How Can a Balloon's Volume Increase When Cooled?

In summary: ERICA!In summary, the balloon will have a larger volume when it is cooled to 30F in a refrigerator than when it is at 90F.
  • #1
nicksim117
23
0

Homework Statement


On a warm day 90F, the air in aballon occupies a volume of .25m3 and exerts a pressure of 20 lb/in2. If the baloon is cooled to 30F in a refrigerator, the pressure drops to 14.2lb/in2. What is the volume of the balloon


Homework Equations


P1xV1/T1=P2xV2/T2-----((P1xV1)/T1)xT2/P2
T1=305.2K
T2=271.88K

The Attempt at a Solution


.31m3 is the answer i keep getting How is this possible when this volume is larger than the original.
 
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  • #2
You've got two competing effects here. First, there's a temperature drop, which tends to make the balloon shrink. But you've also got a pressure drop, which tends to make it expand. It just so happens here that the effect of the pressure drop dominates the effect of the temperature drop, hence you have a larger balloon.
 
  • #3
But the pressure drop is inside the balloon.
 
  • #4
There is no way you are going to have a larger balloon when you cool it down in a freezer.
 
  • #5
What values did you use for T1 and T2? In the ideal gas law, termperatures must be given in degrees Kelvin.
 
  • #6
I listed them in the problem T1 305.2 T2-271.88
 
  • #7
nicksim117 said:
But the pressure drop is inside the balloon.

Yes that's true. And how that pressure drop is achieved, I do not know as the problem doesn't say. All I know is that under the temperature and pressure drops given in the problem statement, you do in fact get a larger volume.

It could be that the data given in the problem is unrealistic, but your calculation is nonetheless correct.
 
  • #8
I suspect this is yet another case of someone making up a problem without checking to see if the given information makes sense in terms of a natural process. I don't really believe that the pressure in the balloon would drop to that value by chilling it in the fridge that way. However, the answer of 0.314 m^3 is what you get from the ideal gas law using that information. (Yeah, right: I want to see how they got a balloon with a diameter over 60 cm. into the refrigerator to start with... Maybe it's a "walk-in"...).

The point is that, for the purposes of this problem [grinds teeth], the balloon would have to have expanded and cooled from its initial state in order to reach that final state. You did the problem right -- it's not your fault it makes little sense...
 
  • #9
nicksim117 said:

The Attempt at a Solution


.31m3 is the answer i keep getting How is this possible when this volume is larger than the original.
Your answer is correct.

The problem here is that you are imagining that this is occurring in the real atmosphere. It isn't.

If the atmospheric pressure was held constant (i.e as in the real atmosphere), when the temperature decreases the internal pressure decreases and the balloon volume is reduced until the pressure inside balances the atmospheric pressure + pressure of the balloon material. This is what you are imagining is occurring. But it isn't. In this problem you end up with a lower pressure.

In the problem, you have a balloon at 90F experiencing a reduction in the external pressure from 20 lb to 14.2 This will result in expansion of the gas and, therefore, cooling. In fact, the expansion due to pressure reduction will result in the gas cooling to even less than 30 F. So in the problem, heat actually flows into the balloon.

AM
 

Related to How Can a Balloon's Volume Increase When Cooled?

1. What is the ideal gas law?

The ideal gas law is a formula that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is written as PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

2. How is the ideal gas law used in science?

The ideal gas law is used in various fields of science, including chemistry, physics, and engineering. It is especially useful in studying the behavior of gases under different conditions, such as changes in pressure, volume, and temperature.

3. What are the units of measurement for the variables in the ideal gas law?

The units for the variables in the ideal gas law are as follows: P is measured in Pascals (Pa), V is measured in cubic meters (m^3), n is measured in moles (mol), R is measured in Joules per mole per Kelvin (J/mol·K), and T is measured in Kelvin (K).

4. Can the ideal gas law be used for any type of gas?

The ideal gas law is most accurate for low pressure and high temperature conditions, and for gases that do not interact with each other. However, it can still provide useful approximations for real gases under different conditions.

5. How can the ideal gas law be manipulated to solve for different variables?

The ideal gas law can be rearranged to solve for any of its variables, depending on the information that is given. For example, to solve for pressure, the formula can be rewritten as P=nRT/V, while to solve for volume, the formula can be written as V=nRT/P. It is important to make sure that all units are consistent when manipulating the equation.

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