Algebraic Improvement needed in Comparing enthelpy and internal energy change

[subhradeep mahata]

MODERATOR'S NOTE: I have been having difficulty explaining to this member the error in his algebraic development, in particular being cavalier in his attentiveness to proper algebraic manipulation of signs. I explained his error in post #7 (and the problems that could arise if he continues to do what he has been doing in more complicated problems), but my message does not seem to be getting through. Any help from the math mavens would be greatly appreciated.

ChesterMiller1. Homework Statement
Suppose the temperature of an ideal gas decreases from T1 to T2 ( T1>T2). Compare the change in enthalpy and change in internal energy.

Homework Equations

Change in enthalpy= n * C(p) * delta T
Change in internal energy U = n* C(v) * delta T[/B]

The Attempt at a Solution

I have almost solved it.
Since temperature is decreasing, we have:
delta H= -n* C(p) * delta T
delta U= -n *C(v)*delta T
C(p) is greater than C(v) , so magnitude wise delta H is greater than delta U. But I am confused with the negative sign. Do I have to compare along with the sign? So in this case, delta U is greater than delta H (which is the correct answer according to my book). I think that the negative sign should not be considered, as it is only showing that the gas does work on the surroundings. Please help me out.

 

[Charles Link]

You should recognize that $H=U+PV$, where $H$ is the enthalpy and $U$ is the internal energy. For an ideal gas $PV=nRT$. That should help relate $\Delta H$ to $\Delta U$.

 

[Chestermiller]

Homework Statement

Suppose the temperature of an ideal gas decreases from T1 to T2 ( T1>T2). Compare the change in enthalpy and change in internal energy.

Homework Equations

Change in enthalpy= n * C(p) * delta T
Change in internal energy U = n* C(v) * delta T[/B]

The Attempt at a Solution

I have almost solved it.
Since temperature is decreasing, we have:
delta H= -n* C(p) * delta T
delta U= -n *C(v)*delta T

These last two equations are written incorrectly. They should read $$\Delta H=nC_p\Delta T$$and $$\Delta U=nC_v\Delta T$$where $\Delta T$ represents the change from the initial state to the final state. In the present case, $\Delta T$ is negative.

C(p) is greater than C(v) , so magnitude wise delta H is greater than delta U. But I am confused with the negative sign. Do I have to compare along with the sign? So in this case, delta U is greater than delta H (which is the correct answer according to my book). I think that the negative sign should not be considered, as it is only showing that the gas does work on the surroundings. Please help me out.

I get what you are saying and I get what the book is saying. In my judgment, this is all just a matter of taste and interpretation. I personally prefer your interpretation in terms of the magnitudes of the changes. But the book's purely mathematical interpretation using "greater than" and "less than" comparisons can be argued also.

As an aside, the gas doing work is irrelevant to the changes in U and H, since these are functions of state, independent of how the change between the initial state and the final state was brought about.

 

[alejandromeira]

You can also use the relationship between Cp and Cv for an ideal gas.

 

[subhradeep mahata]

@Chestermiller Which two equations are incorrect?? Delta T is negative that's why I have given the negative sign.

 

[subhradeep mahata]

@Charles Link That's ok, but I want to know what is wrong in my method.

 

[Chestermiller]

@Chestermiller Which two equations are incorrect?? Delta T is negative that's why I have given the negative sign.

Algebraically, when we use the quantities $\Delta T=(T_2-T_1)$ and $\Delta H=(H_2-H_1)$ in the same equation, we mean that $T_1$ and $H_1$ correspond to the point $(T_1, H_1)$ in the T-H plane and $T_2$ and $H_2$ correspond to the point $(T_2, H_2)$ in the T-H plane. For example, when we write that the slope of a line is $$m=\frac{\Delta y}{\Delta x}=\frac{(y_2-y_1)}{(x_2-x_1)}$$, we mean that $x_1$ and $y_1$ correspond to the point $(x_1, y_1)$ in the x-y plane and $x_2$ and $y_2$ correspond to the point $(x_2, y_2)$ in the x-y plane; this equation is always what we use to calculate the slope, irrespective of the signs of $(x_2-x_1)$ and $(y_2-y_1)$.

