Respuesta :
Answer:
The answers to the questions are;
a. The entropy of sublimation for carbon dioxide (the system) is
134.07 J/Kmol.
b. The entropy of the universe for this reversible process is 376 J/K.
Explanation:
Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at the temperature where the sublimation is taking place
a. We note that the mass of the solid CO₂ = 389 g
Molar mass of CO₂ = 44.01 g/mol
Number of moles of CO₂ in the sculpture = Mass/(Molar mass)
= (389 g)/(44.01 g/mol) = 8.84 Moles
Entropy of sublimation is given by
ΔS[tex]_{sublimation}[/tex] = [tex]S_{vapor}[/tex] - S[tex]_{solid}[/tex] = [tex]\frac{\Delta H_{sublimation}}{T}[/tex]
Where:
ΔH[tex]_{sublimation}[/tex] = 26.1 KJ/mol
T = Temperature = –78.5°C = 194.65 K
Therefore the amount of heat required to cause the 389 g of dry ice to sublime = 26.1 KJ/mol × 8.84 Moles = 230.695 KJ
Therefore the entropy of sublimation = ΔS[tex]_{sublimation}[/tex] = [tex]\frac{230.695 KJ}{194.65 K}[/tex]
= 1.185 KJ/K
= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol
b. The entropy of the universe is given by;
ΔS[tex]_{universe}[/tex] = [tex]\Delta S_{system}[/tex] + ΔS[tex]_{surrounding}[/tex]
If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;
ΔS[tex]_{universe}[/tex] = [tex]\frac{Q}{ T_{system}}[/tex] [tex]-\frac{Q}{T_{surrounding}}[/tex]
=1.185 KJ/K - [tex]-\frac{230.695 KJ}{285.15K}[/tex] = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K
= 376 J/K.
The entropy sublimation will be "134.1 J/mol.K".
Given:
- Mass = 389 g
- Sublimation point = –78.5°C
- Granite temperature = 12.0°C
- Enthalpy of sublimation = 26.1
Now,
The Entropy of sublimation of CO₂ will be:
= [tex]\frac{\Delta H_{sublimation}}{T_{sublimation}}[/tex]
By substituting the values, we get
= [tex]\frac{26100 }{273.15-78.5 }[/tex]
= [tex]134.1 \ J/mol.k[/tex]
Thus the answer above is right.
Learn more about entropy here:
https://brainly.com/question/10936758
