A boy Inflates hls blcycle tire with a hand-operated pump. Kinetic theory helps
ID: 1504402 • Letter: A
Question
A boy Inflates hls blcycle tire with a hand-operated pump. Kinetic theory helps to describe the details of the air in the pump. (OCengoge Learning/George Semple) (a) What is the initial volume of the air in the pump? Enter a number (b) What is the number of moles of air in the pump? mol (c) What is the absolute pressure of the compressed air? Pa (d) What is the volume of the compressed air? (e) What is the temperature of the compressed air? (f) What is the increase in internal energy of the gas during the compression? The pump is made of steel that is 1.85 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 x 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 1/kg °C.) Need Help? Read It Submit Answer Save Progress Practice Another VersionExplanation / Answer
Here ,
temperature, T1 = 30 degree C
T1 = 303 K
inner diameter , di = 2.50 cm
length , L = 52.5 cm
P2 = 8 *10^5 Pa gauge
a) for the initial volume of the air
initial volume of the air = pi * (d/2)^2 * L
initial volume of the air = 3.141* (0.025/2)^2 * 0.525
initial volume of the air = 2.58 *10^-4 m^3
b)
let the number of moles of gas is n
as P1 * v1 = n * R * T1
(1.01 *10^5 )* 2.58 *10^-4 = n * 8.314 * 303
n = 0.0103
the number of moles of air in pump is 0.0103
c)
absolute pressure of compressed air , Pa = Pg + P2
Pa = (8 *10^5 + 1.01 *10^5 )
Pa = 9.01 *10^5 Pa
the absolute pressure of compressed air is 9.01 *10^5 Pa
d)
for the final volume of the compressed air
for abiabatic process
as PV^1.4 = constant
1.01 *10^5 * (2.58 *10^-4)^1.4 = (8 *10^5 + 1.01 *10^5 ) * V2^1.
v2 = 5.405 *10^-5 m^3
the volume of compressed air is 5.405 *10^-5 m^3
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