# How do I find the entropy variation?

• Upupumiau
In summary, the student attempted to find the entropy change for water at different temperatures by using the following equation: ∆S=Q/T+nCpm ln T2/T1. However, they were not able to determine the change in entropy because they were not given the correct data for Cpm. After correcting their mistake, they attempted to solve the equation using the more general equation: ∆S=Q/T+nCpm ln T2/T1+p. Unfortunately, they were not able to obtain the correct answer for p and are still unsure why.
Upupumiau

## Homework Statement

Find the ∆S per mol between liquid water at -5 ºC and ice at -5ºC at 1020hPa
Data:
∆CP,m fusion = 37,3 J K-1 mol-1
∆fusH = 6,01 kJ mol-1
The answer is 21.3 J/K mol

## Homework Equations

Usually I solve these problems by steps when they are at P=1 atm but since its at P=1020 hPa i don't really know how to proceed.
I tried using the following equation:
∆S= Q/T + nCpm ln T2/T1
Since I am given the Cpm I interpret that the pression is constant throughout the process

## The Attempt at a Solution

I first tried to solve it like I normally would, without taking into account that the press ion a was not 1 atm.
I want to try and find the temperature at which water soldifies at 1020 hPa but I don't know how with the data I am given. I thought that since Qp is the enthalpy maybe I could find it using Q = n∆T specific heat of water
but even thought I know what the specific heat of water is (41814 J/K kg) I'm not given this data so I'm unsure if this is what I have to do.

This is what I first tried doing but after I realized my mistake I did this:

I didn't obtain the value for entropy I should have with neither of the methods I tried. Where am I wrong?

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You are correct about thinking of this as a multi-step process.

ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
ice at 0 C and 1 atm to water at 0 C and 1 atm
water at 0 C and 1 atm to water at 0C and 1020 hPa
water at 0C and 1020 hPa to water at 5 C and 1020 hPa

Upupumiau
Chestermiller said:
You are correct about thinking of this as a multi-step process.

ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
ice at 0 C and 1 atm to water at 0 C and 1 atm
water at 0 C and 1 atm to water at 0C and 1020 hPa
water at 0C and 1020 hPa to water at 5 C and 1020 hPa

I tried doing it this way and it really does make more sense but I can't still obtain the 21.3 J/K I should and I really don't know why. I thought maybe it's because of the Cpm I have been given , but it turns out it doesn't really differ from the one I was given, so I thought I might change the sign since one it's fusion and the other one its for solidification. Still, I haven't obtained the result I need and I don't know why. Could anyone please help me?

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You did this mostly correctly in your first attempt. The first thing to recognize is that 1020 hPa is essentially 1 atm, so there is no need to account for a pressure effect. So the change in entropy in going from 1 mole of ice at -5 C to water at - 5 C is given by:
$$\Delta S=C_i\ln(273/268)+\frac{\Delta H(0)}{273}+C_w\ln(268/273)=(C_W-C_i)\ln(268/273)+22.0$$
We know that ##C_w-C_i=37.3\ J/mol.K##

Lord Jestocost
I don't quite understand how you're using logarithmic properties here, could you explain it to me please?

Upupumiau said:
I don't quite understand how you're using logarithmic properties here, could you explain it to me please?
Are you saying that you are not sure how to determine the change in entropy of a solid or liquid when it is heated or cooled from an initial temperature to a final temperature (and why this should involve natural logs of temperatures)?

I do, sorry. I had to revisit the basics. Thank you for your help.

## 1. What is entropy variation?

Entropy variation is a measure of the change in entropy, which is a thermodynamic property that describes the disorder or randomness of a system. It is typically denoted as ΔS and is measured in units of joules per kelvin (J/K).

## 2. How is entropy variation calculated?

The calculation of entropy variation involves determining the change in entropy between two states of a system. This can be done using the formula ΔS = ∑q/T, where ∑q represents the total heat added or removed from the system and T is the temperature in Kelvin.

## 3. What factors affect entropy variation?

Entropy variation is affected by changes in temperature, pressure, volume, and the number of particles in a system. Additionally, the type of process (e.g. reversible or irreversible) and the degree of disorder in the system can also impact entropy variation.

## 4. How do I interpret the sign of entropy variation?

A positive entropy variation (ΔS > 0) indicates an increase in disorder or randomness in the system, while a negative entropy variation (ΔS < 0) indicates a decrease in disorder. A zero entropy variation (ΔS = 0) may indicate a state of equilibrium or no change in disorder.

## 5. What is the significance of entropy variation in thermodynamics?

Entropy variation is a fundamental concept in thermodynamics as it represents the direction and magnitude of change in a system. It is closely related to the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time.

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