Respuesta :
Answer:
The value is the temperature of the air inside the tire [tex]T_{2} =[/tex] 340.54 K
% of the original mass of air in the tire should be released 99.706 %
Explanation:
Initial gauge pressure = 2.7 atm
Absolute pressure at inlet [tex]P_{1}[/tex] = 2.7 + 1 = 3.7 atm
Absolute pressure at outlet [tex]P_{2}[/tex] = 3.2 + 1 = 4.2 atm
Temperature at inlet [tex]T_{1}[/tex] = 300 K
(a) Volume of the system is constant so pressure is directly proportional to the temperature.
[tex]\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }[/tex]
[tex]\frac{T_{2} }{300} = \frac{4.2}{3.7}[/tex]
[tex]T_{2} =[/tex] 340.54 K
This is the value is the temperature of the air inside the tire
(b). Since volume of the tyre is constant & pressure reaches the original value.
From ideal gas equation P V = m R T
Since P , V & R is constant. So
m T = constant
[tex]m_{1} T_{1} = m_{2} T_{2}[/tex]
[tex]\frac{m_{2} }{m_{1} } = \frac{T_{1} }{T_{2} }[/tex]
[tex]\frac{m_{2} }{m_{1} } = \frac{300}{354.54}[/tex]
[tex]\frac{m_{2} }{m_{1} } =0.00294[/tex]
value of the original mass of air in the tire should be released is [tex]\frac{m_{2} - m_{1}}{m_{1}}[/tex]
[tex]\frac{0.00294-1}{1}[/tex]
⇒ -0.99706
% of the original mass of air in the tire should be released 99.706 %.