When you wrote $$\Delta H=-nC_p\Delta T$$ did you really mean that $$(H_2-H_1)=-nC_p(T_2-T_1)=+nC_p(T_1-T_2)$$
I don't think you did. This was just you playing "fast and loose" with the rules of algebra. You might be able to get away with antics like this with a simple problem like the present one, but, if you try the same kinds of antics on more complicated problems, you're going to get the wrong answer.

 

[alejandromeira]

How Chestermiller said, it's not important if delta T is positive or negative.

These last two equations are written incorrectly. They should read $$\Delta H=nC_p\Delta T$$and $$\Delta U=nC_v\Delta T$$where $\Delta T$ represents the change from the initial state to the final state. In the present case, $\Delta T$ is negative.

Then you can use the ideal gas equation like say Charles Link (that's how I teach it to my students).

You should recognize that $H=U+PV$, where $H$ is the enthalpy and $U$ is the internal energy. For an ideal gas $PV=nRT$. That should help relate $\Delta H$ to $\Delta U$.

But there are another way with your relevant equations:$$\Delta H=nC_p\Delta T$$$$\Delta U=nC_v\Delta T$$
Substracting these equations you can write:
$$\Delta H -\Delta U = nC_p\Delta T - nC_v\Delta T$$
Ok. Then if you now know that in a ideal gas $C_p - C_v=R$ what is the right side of the above equation. Can you solve it?
If you solve this question, I'll show you a trick to use that relationship between $\Delta H$ and $\Delta U$.

 

[alejandromeira]

Hello subhradeep mahata, I thought you were looking for the deduction of: $\Delta H = \Delta U + \Delta n_g·R·T$, but reading moore deeply your problem, now I understand that perhaps it is not true. Here there is not a chemical reaction, there is a n mol of gas with change in $\Delta T$ (I automatize the calculations of the above equation because I teach these deduction every year and then I travel in automatic pilot, sorry : -).
Regardless of this confusion of mine, you can use perfectly the equation, what I showed you above: $$\Delta H - \Delta U = nC_p \Delta T - nC_v\Delta T$$ If you solve the right side with the relationship $(C_p - C_v)=R$ you will find what you are looking for.
But in this case sign of $\Delta T$ is relevant.
Ok, If there is something wrong please tell me, I want to learn too.

 

[subhradeep mahata]

@alejandromeira
delta H - delta U = - n*delta T * R (now let me make it clear, delta T is negative that is why I gave the negative sign, ok?)
delta H= -n* delta T* R + delta U
So, delta U is greater than Delta H, which is indeed the correct answer :-)
Now can you tell me your trick?

 

[subhradeep mahata]

@Chestermiller
I can't understand why there is so much confusion in this.
delta U, for example, is negative as delta T is negative
delta U = n* C(v) * delta T
which means delta U = -n * C(v) * delta T
Perhaps we are having the misunderstanding as I jumped the step and wrote it directly.

 

[alejandromeira]

@alejandromeira
delta H - delta U = - n*delta T * R (now let me make it clear, delta T is negative that is why I gave the negative sign, ok?)
delta H= -n* delta T* R + delta U
So, delta U is greater than Delta H, which is indeed the correct answer :-)
Now can you tell me your trick?

Yes but, you did not answer my question, what is the right side of:?
$$\Delta H - \Delta U = nC_p \Delta T - nC_v\Delta T$$
Remember $(C_p-C_v)=R$

Ok. I give you one trick and you solve the right side of above.

