# Charles' law practical questions help

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#1
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
0
3 years ago
#2
(Original post by 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
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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#3
(Original post by 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!
We plotted length of trapped air column (in a capillary tube) against water temperature.
0
3 years ago
#4
(Original post by VioletPhillippo)
We plotted length of trapped air column (in a capillary tube) against water temperature.
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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#5
(Original post by 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?
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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3 years ago
#6
(Original post by 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.
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!
1
#7
(Original post by 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!
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.
0
3 years ago
#8
(Original post by 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.
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.
1. The Gas molecules should obey newtons laws of motion (pretty good approximation)
2. 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)
3. There are no interactions between gas molecules. (this one can cause more problems for some gases)
N2 and O2 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.

CO2 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 CO2. 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, CO2 stops being a gas below -79oC, whilst N2 is still gaseous all the way down to -196oC)
0
#9
(Original post by 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.
1. The Gas molecules should obey newtons laws of motion (pretty good approximation)
2. 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)
3. There are no interactions between gas molecules. (this one can cause more problems for some gases)
N2 and O2 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.

CO2 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 CO2. 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, CO2 stops being a gas below -79oC, whilst N2 is still gaseous all the way down to -196oC)
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.
1
3 years ago
#10
(Original post by 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.
Seemed pretty physicsy! Ideal gas law is certainly physics but its very relevant to lots of chemistry too! Always good to be interdisciplinary!
Happy to help
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