Question

Air (a diatomic ideal gas) at 26.5°C and atmospheric pressure is drawn into a bicycle pump (see figure below) that has a cylinder with an inner diameter of 2.50 cm and length 47.0 cm. The downstroke adiabatically compresses the air, which reaches a gauge pressure of 8.00 105 Pa before entering the tire. We wish to investigate the temperature increase of the pump.

The pump is made of steel that is 1.80 mm thick. Assume 4.00 cm of the cylinder's length is allowed to come to thermal equilibrium with the air.

g. What is the volume of steel in this 4.00-cm length?

h. What is the mass of steel in this 4.00-cm length? (The density of steel is 7.86 103 kg/m3.)

i. Assume the pump is compressed once. After the adiabatic expansion, conduction results in the energy increase in part (f) being shared between the gas and the 4.00-cm length of steel. What will be the increase in temperature of the steel after one compression? (The specific heat of steel is 448 J/kg · °C.)

Answer 1

g )

given inner diameter is 2.5 cm

r₁ = 2.5 /2 = 1.25 cm

= 0.0125 m

and

r₂ = 0.0125 + 0.0018 ( given 1.8 mm thick = 0.0018 m )

= 0.0143 m

and h = 4 cm

h = 0.04 m

using V_steel = h ( r₂² - r₁² )

= 3.14 x 0.04 x ( 0.0143² - 0.0125² )

V_steel = 6.058 x 10^-6 m³

h )

The density of steel is 7.86 x 10³ kg/m³

mass = density x volume

m = 7.86 x 10³ x 6.058 x 10^-6

m = 0.047623 kg

or

m = 47.623 g

i )

using equation

T = E / ( n_air C_v + m_steel C )

here

C_v = 5R/2

R = 8.314

E value needed from previous,

so remaining plug in the values.