A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the process in which a CO2 sculpture, weighing 389 g, sublimes on a granite tabletop. The temperature of the granite is 12.0°C and the process occurs reversibly. Assume a final temperature for the CO2 vapor of –78.5°C. The enthalpy of sublimation of CO2 is 26.1 kJ/mol. a. Calculate the entropy of sublimation for carbon dioxide (the system) ___ J/Kmol b. Calculate the entropy of the universe for this reversible process. Use three significant figures in the answer __J/K

Given:

Now,

The Entropy of sublimation of CO₂ will be:

=

By substituting the values, we get

=

=

Thus the answer above is right.

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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 = - S =

Where:

ΔH  = 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 =

= 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 = + ΔS  

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS =    

                =1.185 KJ/K - = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.