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5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol
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5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol(b)Evaporation of one mole of liquid carbon dioxide at its boiling point, 216.6 K. ΔHvap = 15.326 kJ/mol
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2021-09-05T01:38:30+00:00
2021-09-05T01:38:30+00:00 1 Answers
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Answers ( )
Answer:
a) ΔS = 14.46 J/k
b) ΔS = 70.76 J/k
Explanation:
The general formula to calculate the entropy change accompanied with a process is:
ΔS = ΔQ/T
where,
ΔS = entropy change for the process
ΔQ = Heat Transfer during the process
T = Absolute Temperature during the process
a)
In this case the heat transfer will be given as:
ΔQ = (ΔHfus)(N)
where,
ΔHfus = Molar Heat of Fusion of Tin = 7.029 KJ/mol
N = No. of moles of tin = 1 mol
Therefore,
ΔQ = (7.029 KJ/mol)(1 mol)
ΔQ = 7.029 KJ = 7029 J
and the absolute temperature is:
T = 213°C +273 = 486 k
using these values in the entropy formula, we get:
ΔS = 7029 J/486 k
ΔS = 14.46 J/k
b)
In this case the heat transfer will be given as:
ΔQ = (ΔHvap)(N)
where,
ΔHvap = Molar Heat of Vaporization of Carbon Dioxide = 15.326 KJ/mol
N = No. of moles of Carbon Dioxide = 1 mol
Therefore,
ΔQ = (15.326 KJ/mol)(1 mol)
ΔQ = 15.326 KJ = 15326 J
and the absolute temperature is:
T = 216.6 k
using these values in the entropy formula, we get:
ΔS = 15326 J/216.6 k
ΔS = 70.76 J/k