Here is the trick: I teach a pre-universitary and then is a very low level, perhaps for you then this trick has not importance.
You can find in the books the same relationship in two forms: $$\Delta H = \Delta U + \Delta n_gRT$$and $$\Delta U = \Delta H - \Delta n_gRT$$

I always recommend my students to use the first equation and not the second because is better to use the sign + than the sing -

Also in a lower level is easy to make mistakes in the sings, and then I recommend learn the first equation an NEVER clear variables in the equation. If you need to solve a numerical question you must to have the different values of variables. Then put those values before in the equation and then solve the numerical value. Don't clear fist the variable an put the numerival values later, because is easy to confuse the sings and variables of the first equation and the second equation, and you will always end up mixing H with U and + with -

Ok now, if you solve the righ side of the equation $\Delta H - \Delta U = nC_p \Delta T - nC_v\Delta T$ and clear $\Delta H$, you will see the problem in another perspective, an will see in a powerfull way how the sign of $\Delta T$ influences in the entire result.
I'm sorry from my English, I'm from Spain, Europe.

 

[Chestermiller]

@Chestermiller
I can't understand why there is so much confusion in this.
delta U, for example, is negative as delta T is negative
delta U = n* C(v) * delta T
which means delta U = -n * C(v) * delta T
Perhaps we are having the misunderstanding as I jumped the step and wrote it directly.

You're confused because you don't know how to do algebra properly. Go back and relearn it.

Here is a link to an analysis done by someone who embarrasses himself in front of the entire world by making the exact same bogus algebra error that you made in your post, in determining the equation for the work done by a gas on its surroundings: https://chem.libretexts.org/Core/Ph..._Chemistry/Thermodynamics/Path_Functions/Work
He concludes that the work done by a gas on its surroundings is $-P\Delta V$, and the work done by the surroundings on the gas is $P\Delta V$. This is exactly the opposite of what can be found in each and every textbook on thermodynamics.

 

[subhradeep mahata]

@Chestermiller Please see my 'relevant equations' part...are they correct?
I want to know exactly where i am making the mistake according to you.

 

[Chestermiller]

@Chestermiller Please see my 'relevant equations' part...are they correct?
I want to know exactly where i am making the mistake according to you.

I tried my best to explain this in post #7. Apparently, what I said did not resonate with you. I am going to move this thread to the PreCalculus Math forum to see whether some of the mathematicians there can help you more effectively.

 

[Ray Vickson]

@Chestermiller Please see my 'relevant equations' part...are they correct?
I want to know exactly where i am making the mistake according to you.

It is not just "according to" Chestermiller, it is according to everybody.

Some of your relevant equations are correct, and some are not. In post #11 you wrote $\Delta U = n C_v \Delta T,$ which is correct. Then, immediately below that you wrote $\Delta U =- n C_v \Delta T,$ which is wrong! You have to make up your mind: either $\Delta U = n C_v \Delta T$ or $\Delta U =- n C_v \Delta T$; you cannot have them both in the same problem. For example suppose we had $n C_v = 2,$ hence $\Delta U = 2 \Delta T$. So, if $\Delta T= 1$ we have $\Delta U = 2$, and if $\Delta T = -1$ (NEGATIVE!) we have $\Delta U = -2$ (ALSO NEGATIVE!). The point is that $U$ and $T$ move in the same direction, but at different rates. When $T$ increases ($\Delta T > 0$), $U$ also increases ($\Delta U > 0$), and when $T$ decreases ($\Delta T < 0$), $U$ decreases also ($\Delta U < 0$). When you write $\Delta U = -2 \Delta T$ you have $T$ and $U$ moving in opposite directions: when one increases, the other decreases. You are not allowed to cavalierly insert a minus sign just because $\Delta T$ may be negative, because when you do that you ruin the entire relationship between $U$ and $T$.

You must learn to use mathematics correctly, or at least learn to accept corrections and not stubbornly and defiantly stick to wrong concepts. You do not get to make up our own version of mathematics that is different from what everybody else in the world is doing. If we used your brand of mathematics to design things, motors would not turn, airplanes would not fly and computers would be impossible.