Thermodynamics: Comment on mistake

AI Thread Summary
The discussion centers on a thermodynamics problem involving the calculation of Gibbs free energy change when heating a mole of water from 25°C to 30°C at atmospheric pressure. The initial approach incorrectly equated Gibbs free energy change solely to heat added, neglecting the entropy term. Participants pointed out the importance of considering the relationship between Gibbs free energy, enthalpy, and entropy, especially when temperature is not constant. The conversation also highlighted the potential use of thermodynamic tables for values like entropy to solve the problem correctly. Overall, the key takeaway is the need to account for all relevant thermodynamic variables and relationships in such calculations.
WWCY
Messages
476
Reaction score
14

Homework Statement


Suppose you have a mole of water at 25##^{\circ}## at atmospheric pressure. You then raise its temperature to 30##^{\circ}##. Determine the change in Gibbs free energy. To compensate for the change, once can raise the pressure on water, how much pressure you need to increase?

I know the right way to do this problem, but I would like to know what I did wrong initially with the following steps:

I knew ##\Delta G = \Delta H - T\Delta S = \Delta Q - \Delta W_{other}##, so i thought: To raise temperature I'd probably need to provide heat, and since no mention of any non-mechanical work was made in the problem, let's take ##\Delta W_{other} = 0##

Therefore ##\Delta G = \Delta Q = C_p \Delta T##, where ##C_p## was a given.

I would then use the thermodynamic identity to work out the pressure change.

What were the wrong assumptions made?

Assistance is greatly appreciated

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
Hi,

In my book ##\Delta H = c_p \Delta T##, so you missed the ##T\Delta S## altogether ? Anyway, if T is not a constant, what do you do with a term like ##T\Delta S## ?
 
BvU said:
Hi,

In my book ##\Delta H = c_p \Delta T##, so you missed the ##T\Delta S## altogether ?

Good heavens, I just spotted my mistake. It should have been ##\Delta G = \Delta W_{other}## at the end when ##Q## and ##T\Delta S## cancel. Thanks for pointing it out.

I guess that even this result wouldn't have been the most useful.

I would have made use of the fact that ##\Delta G / \Delta T = -S## to solve the problem by using the ##G## thermo-identity.

BvU said:
Hi,
Anyway, if T is not a constant, what do you do with a term like ##T\Delta S## ?

Could you elaborate on what you mean by this? Apologies.
 
WWCY said:
Could you elaborate
My comment was exclusively triggered by your ##
\Delta G = \Delta H - T\Delta S = \Delta Q - \Delta W_{other}##.

However, the natural variables are ##T,p## and ##{N_i}## so that ##dG = -SdT + V dP +\sum\mu_idN_i## or ##\Delta G = -S\Delta T## as you found for your isobaric scenario.
 
Is it possible you are supposed to solve this problem using thermodynamic tables? Otherwise, what is your relationship for ##\Delta S## at constant pressure?
 
  • Like
Likes BvU
Chestermiller said:
Is it possible you are supposed to solve this problem using thermodynamic tables? Otherwise, what is your relationship for ##\Delta S## at constant pressure?

If what is meant by thermodynamic tables is given values for quantities like entropy then yes, I was. I guess most of my confusion came from not knowing where to start, given the table of values.
 
WWCY said:
If what is meant by thermodynamic tables is given values for quantities like entropy then yes, I was. I guess most of my confusion came from not knowing where to start, given the table of values.
Does your (compressed water) table give values for P, T, v, u, h, and s?
 
Back
Top