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Hi,

Today we did a practical investigating Charles' Law. There are a couple of practical questions that I didn't understand. I'd really appreciate it if someone could explain.

Thanks

Today we did a practical investigating Charles' Law. There are a couple of practical questions that I didn't understand. I'd really appreciate it if someone could explain.

Thanks

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#2

(Original post by

Hi,

Today we did a practical investigating Charles' Law. There are a couple of practical questions that I didn't understand. I'd really appreciate it if someone could explain.

Thanks

**VioletPhillippo**)Hi,

Today we did a practical investigating Charles' Law. There are a couple of practical questions that I didn't understand. I'd really appreciate it if someone could explain.

Thanks

More detail of the experiment, and what you plotted please!

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(Original post by

I'm afraid that, to tell you how the graph would change, I would need some idea of what graph you plotted!

More detail of the experiment, and what you plotted please!

**MexicanKeith**)I'm afraid that, to tell you how the graph would change, I would need some idea of what graph you plotted!

More detail of the experiment, and what you plotted please!

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#4

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We plotted length of trapped air column (in a capillary tube) against water temperature.

**VioletPhillippo**)We plotted length of trapped air column (in a capillary tube) against water temperature.

V/T = k is just telling you that, according to Charles' law you should get a straight line graph because V= kT so V and T are proportional to one another.

If k increases, then for a given value of T, V (and hence length of air) will be larger, but T and V must still be proportional so you would observe a graph with a straight line but now with a steeper gradient, the reverse is true for k decreasing.

That answers question 7, just wondering, how did you calculate the volume at absolute zero? did you simply extrapolate the line through the points?

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(Original post by

So the length of the trapped air is proportional to its volume.

V/T = k is just telling you that, according to Charles' law you should get a straight line graph because V= kT so V and T are proportional to one another.

If k increases, then for a given value of T, V (and hence length of air) will be larger, but T and V must still be proportional so you would observe a graph with a straight line but now with a steeper gradient, the reverse is true for k decreasing.

That answers question 7, just wondering, how did you calculate the volume at absolute zero? did you simply extrapolate the line through the points?

**MexicanKeith**)So the length of the trapped air is proportional to its volume.

V/T = k is just telling you that, according to Charles' law you should get a straight line graph because V= kT so V and T are proportional to one another.

If k increases, then for a given value of T, V (and hence length of air) will be larger, but T and V must still be proportional so you would observe a graph with a straight line but now with a steeper gradient, the reverse is true for k decreasing.

That answers question 7, just wondering, how did you calculate the volume at absolute zero? did you simply extrapolate the line through the points?

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#6

(Original post by

Thanks so much, that's really helpful For Q7. I extrapolated back to the x axis to find an approx value for absolute zero.

**VioletPhillippo**)Thanks so much, that's really helpful For Q7. I extrapolated back to the x axis to find an approx value for absolute zero.

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(Original post by

Fair enough, Question 9 is more difficult. However, if we can assume that each reading was wrong by the same amount (eg by 2mm), then its easy, it would simply mean that the line was translated upwards by 2mm and so the value at absolute zero would be higher!

**MexicanKeith**)Fair enough, Question 9 is more difficult. However, if we can assume that each reading was wrong by the same amount (eg by 2mm), then its easy, it would simply mean that the line was translated upwards by 2mm and so the value at absolute zero would be higher!

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#8

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Thanks There's just one more question which I can't find the answer to which is why nitrogen and oxygen provide good experimental data but carbon dioxide doesn't for gas law experiments.

**VioletPhillippo**)Thanks There's just one more question which I can't find the answer to which is why nitrogen and oxygen provide good experimental data but carbon dioxide doesn't for gas law experiments.

What it boils down to is the fact that Charles' law (which is an experimental demonstration of the ideal gas equation) only applies for an

**ideal**gas.

Ideal models always involve simplifying the situation so that it is easily modelled.

For the ideal gas model we assume a few things to simplify the problem.

- The Gas molecules should obey newtons laws of motion (pretty good approximation)
- The volume of gas molecules should be negligible compared to volume of the container they are in (pretty good providing the gas isn't close to the temperature at which it would condense)
- There are no interactions between gas molecules. (this one can cause more problems for some gases)

_{2}and O

_{2}have bonds which aren't at all polar. So no atoms in these molecules have any sort of charge and the interactions between molecules are minimised, that means that the third assumption of the ideal gas model is pretty good and so these are good gases with which to demonstrate ideal gas behaviour.

CO

_{2}is symmetrical and so

**overall**has no dipole, however each C=O double bond is polarised, this means that the Oxygen atoms have partial negative charge and the Carbon has partial positive charge.

Opposite charges attract and like charges repel and so there is a significantly larger degree of intermolecular interaction in CO

_{2}. This makes it less ideal and so it is a worse choice of gas to demonstrate ideal behaviour.

(This extra intermolecular interaction is also the reason that, at atmospheric pressure, CO

_{2 }stops being a gas below -79

^{o}C, whilst N

_{2 }is still gaseous all the way down to -196

^{o}C)

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(Original post by

That's an interesting question.

What it boils down to is the fact that Charles' law (which is an experimental demonstration of the ideal gas equation) only applies for an

Ideal models always involve simplifying the situation so that it is easily modelled.

For the ideal gas model we assume a few things to simplify the problem.

CO

Opposite charges attract and like charges repel and so there is a significantly larger degree of intermolecular interaction in CO

(This extra intermolecular interaction is also the reason that, at atmospheric pressure, CO

**MexicanKeith**)That's an interesting question.

What it boils down to is the fact that Charles' law (which is an experimental demonstration of the ideal gas equation) only applies for an

**ideal**gas.Ideal models always involve simplifying the situation so that it is easily modelled.

For the ideal gas model we assume a few things to simplify the problem.

- The Gas molecules should obey newtons laws of motion (pretty good approximation)
- The volume of gas molecules should be negligible compared to volume of the container they are in (pretty good providing the gas isn't close to the temperature at which it would condense)
- There are no interactions between gas molecules. (this one can cause more problems for some gases)

_{2}and O_{2}have bonds which aren't at all polar. So no atoms in these molecules have any sort of charge and the interactions between molecules are minimised, that means that the third assumption of the ideal gas model is pretty good and so these are good gases with which to demonstrate ideal gas behaviour.CO

_{2}is symmetrical and so**overall**has no dipole, however each C=O double bond is polarised, this means that the Oxygen atoms have partial negative charge and the Carbon has partial positive charge.Opposite charges attract and like charges repel and so there is a significantly larger degree of intermolecular interaction in CO

_{2}. This makes it less ideal and so it is a worse choice of gas to demonstrate ideal behaviour.(This extra intermolecular interaction is also the reason that, at atmospheric pressure, CO

_{2 }stops being a gas below -79^{o}C, whilst N_{2 }is still gaseous all the way down to -196^{o}C)
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#10

(Original post by

Thanks for all your help, I really appreciate it. This was a physics practical but I was trying to think it through using more chemistry based ideas.

**VioletPhillippo**)Thanks for all your help, I really appreciate it. This was a physics practical but I was trying to think it through using more chemistry based ideas.

Happy to help

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