What are you doing wrong? You're using gauge pressure (psig) and not "real" pressure (psia). What you want to do is to convert gauge pressure to absolute (real) pressure, use P1V1/T1 = P2V2/T2, solve for new P2, convert back to gauge.
pgauge + 14.7 = pabsolute
You start with 36.0+14.7 = 50.7 for P1.
V1 is 11.8L and T1 is 285k.
P2 is ?, V2 is 12.2L and T2 is 338K. Solve for P2(absolute) and I get something like 58.2, then 58.2-14.7 = 43.5 psig gauge and that exceeds 38.0 psig.
An automobile tire has a maximum rating of 38.0 psi (gauge pressure). The tire is inflated (while cold) to a volume of 11.8 Liters and a gauge pressure of 36.0 psi at a temperature of 12.0 Celsius. While driving on a hot day, the tire warms to 65.0 Celsius and it's volume expands to 12.2 Liters. Does the pressure in the exceed it's maximum rating? (Note: the gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi.)
I have tried over again to figure out the problem. I have used the formula
(P1)(V1) / (n1)(R)(T1) = (P2)(V2) / (n2)(R)(T2)
R= 0.08206 (liters)(ATM) / (mol)(kelvin)
I have converted the Celsius to kelvin, and the psi to ATM, but the answer I have in the book. Which is "Yes, the final gauge is 43.5 psi which exceeds the maximum rating."
I have tried to figure it out and have tries to do all the work, but my answer doesn't match the books answer. If you can please help me that would be much appreciated. Thanks coco😃
3 answers
Coco, sorry it's a "little" late, but did you ever find an answer to this question? Can I ask where you came across this question? I just came across it in a textbook and discovered you need to use the Combined Gas Law (Gay-Lussac's Law) which isn't even taught in the textbook.
Hello, I have the same problem with identical units in my textbook. 43.5 psi is the answer at the back of the text book. Thank you!
1 answer