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Pump add air in to the flat tire in the gas station

  1. Nov 8, 2007 #1
    pump add air in to the flat tire in the gas station. The compressed air is at 180 psia and 60F(+460). the tire is initially at vaccum, therefore p0=0 psia. Air is added in to the tire until it reaches Pf =32.9 psig which is (23.9+14.7=47.6 psia), then stopped . this process is adiobatic which mean Q=0. find the final tempreture Tf in R degree. then find the entropy change and internal changed for a rigid tire which has volume of 0.6 cuft. what is the system (control mass or control volume . In this cas Air is an ideal gas with cp=7/2 R
  2. jcsd
  3. Nov 8, 2007 #2


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    This appears to be a homework problem, so please in future post such questions in the HW forums, in this case Introductory Physics.

    We also ask students to show effort and their work when asking for assistance.

    Try to write the appropriate equations based on the variables know and what one is trying to calculate.
  4. Nov 8, 2007 #3
    from the notes i know its adiabatic, and when we pumping the air quickly therefore its getting adiabatic and the pressure of system and surrounding are not same so its also ireversible.
    to find delta U we can use delaU=ncvdelta T

    but when its asking to find final temperature , it mean temperature of the tire which has air?
    then what would be our p initial??
  5. Nov 8, 2007 #4


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    Is one sure the tire is at a vacuum?! Or 1 atm abs (14.7 psia), or 0 psig?

    It is then inflated to 32.9 psig (47.6 psia), we assume a rigid tire so the volume does not change, but the pressure changes so the moles of gas much change.

    What formula would one propose?
  6. Nov 8, 2007 #5
    i was just confused which equation should i use? is V really constant or we have change of volume equal to 0.6?initially tire is in vaccum(Po=o psia).work is done by pump,but how we should apply pressure of pump and its temp. in our equations?
  7. Nov 8, 2007 #6


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    OK - thinking about this a little. Assume the volume of the tire is fixed at 0.6 cu ft, and it is rigid.

    What one has is compressed air at 180 psia and 60°F (520 R) expands adiabatically into the volume (initially a vacuum) to a final state of 47.6 psia.

    In this case, assume Air is an ideal gas with cp=7/2 R

    One is to find:

    1. the final temperature Tf in R degree.

    2. the entropy change and internal change.

    and then identify the system (control mass or control volume).

    Can one identify a formula for adiabatic expansion of an ideal gas?
  8. Nov 8, 2007 #7
    so basicllay when its adiabatic, and this is a close system we have DeltaU=Q-w but Q=0 and we have W in positive becase the pump is working on system. so it is reviersibel?? or ireversible?
    if its reversible we have pv^r= constant so then how can we find the Tf? if we have Pf but about p initial? . the system is control mass becasue is has air .. we can t use the p1v1/T1=p2v2/T2 then which equasion??
  9. Nov 8, 2007 #8


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    If we allow [tex]\frac{PV}{T}[/tex] = constant, and [tex]PV^{\gamma}[/tex] = constant,

    then [tex]V^{\gamma-1}T[/tex] = constant (for adiabatic)

    and this process is irreversible.

    And hear is a reference on an adiabatic process.


    For a discussion of an adiabatic irreversible process see
    http://classweb.gmu.edu/sdavis/chem331/adiabatic.htm [Broken]

    Assuming [tex]PV^{\gamma}[/tex] = constant applies, one knows the final volume and pressure, and the initial pressure, so one can compute the initial volume of the amount of gas which flows into the tire. Then knowing the initial P, T, V and the final P, V, the final T can be computed.
    Last edited by a moderator: May 3, 2017
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