You fill a balloon with helium gas to a volume of 2.96 L at 23 °C and 758 mmHg. Now you release the balloon. What would be the volume of the helium if its pressure changes to 636 mmHg but the temperature is unchanged?
2007-10-16 02:25:59 · 7 answers · asked by Bernie 1 in Science & Mathematics ➔ Chemistry
Boyle's law: P1V1 = P2V2
V2 = P1V1/P2
As you point out, T doesn't change, so it doesn't matter that 23C = 296K
2007-10-16 02:44:14 · answer #1 · answered by steve_geo1 7 · 1⤊ 0⤋
Ideal gas law. PV=nRT.
Since the amount of gas and temperature remains the same, PV= constant, ie the pressure is inversely proportional to the volume.
So, P1*V1 =P2*V2 (should use subscrpit numbers)
P1= 758mmHg, V1= 2.96L, P2= 636mmHg
V2= (758/636)*2.96L
2007-10-16 09:42:31 · answer #2 · answered by BotsMaster 3 · 0⤊ 0⤋
Look up Boyle's Law T does not change. Get yourself a tutor if your in the dark because it only gets harder.
2007-10-16 10:07:25 · answer #3 · answered by Veritas 7 · 0⤊ 0⤋
use the ideal gas law. PV=nRT solve for n, then use that value in the second part where you substitute the new pressure in.
2007-10-16 09:53:58 · answer #4 · answered by Brent 3 · 0⤊ 0⤋
P1V1 = P2V2
since temp is same, T gets canceled...
2007-10-16 10:24:33 · answer #5 · answered by Kay :) 3 · 0⤊ 0⤋
cause the temperature remain unchange. cause they are the same throughout. lol, i don't know what i'm talking about. but isn't it common sense?
2007-10-16 09:29:10 · answer #6 · answered by V L 3 · 0⤊ 0⤋
P1V1 = P2V2
2007-10-16 09:41:26 · answer #7 · answered by gelo 2 · 1⤊ 0⤋