Asked by Nik
A bicycle pump is a cylinder 20 cm long and 3.0 cm in diameter. The pump contains air at 25.0 C and 1.0 atm. If the outlet at the base of the pump is blocked and the handle is pushed in very quickly, compressing the air to half its original volume, how hot does the air in the pump become?
Answers
Answered by
drwls
In an isentropic adiabatic process such as this, P*V^1.4 is a constant for diatomic gases such as air.
Since V decreases by a factor of 2,
P2/P1 = (V1/V2)^1.4 = 2.639
Use the perfect gas law to compute T2
P1*V1/T1 = P2*V2/T2
T2/T1 = (P2V2)/(P1V1) = 2.639*(1/2) = 1.3195
T2 = 1.3195*298.2 K = 393.5K = 120.3 C
Since V decreases by a factor of 2,
P2/P1 = (V1/V2)^1.4 = 2.639
Use the perfect gas law to compute T2
P1*V1/T1 = P2*V2/T2
T2/T1 = (P2V2)/(P1V1) = 2.639*(1/2) = 1.3195
T2 = 1.3195*298.2 K = 393.5K = 120.3 C
Answered by
Nik
thanks. i guess the dimensions in the beginning was supposed to throw me off
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.