What is the temperature and mass of air in a tire after a change in pressure?

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Homework Help Overview

The discussion revolves around a physics problem involving the behavior of air in a car tire under changing pressure conditions. The problem specifically addresses the temperature of the air after an increase in pressure and the mass of air that needs to be removed to return to the original pressure, while considering the tire's volume remains constant.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between pressure, volume, and temperature using the ideal gas law. There are attempts to calculate the new temperature and the mass of air required at different pressures. Some participants question whether the calculations reflect the mass of air that needs to be removed rather than the total mass at the higher pressure.

Discussion Status

Several participants have provided feedback on the calculations, with some suggesting that the original poster may have miscalculated the mass of air to be removed. There is a productive discussion about calculating the mass of air at both pressures to find the difference, indicating a collaborative effort to clarify the problem.

Contextual Notes

Participants are working under the assumption that the volume of the tire remains unchanged and are using specific values for pressure and temperature in their calculations. The molecular mass of air is also specified as part of the problem constraints.

chawki
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Homework Statement


A car tyre with an intemal volume of 0,02m3 is filled to a pressure of 310 kPa at 15C.
After a long run, the pressure rises to 340 kPa.

Homework Equations


a) What is the temperature of the air in the tyre, assuming that the volume of the tyre is unchanged?
b) What mass of air must be removed for the pressure to retum to 310 kPa ? Take the
molecular mass of molecules in air to be 29 g/mol

The Attempt at a Solution


a) volume unchanged, means: P2/P1 = T2/T1
340000/310000 = T2/(273.15+15)
T2 = 316.03 K.

b) PV = nRT = (mRT/M)
m = (P1*V*M)/R*T2
m = (310000*0.02*29)/(8.314*316.03)
m = 68.43g
 
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Anyone replying?
 


chawki said:

Homework Statement


A car tyre with an intemal volume of 0,02m3 is filled to a pressure of 310 kPa at 15C.
After a long run, the pressure rises to 340 kPa.

Homework Equations


a) What is the temperature of the air in the tyre, assuming that the volume of the tyre is unchanged?
b) What mass of air must be removed for the pressure to retum to 310 kPa ? Take the
molecular mass of molecules in air to be 29 g/mol

The Attempt at a Solution


a) volume unchanged, means: P2/P1 = T2/T1
340000/310000 = T2/(273.15+15)
T2 = 316.03 K.

b) PV = nRT = (mRT/M)
m = (P1*V*M)/R*T2
m = (310000*0.02*29)/(8.314*316.03)
m = 68.43g

Your work looks okay to me. Do you know if it is correct?
 


It would appear that you've calculated the total mass of the air required in the tire at the higher temperature, rather than the amount of air that must be removed to reach that total.
 


I think it should be correct.
 


gneill said:
It would appear that you've calculated the total mass of the air required in the tire at the higher temperature, rather than the amount of air that must be removed to reach that total.

ahhhh i think you're right.
maybe we should calculate the mass required for 340kpa and for 310kpa, and then the difference of those masses is the mass to be removed ?
 


gneill said:
It would appear that you've calculated the total mass of the air required in the tire at the higher temperature, rather than the amount of air that must be removed to reach that total.

Thanks for catching that, gneill.
 


m2=340000*0.02*29/8.314*316.03
m2=75.05g

m=310000*0.02*29/8.314*316.03
m=68.43g

75.05-68.43=6.62g and that's the mass we must remove to get 310000Pa ?
 


Looks reasonable.
 
  • #10


Thank you gneill.
 

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