In a common classroom demonstration, a balloon is filled with air and submerged in liquid nitrogen. The balloon contracts as the gases within the balloon cool. Suppose the balloon initially contains 3.00L of air at a temperature of 29.5∘C and a pressure of 0.999atm. Calculate the expected volume of the balloon upon cooling to -196 ∘C (the boiling point of liquid nitrogen). When the demonstration is carried out, the actual volume of the balloon decreases to 0.61L.
I'm trying to understand the difference between the observed volume of the balloon and the expected volume (which is 0.76 L).

Question

In a common classroom demonstration, a balloon is filled with air and submerged in liquid nitrogen. The balloon contracts as the gases within the balloon cool. Suppose the balloon initially contains 3.00L of air at a temperature of 29.5∘C and a pressure of 0.999atm. Calculate the expected volume of the balloon upon cooling to -196 ∘C (the boiling point of liquid nitrogen). When the demonstration is carried out, the actual volume of the balloon decreases to 0.61L.
I'm trying to understand the difference between the observed volume of the balloon and the expected volume (which is 0.76 L).

kc_kennylau · Answer

I don't know much, but I think the Ideal Gas Law will be useful here?

anonymous · Answer

I was able to use Charle's law to find the expected volume. I'm just not sure how come the expected volume is different from the observed volume (0.16L). Does liquid nitrogen temp fluctuate? Is that why the volumes were different?

aaronq · Answer

it's likely due to the attractive (i.e. intermolecular) forces that, when gases are treated as ideal, are ignored. I can also be accounted for by the solidification of certain molecules with higher melting points than $N_2$ (e.g $CO_2$ MP= ~80 celsius).

anonymous · Answer

That makes sense. Thank you for explaining! :)

aaronq · Answer

glad it makes sense. no problem !