Enthalpy and Internal energy

[Scienc]

I have a question about the resolution of this exercise: 

 

I tried to solve this problem with the following steps:

 

For the 1º law of thermodynamics, 𐤃U = q+w.

As q = 𐤃H at constant pressure, I assumed that in this process q = - 15 kJ.

 

As a result, 𐤃U = -15 - 22 (because was an expansion) and for this reason, 𐤃U = -37 kJ.

 

However, the book has this resolution

 

I understood why the book used this method, but I did not understand why my method is wrong.

[joigus]

Think about it. You must provide heat to make the pressure constant during an expansion.

Hm. There's something more to it. I can't read the image very well..

Sorry, I misread. It's a chemical reaction.

OK. I think I see what the problem is. The criterion that you use is consistent with \( w = -P\Delta V \), right?

Not with \( w = +P\Delta V \)

Some people write \( \Delta U=q-w \), and other people prefer \( \Delta U=q+w \), but when expressed in terms of P, V, in the 1st principle, it must always be, \( \delta U = q-p\Delta V \), because if it's an expansion, the contribution to the internal energy must be negative. That's what's written in stone.

Maybe you got confused with a sign criterion you saw somewhere else?

Sorry for the initial confusion. Was that helpful?

Edited October 29, 2022 by joigus
correction

[exchemist]

1 hour ago, Scienc said:

I have a question about the resolution of this exercise: 

 

I tried to solve this problem with the following steps:

 

For the 1º law of thermodynamics, 𐤃U = q+w.

As q = 𐤃H at constant pressure, I assumed that in this process q = - 15 kJ.

 

As a result, 𐤃U = -15 - 22 (because was an expansion) and for this reason, 𐤃U = -37 kJ.

 

However, the book has this resolution

 

I understood why the book used this method, but I did not understand why my method is wrong.

I wonder if this is to do with  the wording. They say 22kJ is expansion work done "on the system". They don't say 22kJ of expansion work is done by the system. 

Presuming "the system" means the reactant and products, I'm confused as to whether the atmosphere does work on the system, i.e. the volume decreases in the course of the reaction, or whether the system does work pushing back the atmosphere, i.e. the volume increases. 

If the volume increases, then I agree the internal energy change must be enough to supply that work AND still give you a measured heat output of 15kJ. 

[sethoflagos]

3 hours ago, Scienc said:

I have a question about the resolution of this exercise: 

 

I tried to solve this problem with the following steps:

 

For the 1º law of thermodynamics, 𐤃U = q+w.

As q = 𐤃H at constant pressure, I assumed that in this process q = - 15 kJ.

 

As a result, 𐤃U = -15 - 22 (because was an expansion) and for this reason, 𐤃U = -37 kJ.

 

However, the book has this resolution

 

I understood why the book used this method, but I did not understand why my method is wrong.

The exercise is misleadingly worded.

If a +ve amount of work is 'done on the system' then this must be understood as energy being added to the system whatever sign convention you are following, and therefore the internal energy change arising from that work must increase. This is where your method gave the wrong answer.

The term 'expansion work' is commonly used as a synonym for any PdV process even when, as in your example, the system is being compressed. We know that the book intends compression because their PdV term is negative, and therefore dV is negative.

 

[Scienc]

6 hours ago, sethoflagos said:

The exercise is misleadingly worded.

If a +ve amount of work is 'done on the system' then this must be understood as energy being added to the system whatever sign convention you are following, and therefore the internal energy change arising from that work must increase. This is where your method gave the wrong answer.

The term 'expansion work' is commonly used as a synonym for any PdV process even when, as in your example, the system is being compressed. We know that the book intends compression because their PdV term is negative, and therefore dV is negative.

 

So, is the true value for work +22, because work is done on the system?

9 hours ago, joigus said:

Think about it. You must provide heat to make the pressure constant during an expansion.

Hm. There's something more to it. I can't read the image very well..

Sorry, I misread. It's a chemical reaction.

OK. I think I see what the problem is. The criterion that you use is consistent with w=−PΔV , right?

