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B Confused about the ideal gas law

  1. Feb 9, 2019 #1
    Ok, i am struggling to figure something out. I dont know why math is so much easier than physics haha.

    ok, here is my struggle. I have two states, state 1 and 2, which i will call just 1 and 2.

    1:
    T=298kelvin
    V=0.025m(cubed)
    P=310Kpa
    Mass1=Mass2
    R=0.2870

    2:
    T=323kelvin
    V=0.025m(cubed)
    P=??
    Mass2 = Mass1
    R=0.2870

    This is how i solved for P2 and it was WRONG!!!!!

    P2V2=mR2T2 where m is mass

    mass=V(rho) = 0.025(1.22) = 0.0305kg of air in the closed system

    P2= mRT divided by volume. I get 113kpa. Book says 336kpa.

    IF I USE P1V1/T1=P2V2/T2, i get the right answer. BUT it should have worked the way i did solving for mass because mass IS volume multiplied by density and it doesnt change..

    Thanks
     
  2. jcsd
  3. Feb 9, 2019 #2

    256bits

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    Under what conditions of pressure and temperature is ρair = 1.22 kg/m3 ?
    It is apparent that an incorrect assumption is being made here.

    Find the actual density from the information given for State 1.
     
  4. Feb 9, 2019 #3

    to answer your first question, i thought density never changes for a substance. that is why it is called an intensive property? its ALWAYS 1000kg/m3 or 1.22kg/m3, etc etc. because if it were to change, then all my problems i did for finding pressure in a multifluid manometer, would be wrong because we assumed rho of water to be 1000kg/m3, air to be 1.22kg/m3 even when Patm was lower or higher than 1atm

    How would i find density in state one if all i have is volume? m=PV/RT so rho = PV/RT divided by Volume? mass = 0/0906kg divided by 0.025 = 3.62kg/m3.. does not make sense to me. I have always been told that air is 1.22kg/m3 no mater what. thats how we did our multifluid manometers and all of my water pressure problems that included a Patm different than 1atm

    thanks for helping me understand this.

    EDIT:

    I should add. I am not saying you are wrong at all. Just saying it makes no sense to me haha.
     
  5. Feb 9, 2019 #4

    256bits

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    If you have a bicycle pump, close off the valve end with your finger, and press down on the handle.
    The mass in the cylinder doesn't change, but the volume does.
    The density of the air within the cylinder does what?

    another example,
    The atmosphere ( of the earth ) as one goes farther up from that at the surface of the earth to the vacuum of space.
    Density of the air does what as one increases in altitude?


    "No matter what" is a very strong statement to make.
     
  6. Feb 9, 2019 #5

    256bits

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    That's not the meaning of an intensive property.

    Temperature is an intensive property, but we all know that the day to day outside temperature changes.
    A review of this subject t- intensive and extensive properties - might help.
     
  7. Feb 9, 2019 #6
    yeah, the volume of air changes(decreases), and not the mass since none escaped. so density would be different then.

    yes, "no matter what" is a strong statement and that is why i am confused haha. because all the basic physics stuff i took is now being challenged in thermo haha
     
  8. Feb 9, 2019 #7

    LURCH

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    I think you might be confusing “intensive property” with “intrinsic property”, maybe?
     
  9. Feb 9, 2019 #8
    true, but if i take a room at 72F and put a wall in the center to divide it in half, the temp on both sides would be the same 72F. the volume of air would be diffeernet than the original value, but the density would be the same and so would the specific volume as well as the mass.
     
  10. Feb 9, 2019 #9
    Since you think math is so much easier than physics, here is the math equation for the density of an ideal gas:
    $$\rho=\frac{PM}{RT}$$where M is the molar mass of the gas.
     
  11. Feb 9, 2019 #10
    Thank you:) this is one of yhe equations i have in my notes and book. I guess where my confusion was is the fact that density of air is not always 1.22kg/m3 haha
     
  12. Feb 10, 2019 #11

    scottdave

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    Unless they want you to solve for the mass of the air, then you can get by without solving for it.
    With ## P*V = m*R*T ## and you have established that mas is constant, so ##m*R## is also a constant.
     
  13. Feb 10, 2019 #12
    Correct. I can use pv/t for my calculations. I was just stuck on the fact that air is always 1.22kg/m3 but it aint haha
     
  14. Feb 10, 2019 #13

    scottdave

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    If you need the mass, you could just solve for it from the equation P1V1 = m r T1, where you are given everything except m. Also, it is a good idea to include units in all of your calculations, to help keep yourself straight. This one is not too complex, but you may run into ones in the future which get pretty complex
     
  15. Feb 10, 2019 #14

    scottdave

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    The density of a liquid such as water doesn't change by "much" with temperature compared to gasses. But they do change. Even solids expand and contract (same mass but different volume) some with temperature. Any density you see for a material should list the conditions which this value is.
     
  16. Feb 10, 2019 #15
    I tried that and ended up with a number not close at all to what was needed haha
     
  17. Feb 10, 2019 #16
    This is all good to know haha. Breaking the old indoctrinated info with new info haha
     
  18. Feb 12, 2019 #17
    You might try this. p(1)V(1) = n(1)RT(1) and p(2)V(2) = n(2)RT(2) and V(1) = V(2) and n(1) = n(2). Divide eqn, 2 by eqn. 1 and solve for T(2).
     
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