Mole balance issue with volume division

In summary, the conversation discusses the process of dividing the mole balance by the volume and taking the limit as the variable approaches 0. This is a standard procedure, but there is confusion about the transition from equation 6 to 7 and why the constant 4pi is sometimes factored out and sometimes left inside the parentheses. The expert is unable to provide an exact answer to this question, as it is up to the preference of the writer.
  • #1
williamcarter
153
4

Homework Statement


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Capture2.JPG


Here we divide the mole balance by the volume ##4*\pi## ##*## r2##*##dr and take lim as dr->0(standard procedure)

1) How exactly he makes the transition from equation from 6 to 7?

Exercise #2(see below please)
Capture3.JPG


Solution for Problem#2
Capture4.JPG


2)Why does he divide here by ##4*\pi## ##*Δr##? and not by ##4*\pi## ##*## r2##*##dr ??

Thank you very much in advance.
 
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  • #2
williamcarter said:

Homework Statement


View attachment 195716 View attachment 195717

Here we divide the mole balance by the volume ##4*\pi## ##*## r2##*##dr and take lim as dr->0(standard procedure)

1) How exactly he makes the transition from equation from 6 to 7?
What is your problem with this?
Exercise #2(see below please)
View attachment 195718

Solution for Problem#2
View attachment 195720

2)Why does he divide here by ##4*\pi## ##*Δr##? and not by ##4*\pi## ##*## r2##*##dr ??

Thank you very much in advance.
Because the r^2 is inside the parenthesis.
 
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  • #3
Chestermiller said:
What is your problem with this?
I don't get exactly how he got equation 7 from 6?

Because the r^2 is inside the parenthesis.

In the first problem he didnt subsitute S(r) with 4pi*r2, he let it just S(r), but why in second problem he put it inside the ( ) as 4pi*r2?

I don't get why he divided in 2nd example just by 4pi*Δr, because I can't see the difference in procedures, between the 2 examples.
 
  • #4
williamcarter said:
In the first problem he didnt subsitute S(r) with 4pi*r2, he let it just S(r), but why in second problem he put it inside the ( ) as 4pi*r2?

I don't get why he divided in 2nd example just by 4pi*Δr, because I can't see the difference in procedures, between the 2 examples.
In Eqn. 7, with ##S=4\pi r^2##, he would have:$$\frac{d(4\pi r^2)N}{dr}=0$$ Factoring out the ##4\pi## leaves: $$\frac{d(r^2N)}{dr}=0$$
 
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  • #5
Chestermiller said:
In Eqn. 7, with ##S=4\pi r^2##, he would have:$$\frac{d(4\pi r^2)N}{dr}=0$$ Factoring out the ##4\pi## leaves: $$\frac{d(r^2N)}{dr}=0$$
Thank you for your reply !
But why does he factor out the 4*pi? I want to understand the principle
 
  • #6
williamcarter said:
Thank you for your reply !
But why does he factor out the 4*pi? I want to understand the principle
Are you saying that you don't think it is valid to factor out the 4 pi, or are you asking "why would anyone want to factor out the 4 pi?"
 
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  • #7
Chestermiller said:
Are you saying that you don't think it is valid to factor out the 4 pi, or are you asking "why would anyone want to factor out the 4 pi?"
I am going for the 2nd choice ''why would anyone want to factor out the 4 pi?'', how to know when to factor it out or when to leave it there?
 
  • #8
williamcarter said:
I am going for the 2nd choice ''why would anyone want to factor out the 4 pi?'', how to know when to factor it out or when to leave it there?
Factoring out the constant (4 pi) simplifies the equation. It's analogous to what we do when we reduce a fraction to its least common denominator.
 
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  • #9
Chestermiller said:
Factoring out the constant (4 pi) simplifies the equation. It's analogous to what we do when we reduce a fraction to its least common denominator.
Thank you for your answer, but why in the first example he did not factor the 4pi out?

How to know when to factor it out and when not to?
 
  • #10
williamcarter said:
Thank you for your answer, but why in the first example he did not factor the 4pi out?

How to know when to factor it out and when not to?
I have no answer for this. It is not addressing any specific problem, so there is no one right way of doing it. It is up to the taste of the writer.
 

1. What is a mole balance issue with volume division?

A mole balance issue with volume division refers to a problem encountered in chemical reactions where the volume of the system changes during the reaction. This can make it difficult to accurately measure and track the number of moles of each substance involved.

2. How does a mole balance issue with volume division affect the accuracy of calculations?

A mole balance issue with volume division can significantly impact the accuracy of calculations because it introduces an additional variable (volume) that must be taken into account. This can lead to errors and discrepancies in the final results.

3. What are some strategies for addressing a mole balance issue with volume division?

One strategy for addressing a mole balance issue with volume division is to use the ideal gas law to account for the change in volume. Another approach is to measure the volume at several different points during the reaction and use an average value in calculations.

4. How do you determine the volume of a system in a mole balance issue with volume division?

The volume of a system can be determined by measuring the change in volume over time using a graduated cylinder or other measuring device. Alternatively, the volume can be calculated using the ideal gas law if the temperature and pressure of the system are known.

5. Can a mole balance issue with volume division be completely eliminated?

No, a mole balance issue with volume division cannot be completely eliminated. However, by carefully measuring and tracking the volume changes, and using appropriate calculations and equations, the impact of this issue can be minimized and the accuracy of results can be improved.

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