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Help with ideal gases (simple thermo question)?
So, an ideal gas is contained in a bottle (pressure in the bottle is 1.29 times the atmospheric pressure). When the bottle is opened, the gas is expelled and the pressure inside the bottle is reduced to the atmospheric pressure. What is the temperature of the gas inside the bottle immediately after the pressure is released?
1 Answer
- ?Lv 78 years agoFavourite answer
Gay-Lussac's law :
(P1 / T1) = (P2 / T2)
P1 = 1.29 atm
P2 = 1.00 atm
T1 (is not given, so just call it T1)
T2 = ?
Rearranging the Gay-Lussac law to solve for T2 gives
T2 = [T1 x (P2 / P1)]
= [T1 x (1.00 atm / 1.29 atm)]
= 0.775 T1
Whatever the initial temperature was, (T1), after opening the bottle the temperature is theoretically (0.775 x T1). However, I have a little trouble believing that this temperature change happens 'immediately'.
