cdm1a23
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A tire is filled with air at 15 C and is at a gauge pressure of 220 kPa. If the air temperature goes to 46 C, what fraction of the original air must be released to maintain the original pressure?
Here is what I did:
15 + 273 = 288
46 + 273 = 319
PV = nRT
P1V1 = (n1)288
P2V2 = (n2)319
(n1)288 = (n2)319
(n1) * 288/319 = (n2)
288/319 = .90 so (n2) is 90 percent of (n1), and 10 percent should be removed.
Does this look right?
Thanks Very Much!
Here is what I did:
15 + 273 = 288
46 + 273 = 319
PV = nRT
P1V1 = (n1)288
P2V2 = (n2)319
(n1)288 = (n2)319
(n1) * 288/319 = (n2)
288/319 = .90 so (n2) is 90 percent of (n1), and 10 percent should be removed.
Does this look right?
Thanks Very Much!