# Problem on limits from chemistry

• Bipolarity
In summary, the problem being discussed is a calculus question in relation to the Gibbs Free Energy and the finite values of \Delta G and \Delta S. The goal is to prove that \Delta G ≤ 0, and through a proof by contradiction, it is shown that this is indeed the case when the value of \epsilon, an infinitely small positive quantity, is smaller than a specific number calculated from the given constraints. This problem has been reduced to a purely mathematical question and further efforts are appreciated.
Bipolarity
I've been doing some chemistry, and come up with a confusing problem that I can't seem to solve. It's a problem I made up while trying to understand the Gibbs Free Energy, but I will try to phrase it solely as a calculus question.

It is known that

$\Delta G + \epsilon \Delta S < 0$ and that $\epsilon$ is an infinitely small positive quantity. Nothing is known about the sign of $\Delta S$.

Both $\Delta G$ and $\Delta S$ are finite.

Can I prove that $\Delta G ≤ 0$ ?

I appreciate all help on this matter. Thanks!

BiP

Last edited:
By infinitely small do you mean that $\epsilon$ can be made as small as we desire? If so:

Suppose that $$\Delta G > 0$$. Then we can find $\epsilon > 0$ such that $$\Delta G + \epsilon \Delta S = 0$$, namely $$\epsilon = -\Delta G/\Delta S$$

Well, this is positive as long as $\Delta S$ is negative . What if $\Delta S \geq 0$? In that case, $\epsilon \Delta S \geq 0$ so if $\Delta G > 0$ we're just adding positive numbers together and of course it can't be negative

Office_Shredder said:
By infinitely small do you mean that $\epsilon$ can be made as small as we desire? If so:

Suppose that $$\Delta G > 0$$. Then we can find $\epsilon > 0$ such that $$\Delta G + \epsilon \Delta S = 0$$, namely $$\epsilon = -\Delta G/\Delta S$$

Well, this is positive as long as $\Delta S$ is negative . What if $\Delta S \geq 0$? In that case, $\epsilon \Delta S \geq 0$ so if $\Delta G > 0$ we're just adding positive numbers together and of course it can't be negative

By infinitely small, I mean that $\epsilon > 0$ is less than any positive real number.

BiP

Bipolarity said:
By infinitely small, I mean that $\epsilon > 0$ is less than any positive real number.

BiP

No such epsilon exists in the real numbers (how can $\epsilon/2 > \epsilon$)? Is this supposed to be a problem in the hyperreals or something?

Hmm well I don't know to say it, all I mean is that $\epsilon$ is a positive infinitesimal.

I reasoned that because of its being infinitesimal it can be neglected from the equation, yielding $\Delta G < 0$. But I don't know if that's correct.

Also your original proof does not satisfy the main constraint of the problem, i.e. $\Delta G - \epsilon \Delta S < 0$

But I think we're getting somewhere! Thanks!BiP

Bipolarity said:
Hmm well I don't know to say it, all I mean is that $\epsilon$ is a positive infinitesimal.

I reasoned that because of its being infinitesimal it can be neglected from the equation, yielding $\Delta G < 0$. But I don't know if that's correct.

Can you describe a bit the setup that led you to this equation? It will help sort out exactly what kind of requirements you really have on $\epsilon$ - since if you know it's smaller than all positive numbers, and non-negative for example, then $\epsilon=0$ necessarily.
Also your original proof does not satisfy the main constraint of the problem, i.e. $\Delta G - \epsilon \Delta S < 0$

This differs by a minus sign from the original post

Office_Shredder said:
Can you describe a bit the setup that led you to this equation? It will help sort out exactly what kind of requirements you really have on $\epsilon$ - since if you know it's smaller than all positive numbers, and non-negative for example, then $\epsilon=0$ necessarily.This differs by a minus sign from the original post

Sorry I meant $\Delta G + \epsilon \Delta S$ from the original post. The second post is a typo.

