Thermal expansion of air inside a bottle -- some questions....

In summary, the pressure exerted on the inside of the bottle due to the air inside expanding will be the same everywhere else inside the bottle according to the perfect gas law.
  • #1
James124900
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For thermal expansion in a bottle - is there a theoretical model on the pressure exerted on the inside of the bottle due to the air inside expanding when the bottle heats up.
The air will be expanding randomly in all directions so the pressure exerted at any point on the inside of the bottle walls will be the same everywhere else inside the bottle.
 
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  • #3
anorlunda said:
Yes, it is called the perfect gas law. https://en.wikipedia.org/wiki/Ideal_gas_law
Ok I do not really understand this law and how I will use it in a practical situation with air in a bottle where I will need the molar mass to calculate the pressure.
 
  • #4
All you need to know is the change Use PV=nRT

Let's say you know P, V and T initially Then nR/V=P/T.

Now increase the temperature from ##T_1## to ##T_2##. nR remains constant. The volume of the bottle remains constant. That means

##\frac{P_1}{T_1}=\frac{P_2}{T_2}##

(be careful T is in degrees Kelvin)

Does that help with your problem?
 
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  • #5
Yes. Thank you. But how will I know nR? and just to be sure I will calculate initial pressure with PV=nRT (or is it just atmospheric pressure?).

Edit: Did you mean to calculate nR with initial P, V and T?
(I do not have P - unless it is just atmospheric pressure. However I am not sure of this since the bottle will be cooled to a low temperature before being heated up).
 
  • #6
James124900 said:
Did you mean to calculate nR with initial P, V and T?

Yes, that's what I meant. You can't solve the problem unless you know initial P and T.
 
  • #7
anorlunda said:
Yes, that's what I meant. You can't solve the problem unless you know initial P and T.
Ok, is initial P atmospheric pressure, or would it be different since the bottle will have been cooled?
 
  • #8
James124900 said:
Ok, is initial P atmospheric pressure, or would it be different since the bottle will have been cooled?

I can't answer that because I don't know the circumstances. It is a good guess that at some point in history it was at atmospheric presssure. If you know the temperature and volume at that time, then that is your initial point.
 
  • #9
James124900 said:
Ok, is initial P atmospheric pressure, or would it be different since the bottle will have been cooled?
It's your scenario, so you are going to have to provide the details.
 
  • #10
russ_watters said:
It's your scenario, so you are going to have to provide the details.
A bottle is cooled, I know the temperature and volume of the bottle. How can I calculate the pressure in the bottle to calculate nR?
 
  • #11
Before the bottle was sealed, you need the P and T. Possibly atmospheric pressure and room temperature. Then you have P1 and T1. You can then calculate a P2 for any other T2.

If you don't know anything about the conditions when the bottle was sealed, then you're stuck.
 
  • #12
anorlunda said:
Before the bottle was sealed, you need the P and T. Possibly atmospheric pressure and room temperature. Then you have P1 and T1. You can then calculate a P2 for any other T2.

If you don't know anything about the conditions when the bottle was sealed, then you're stuck.
Ok thanks.
 
  • #13
Hey I got another question related to the above concepts:
I have two closed bottles one has a smaller volume that the other. (initially atmospheric pressure in both and then closed with a cap)
If I heat both bottles up, which one will increase in pressure more the larger bottle or smaller one.
 
  • #14
James124900 said:
Hey I got another question related to the above concepts:
I have two closed bottles one has a smaller volume that the other. (initially atmospheric pressure in both and then closed with a cap)
If I heat both bottles up, which one will increase in pressure more the larger bottle or smaller one.

You can use the answers you already have.
 
  • #15
How? because both bottles will have the same pressure to start with (it it not like the bottle decreases in volume thus when smaller will have a larger pressure)
so I am confused which one will increase in pressure more.
 
  • #16
James124900 said:
How? because both bottles will have the same pressure to start with (it it not like the bottle decreases in volume thus when smaller will have a larger pressure)
so I am confused which one will increase in pressure more.

you have the initial P, V and T for each bottle.
you know the volume for each bottle isn't going to change so use the formula you were shown to work out change in P for changes in T for each bottle
 
  • #17
davenn said:
you have the initial P, V and T for each bottle.
you know the volume for each bottle isn't going to change so use the formula you were shown to work out change in P for changes in T for each bottle
I don't understand why the smaller bottles have a larger pressure than the smaller bottles result in a larger pressure.
 
  • #18
James124900 said:
I don't understand why the smaller bottles have a larger pressure than the smaller bottles result in a larger pressure.
that didn't make sense ... did you do the working out ?

show us your working so somebody can check it :smile:
 
  • #19
no calculations. I just don't understand why a closed smaller bottle will increase in temperature more than a larger bottle when both are heated by the same amount.
 
  • #20
James124900 said:
no calculations. I just don't understand why a closed smaller bottle will increase in temperature more than a larger bottle when both are heated by the same amount.
Do you understand why heating a swimming pool for one hour under a single heat lamp will result in less temperature rise than doing the same with your kitchen sink?

It appears that the intended query had to do with the increase in pressure when a smaller bottle has the same increase in temperature as a larger bottle. But I see no foundation for such a question. Who is trying to claim that such a difference in pressure increase will be seen?
 
Last edited:
  • #21
jbriggs444 said:
Do you understand why heating a swimming pool for one hour under a single heat lamp will result in less temperature rise than doing the same with your kitchen sink?
I meant both bottle's will be heated up to the same temperature.

Edit: Sorry for my other post I meant increase in pressure not temperature
 
  • #22
James124900 said:
I meant both bottle's will be heated up to the same temperature.

Edit: Sorry for my other post I meant increase in pressure not temperature
James124900 said:
I meant both bottle's will be heated up to the same temperature.

Edit: Sorry for my other post I meant increase in pressure not temperature
Now you have me confused. What are you trying to ask and why?
[My reply has been editted]
 
  • #23
Part of this is the multiple uses for the word "heating". With a slight difference in wording it can mean "to add heat" or "to increase temperature."

I think what is intended here is both bottles start at the same temperature and pressure and end at the same temperature...so they must end at the same pressure.
 
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Related to Thermal expansion of air inside a bottle -- some questions....

1. What causes thermal expansion in the air inside a bottle?

Thermal expansion in the air inside a bottle is caused by an increase in temperature. As the temperature of the air increases, the molecules gain kinetic energy and move farther apart, causing the air to expand.

2. How does the expansion of air affect the bottle?

The expansion of air inside a bottle can cause an increase in pressure inside the bottle, which can lead to the bottle becoming more rigid and potentially causing it to burst if the pressure becomes too great.

3. What factors can affect the thermal expansion of air inside a bottle?

The thermal expansion of air inside a bottle can be affected by the volume of the bottle, the initial temperature of the air, and the material of the bottle itself. The type of gas inside the bottle can also play a role in the expansion.

4. How does the thermal expansion of air inside a bottle relate to the ideal gas law?

The ideal gas law, which states that pressure is directly proportional to temperature and volume, can be used to explain the thermal expansion of air inside a bottle. As the temperature increases, the pressure and volume of the air also increase.

5. Can the thermal expansion of air inside a bottle be controlled or prevented?

The thermal expansion of air inside a bottle can be controlled or prevented by keeping the bottle at a constant temperature, using materials that have a low coefficient of thermal expansion, and ensuring that the bottle is not completely filled with air to allow for some expansion. Additionally, using a pressure release valve or venting the bottle can also help prevent excess pressure buildup.

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