number of microstates

1) Number of adoption sites = 484

w = 484

[tex]S_1=1.38\times 10^{-23} J/K\ln 484=8.5312\times 10^{-23} J/K[/tex]

2) Number of adoption sites = 729

w =729

[tex]S_2=1.38\times 10^{-23} J/K\ln 729=9.0965\times 10^{-23} J/K[/tex]

Change in entropy =[tex]S_2-S_1[/tex]

[tex]\Delta S=9.0965\times 10^{-23} J/K-8.5312\times 10^{-23} J/K=5.653\times 10^{-24} J/K[/tex]

The change in entropy is [tex]5.65\times 10^{-24} J/K[/tex].