physics - another thermo problem
posted 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?
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
thanks. i guess the dimensions in the beginning was supposed to throw me off