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I'd tried to put a balloon in the freezer and after enough time (several hours), I took it out and noticed that as I touch it that it's temperature did not drop down . I took an empty balloon ( not blown) and put it inside the same freezer and it did not take long time for it's temperature to drop down !. To measure any changes in the volume, I tied a thin thread around the circumference of the balloon. After several hours , I'd took out and noticed that this thread was now loose around the balloon not like how I'd left it when I inserted the balloon in the freezer.

Note : I know that my experiments are not fully scientific and accurate,but there is something that is going on ! .

Fadi
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2 Answers2

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Let me quote this line which says that:

I touch it that it's temperature did not drop down

Its better to use thermometer to check the readings as it gives you accurate reading. Please check this link as it shows what you did wrong:

Why does cold metal seem colder than cold air?

The process of touching and determining its temperature is wrong.

There is an answer mentioned in the above link:

"The thermometer measures actual temperature (which is the same for both), while your hand measures the transfer of energy (heat), which is higher for the pot than the air."

So there you were not able to determine the temperature of air correctly due to which you were not able to guess if the temperature decreases or not. So its recommended to use thermometer. I guess now you know what's happening behind that. So coming back to the answer we know that: $$PV = nRT$$

Since you kept balloon in freezer so temperature will decrease. But we know that

$$ V \propto T$$

So if T will decrease then for sure volume will also decrease. And as you mentioned that

this thread was now loose around the balloon

So we can for sure say that there is decrease in temperature that's why there is change in volume. So the process where you touch and determined the temperature was not appropriate.

Community
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Shashank
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  • alright then I will use a thermometer.I know what you are trying to say about my way of determining the temp. changes. But, Comparatively speaking,the drop is much lower when the same type of balloon is blown !. I guess that the thickness of the balloon when changed , the heat transfer also is changed. – Fadi Jun 17 '15 at 21:19
  • That's also right. Now I guess you got your answer now. – Shashank Jun 18 '15 at 04:32
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I think you might have actually touched on something interesting here. One explanation for the difference in temperature is simply that part of the energy removed from the balloon + air system comes from the air, so the balloon will cool more slowly when filled with air.

But there might be more to the story. I haven't done any calculations on this, so it's very speculative. I would like to hear other people's thoughts.

Using some jargon, my hypothesis is this: the inflated balloon takes longer to cool partly because the heat capacity of the balloon in the high tension = low entropy state is higher than in the deflated state. You could verify this using rubber band thermo, and showing that the heat capacity at constant length increases with increasing tension: $(\frac{\partial C_{L}}{\partial \tau})_L = (\frac{\partial^2 U}{\partial T \partial \tau})_L > 0$. Though this won't give any info on order of magnitude, so this effect might be negligible. Could someone comment?

Kyle Arean-Raines
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  • To test your hypothesis on heat capacity-tention , How about stretching a rubber piece of balloon on some kind of plate ? . This will exclude any effects of pressure and heat of blown air !. – Fadi Jun 19 '15 at 08:20