Question

A bicycle pump is a cylinder 20 cm long and 3.0 cm in diameter. The pump contains air at 21.0 ∘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?

Answer 1

222.8 K

Step-by-step explanation:

length of cylinder, l = 20 cm = 0.2 m

diameter of cylinder = 3 cm

radius of cylinder, r = 1.5 cm = 0.015 m

Volume of air initially, V1 = πr²l

V1 = 3.14 x 0.015 x 0.015 x 0.2 = 1.413 x 10^-4 m^3

V2 = V1/2

P1 = 1 atm = 1.01 x 10^5 Pa

T1 = 21°C = 21 + 273 = 294 K

γ = 1.4

This is the adiabatic compression

`T_(1)V_(1)^(1-\gamma )=T_(2)V_(2)^(1-\gamma )`

`(T_(2))/(T_(1))=\left ( (V_(1))/(V_(2)) \right )^(1-\gamma )`

`(T_(2))/(294)=2^(1-1.4 )`

T2 = 222.8 K

Thus, the temperature of air is 222.8 K.

Answer 2

388 K

Step-by-step explanation:

`T_(i)` = initial temperature before handle is pushed = 21 C = 21 + 273 = 294 K

`T_(f)` = final temperature after handle is pushed = ?

`V_(i)` = initial Volume

`V_(f)` = final Volume

Given that :

`V_(f) = (0.5) V_(i)`

For adiabatic process we have

`T_(i) V_(i)^(\gamma -1) = T_(f) V_(f)^(\gamma -1)\\(294) V_(i)^(1.4 -1) = T_(f) ((0.5) V_(i))^(1.4 - 1)\\(294) V_(i)^(1.4 -1) = T_(f) (0.5)^(0.4) V_(i)^(1.4 -1)\\(294) = T_(f) (0.5)^(0.4) \\T_(f) = 388 K`