Armand explained to his friend Bertrand that, in general, if you have a gas container and the amount of gas is doubled, the new pressure will be double the old pressure. The term "general" is employed because this is not exactly accurate, but it is close enough for the purposes of answering his next query.
Armand now posited this question to Bertrand, "If you have a tire filled with the standard 32 psi and you double the amount of air molecules in the tire, what pressure will your tire gauge now read? Assume that the tire does NOT expand, and that the first paragraph containing my initial information is exactly true."
Bertrand immediately answered, "64 psi."
But, Armand replied back to him, "Your answer is incorrect!"
What is the correct answer and why?
When the air is pumped into the tire it is going to heat up as well as be confined to the same small space as the original air occupied, so in
pV = nRT (p = pressure, V = Volume, nR proportional to # of molecules, T = absolute temperature)
both n*R and T are increasing, and since n is doubling while T also increases (R is a constant), and V remains the same small volume, p must more than double.
As to how much it doubles, that depends on the gas used--in this case air. Air is not a good refrigerant--It doesn't do as much heating and cooling on compression and expansion respectively as say freon, so the exact answer must depend on some thermodynamic value inherent in air.
Addendum: now, having read Kenny M's comment, I see I neglected the major contribution: the fact that a reading of 32 psi really represents 32+14.7 = 46.7 psi and that doubles to 93.4 psi which needs to have 14.7 subtracted making the reading 78.7 psi, plus whatever amount is added by the heating of compression.
Edited on September 5, 2022, 10:05 am
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Posted by Charlie
on 2022-09-05 09:57:18 |
inexact solution
