- #1
Ryaners
- 50
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Hi folks,
This is a question about how to calculate entropy change when there is a temperature change involved. I got the correct answer to this, but I don't actually understand why it's correct..!
Any help is much appreciated.
One mole of liquid bromine is heated from 30 ºC to 50 ºC at constant pressure. Calculate the change in entropy (in kJ⋅K-1) associated to the process:
Br2 (liq, 30 ºC) → Br2 (liq, 50 ºC)
knowing that the heat capacity of Br2 at constant pressure in the range 30-50 ºC is 75 kJ⋅mol-1⋅K-1.
q = m⋅C⋅ΔT
ΔS = qrev / T (for a reversible process)
First I calculated q:
(1 mol)(75 kJ⋅mol-1⋅K-1)(20°K) = 1500 kJ
Then I used the ΔS equation above, with T = 298.15 K:
ΔS = 1500 kJ / 298.15 K = 5 kJ⋅K-1
The thing is, this is the correct answer, but I have no idea why you should know to use 298.15 K as the temperature. It's not given in the question, and I would have thought that the above equation for ΔS could only be used in cases where the temperature remains constant, not like in this example where it increases by 20°.
Can anyone explain why?
This is a question about how to calculate entropy change when there is a temperature change involved. I got the correct answer to this, but I don't actually understand why it's correct..!
Any help is much appreciated.
Homework Statement
One mole of liquid bromine is heated from 30 ºC to 50 ºC at constant pressure. Calculate the change in entropy (in kJ⋅K-1) associated to the process:
Br2 (liq, 30 ºC) → Br2 (liq, 50 ºC)
knowing that the heat capacity of Br2 at constant pressure in the range 30-50 ºC is 75 kJ⋅mol-1⋅K-1.
Homework Equations
q = m⋅C⋅ΔT
ΔS = qrev / T (for a reversible process)
The Attempt at a Solution
First I calculated q:
(1 mol)(75 kJ⋅mol-1⋅K-1)(20°K) = 1500 kJ
Then I used the ΔS equation above, with T = 298.15 K:
ΔS = 1500 kJ / 298.15 K = 5 kJ⋅K-1
The thing is, this is the correct answer, but I have no idea why you should know to use 298.15 K as the temperature. It's not given in the question, and I would have thought that the above equation for ΔS could only be used in cases where the temperature remains constant, not like in this example where it increases by 20°.
Can anyone explain why?