The setup for my problem is a little complicated thermodynamics. I don't wish to go into it, but the epsilon is basically the infinitesimal difference in temperature between two reservoirs that are basically the same temperature, except one is higher in temperature by an infinitesimal amount epsilon. Heat will transfer due to that infinitesimal difference. Though the heat transfer is also infinitesimal, it is important in the calculations, whereas the epsilon reduces to the inequality above. I would love to see this problem be solved, seeing as it has been reduced to a purely mathematical problem.

I thank you for your time and efforts, they have made me rethink infinitesimals.

BiP

It sounds like you really want to assume that epsilon is an arbitrarily small positive number (i.e. small enough that only first order effects occur). In this case the solution in my previous post essentially does it for us. There are two cases

1) $\Delta S \geq 0$. Then $\epsilon \Delta S \geq 0$ as well, and if $\Delta G \geq 0$ adding these two inequalities yields $\Delta G + \epsilon \Delta S \geq 0$ which we know isn't true. So it must be $\Delta G < 0$

2) $\Delta S < 0$. We will do a proof by contradiction: suppose that $\Delta G > 0$. Then we will show that if $\epsilon$ is small enough (smaller than a specific number which we will calculate), that $\Delta G + \epsilon \Delta S > 0$ which is a contradiction.

If $\epsilon < -\Delta G / \Delta S$ (which is a positive number since by assumption, delta S is negative and delta G is positive), then $\Delta G + \epsilon \Delta S > \Delta G + (-\Delta G / \Delta S) \Delta S$ where the inequality is > because we're increasing the coefficient of delta S, which is a negative number by assumption. Of course, cancelling the fractions then gives us $\Delta G + \epsilon \Delta S > \Delta G - \Delta G = 0$ which is a contradiction.Note that in this case we had to assume delta G was non-zero, otherwise epsilon would have been exactly zero. But if delta G is exactly zero there's nothing to prove.

So we had two cases: one where delta S was non-negative, and one where delta S was negative. In both cases we proved that delta G must be negative, as long as we can assume epsilon is small enough.

Great! Thanks! I guess I must've just not understood your original solution, but I get this one!

BiP

Office_Shredder, all the time in the sciences like physics and chemistry, we do calculus involving infinitesimals. This can of course be put on a firm rigorous foundation in any number of ways, including hyperreals and differential forms, but we often just use techniques without always referring to their foundations. Just like you take limits all the time without writing out the epsilon-delta definition, we can write things like dA=rdrdθ without writing out the definition of the wedge product or proving Jacobi's theorem. In thermodynamics, things like δW=-PδV is often more useful for our purposes than taking an explicit derivatives, but we don't lay out the theory of inexact differentials in our calculations.

## 1. What is a limit in chemistry?

A limit in chemistry refers to the maximum or minimum amount of a substance that can be present in a reaction or solution before a change in the reaction or solution occurs. This can also refer to the maximum or minimum concentration of a substance that can be achieved in a given solution.

## 2. Why are limits important in chemistry?

Limits are important in chemistry because they help us understand the behavior of substances in chemical reactions and solutions. By determining the limits, we can predict the outcome of a reaction or the solubility of a substance in a given solution.

## 3. How are limits determined in chemistry?

Limits in chemistry are determined through experiments and calculations. The experimental method involves gradually increasing or decreasing the amount of a substance in a reaction or solution until a change is observed. The calculated method involves using mathematical equations and principles to predict the limit based on the properties of the substances involved.

## 4. What factors can affect limits in chemistry?

The factors that can affect limits in chemistry include temperature, pressure, concentration, and the nature of the substances involved. These factors can influence the rate of a reaction, the solubility of a substance, and the equilibrium point of a reaction.

## 5. How can knowing limits help in practical applications of chemistry?

Knowing the limits in chemistry is crucial for practical applications such as drug development, water treatment, and food production. By understanding the limits, chemists can optimize reactions and solutions to achieve desired outcomes, and prevent unwanted changes or reactions. This helps in creating efficient and effective processes for various industries.

Replies
4
Views
1K
Replies
2
Views
2K
Replies
13
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
8
Views
2K
Replies
9
Views
365
Replies
4
Views
2K
Replies
7
Views
2K