a) For adiabatic processes we know PVg = constant. Hence, for two states, a and

Question

a) For adiabatic processes we know PVg = constant. Hence, for two states, a and b, on the same adiabat we have PaVag = PbVbg. Use the ideal gas law to show that the temperatures of the states are related by Ta/Tb = (Va/Vb)1-g

b) When the handle of a bicycle pump is pushed in quickly while the air outlet at the base is blocked, the air is compressed adiabatically. Calculate the final temperature of the air inside the pump, when the handle of a pump with a cylinder of length 38 cm and diameter 3 cm is pushed quickly halfway down. Assume the air is initially at a temperature of 20oC and at atmospheric pressure.

Explanation / Answer

PaVa^g = PbVb^g a) PV = nRT P = nRT/V (nRTa/Va)*Va^g = (nRTb/Vb)*Vb^g (Ta/Va)*Va^g = (Tb/Vb)*Vb^g TaVa^(g-1) = TbVb^(g-1) Ta/Tb = (Vb/Va)^(g-1) = (Va/Vb)^(1-g) b) Vb = 0.5Va 293/Tb = 2^(1-g) Tb = 293/2^(1-g) for ideal air, g = 7/5 Tb = 293/2^(1 - 7/5) Tb = 293*2^(2/5) = 386.61 K = 113.61 degrees celsius

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