Not with w=+PΔV

Some people write ΔU=q−w , and other people prefer ΔU=q+w , but when expressed in terms of P, V, in the 1st principle, it must always be, δU=q−pΔV , because if it's an expansion, the contribution to the internal energy must be negative. That's what's written in stone.

Maybe you got confused with a sign criterion you saw somewhere else?

Sorry for the initial confusion. Was that helpful?

I used the 𐤃U = q+w. because I studied in a chemical class. However, I considered the contribution to work as negative. The book on the other hand, used a positive signal for work.

The images is from the book.

Edited October 30, 2022 by Scienc

[sethoflagos]

7 hours ago, Scienc said:

So, is the true value for work +22, because work is done on the system?

It depends on your sign convention. For myself, work performed on the system by the environment is -ve, because that's how I was taught. And that's the convention adopted by your book. But if energy is added to a system, the energy of that system must increase, and that's the mental picture that it's important to hold on to.

[exchemist]

8 hours ago, Scienc said:

So, is the true value for work +22, because work is done on the system?

I used the 𐤃U = q+w. because I studied in a chemical class. However, I considered the contribution to work as negative. The book on the other hand, used a positive signal for work.

The images is from the book.

I think it best to keep in mind what is happening physically, to ensure your answer makes sense. If there is a volume reduction, the atmosphere is compressing the system. Then as @sethoflagossays, that adds internal energy, so if what you get out as heat is less than this, the net internal energy must have gone up, so ΔU has to be +ve.   

But I confess I too was thrown by the term "expansion work" when in fact what happens is the opposite, viz. compression work - at least from the system's point of view. It may be that the learning point for both of us is what @sethoflagos says about "expansion work" being used to denote any form of PdV work, regardless of whether it is expansion or compression of the system. I was never taught this terminology. We always spoke of "PdV work" and left it at that, which is far less confusing in my estimation. 

 

 

Edited October 30, 2022 by exchemist

[studiot]

9 hours ago, Scienc said:

So, is the true value for work +22, because work is done on the system?

I used the 𐤃U = q+w. because I studied in a chemical class. However, I considered the contribution to work as negative. The book on the other hand, used a positive signal for work.

The images is from the book.

 

I have not posted in this thread before as exchemist and seth were doing a grand job of sorting out the sign conventions and the lax wording in your book.

I think it is worth noting that 7 + 15 = 22.

 

I will just add a comment on sign conventions here as some just take this in ther stride and never have a problem , whereas others stumble over the issues.

The first Law of Thermodynamics is one of at least half a dozen cases in Physical Science where two separate and independent sign conventions are simultaneously in play, with sometimes confusing results.

The first is the case of positive and negative numebrs combined with addition and subtraction in arithmetic.
I have seen at least one Professor of Maths get his sums wrong as a result.

Then we have your example from the first law.

Chemists use   ΔU  =  q + w

Physicists use ΔU  =  q - w  (engineers use this too)

Both are correct in their own way.

I think the Chemists' version is more logical because on the the right hand side refers to energies crossing the system boundary.
Everything that goes in is regarded as positive and everything that comes out is regarded as negative and all you have to do is add them up.
This becomes even more useful when there are (lots) more terms on the RHS for electrical work etc etc.

This does mean we have to adopt 'work done on the system'  as our variable w.

However Thermodynamics was developed in the days of steam and other machinery.
And the purpose of machinery is to do work.
Since the machine constitutes the system in most cases and work is the desired output, it makes sense to adopt the convention that w is 'work done by the system'.
Equally you add heat in the form of coal to a steam engine to get that output work. So q is regarded as heat added to the system.
So the Physicist's first law has to be written differently to take account of the mixed up sign conventions on its RHS.

 

Other scientific examples you might encounter include

In Electricity the sign conventions for voltage and current have the opposite sense

In Optics there are various sign conventions for lenses and mirrors, in addition to the cartesian conventions for spatial layout.

In Mechanics there are again paired sign conventions for loads, deflections and so on, particularly in Beam theory.

 

I used to find it worthwhile to make a note or add a note page at the bginning of any new tech book about its sign conventions to clarify what sign conventions are used in that book.
The best books even do this for you at their beginning.

This saves worry when you see in one book it says +w and in another it says -w.