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Ideal Gas Equation problem

  1. Aug 27, 2015 #1
    Hey guys, I just want to make sure I went about this the right way...your input would be greatly appreciated.
    Thank you in advance.

    1. The problem statement, all variables and given/known data

    A tire is checked before a road trip and the gauge pressure reads 220 kPa (gauge) - State 1
    The same tire is checked after the trip and the gauge pressure reads 240 kPa (gauge) - State 2
    The temperature on the day is 25 degrees Celsius and the pressure is 101.325 kPa (absolute)
    The tire has an approximate inner volume of 25L and can be assumed to be rigid
    Gas Constant R for air = 0.2870

    It asks for:
    A.) Calculate the abs pressure in the tire at state-1.
    B.) Using the ideal gas equation, find the mass in grams of the air inside a single tire at state-1.
    C.) Using the ideal gas equation, find the temperature of the air inside a single tire at state-2
    D.) If having attained state-2, the driver wants to bleed-off some ‘hot’ air from each tire in order to restore the (gauge) pressure to 220kPa, calculate the remaining mass of air inside a single tire after some air is removed

    2. Relevant equations
    - PV = mRT
    - P1*V1/ T1 = P2*V2/ T2
    - Pabs=Pg+Patm


    3. The attempt at a solution
    A.) Pabs=Pg+Patm

    P1 = 220kPa + 101.325kPa = 321.325kPa
    P2 = 240kPa + 101.325kPa = 341.325kPa


    B.) PV=mRT ∴ m= PV/RT
    321.325kPa * 25L / 0.2870 *(25+273) = 939261.11 grams

    C.) P1*V1/ T1 = P2*V2/ T2
    Since V = cosntant the equation is then P1/ T1 = P2/ T2
    T2= (P2/P1) * T1
    T2
    = (341.325/321.325) * (25+273) = 43.55 Degrees Celsius

    D.) PV=mRT ∴ m= PV/RT
    321.325kPa * 25L / 0.2870 *(43.55+273) = 88421.99 grams
     
  2. jcsd
  3. Aug 27, 2015 #2

    DrClaude

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    Staff: Mentor

    Units?

    What does part A ask for?

    939 kg??? Hope you don't need to change a tire!
     
  4. Aug 27, 2015 #3
    I decided to do P2 while I was at it since I would have needed it later on.

    Sorry, 0.2870 Kj/kg.Kelvin **

    As for the the 939kg, I wasn't sure if I was meant to leave 321.325kPa as just that "321.325" or 321.325*10^3..
    Otherwise it would be 93.92 grams
    Thank you again :)
     
  5. Aug 27, 2015 #4

    DrClaude

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    Staff: Mentor

    The mistake was not in the pressure. That's one of the reasons I asked you to specify the units: make sure they are consistent.
     
  6. Aug 27, 2015 #5
    Unfortunately I'm terrible when it comes to differing units.. I didn't see any problem with the R constant...however I did forget to convert the 25L to m^3. Was that it?
     
  7. Aug 27, 2015 #6

    DrClaude

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    Staff: Mentor

    Yes.
     
  8. Aug 27, 2015 #7
    Thank you! :)
     
  9. Aug 27, 2015 #8

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Or it could be that pressure needed to have units of kPa instead of just Pa.

    The tricky thing about R is that it can come in a variety of units, which is why it is important to establish these before doing the arithmetic.

    Take a look at R in this article:

    https://en.wikipedia.org/wiki/Gas_constant

    R can have the same numerical value (8.314), but the units could be L⋅kPa⋅K-1⋅mol-1 or m3⋅Pa⋅K-1⋅mol-1
     
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