A large balloon has a volume of 2.5 L at a temp of 0 degrees cels. what is the new volume of the balloon when the temp rises to 120 degrees cels. when P remains constant?
2007-03-04 01:44:32 · 3 answers · asked by ? 1 in Science & Mathematics ➔ Chemistry
P1.V1/T1 = P2.V2/T2
V1/T1 = V2/T2 when P remains constant (P1 = P2)
Charles Law
V1 = 2.5 L
T1 = 273.15 KELVIN
T2 = 393.15 K
V2 = V1.T2/T1
= 2.5x273.15/393.15 L
[EDIT] Bugger - got the T1 and T2 mixed
V2 = 2.5 X 393.15 / 273.15 L
(balloon gets bigger as the temperature rises)
WTF is this balloon made of, btw?
2007-03-04 01:51:38 · answer #1 · answered by Orinoco 7 · 0⤊ 0⤋
P1V1/T1 = P2V2/T2 P1 P1=760mm, V1=2.5L,T1=0c., P2=760 mm, V2=?,T2=120c solving it 760mm x 2.5L/ 0c =760mm x V2/120c Thus V2+ 2.5Lx120=300L
2007-03-04 01:54:04 · answer #2 · answered by gangico 3 · 0⤊ 0⤋
Charles law: V1T2 = V2T1. Remember to convert to Kelvins
V2 = V1T2/T1 =(2.5L x 393K)/273K = 3.598L or 3.6L (sig fig)
2007-03-04 01:51:14 · answer #3 · answered by docrider28 4 · 0⤊ 0⤋