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Gas Law Problem

  1. Nov 19, 2005 #1
    The Problem" If a scuba driver fills his lungs to full capacity of 5.5L when 10 m below the surface, to what volume would his lungs expand if he rose to the surface? Is this advisable?

    I really don't know where to go with this. I started with pv=nRT so rho(v)gh=nRT but there are still too many unknowns. How do I go about solving or starting this problem???

    Answer is 11L and not advisable.
     
    Last edited: Nov 19, 2005
  2. jcsd
  3. Nov 19, 2005 #2
    any idea anyone? I really need help.
     
  4. Nov 19, 2005 #3

    Astronuc

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    Hydrostatic pressure in a fluid is just [itex]\rho[/itex]gh, where [itex]\rho[/itex] is the density, g is acceleration of gravity, and h is height from some reference.

    So going from h to the surface will represent a change in pressure.

    At the surface, assume pressure is 1 atm = 14.7 psia = 0.101325 N/m2.
     
  5. Nov 19, 2005 #4
    i know what those stand for and u didn't mention "n." The problem is HOWto start the problem or figure the numbers given what I posted. You also don't know temperature "T" or should I assume it is all at STP?

    How come no one know this one to at least help me get started? This only Physics 102 algebra material.
     
    Last edited: Nov 19, 2005
  6. Nov 19, 2005 #5

    Astronuc

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    The number of moles does not change, unless the diver releases the air.

    If n is constant since PV=nRT, then n = PV/RT = P1V1/RT1 = P2V2/RT2.

    One may assume the temperature does not change, or T1 = T2, and with R also a constant, the above equality reveals,

    P1V1 = P2V2, which is known as Boyle's Law. See also http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html#c6
     
  7. Nov 19, 2005 #6
    so P1/P2=V2/V1. Assuming, at the top it is atmospheric pressure then it is 1 atm. Underwater P1=rho(g)(h)=(1*10^3kg/m^3)(9.8m/s^2)(10m) so that gives 98000 atm. V1=5.5L so solving for V2 we get 53900L which is not possible what am I doing wrong?
     
  8. Nov 19, 2005 #7

    Astronuc

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    (1*10^3kg/m^3)(9.8m/s^2)(10m) does not give 98000 atm.

    Look at the units. kg/m3*m/s2*m = (kg-m/s2)/m2= N/m2 = Pa.

    1 atm = 101325 Pa.

    Also realize that the pressure at the surface is 1 atm, while the pressure 10 m below the water is 1 atm + pressure from the depth of water.
     
  9. Nov 19, 2005 #8
    Thanks. That really helped. I just have 1 more physic question from the homework that couldn't get but I made an effort on it so if you have any idea what I'm doing wrong on this please tell me.

    The question states: A hot air ballown achieves its buoyant force lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1800 m^3 and the required lift is 2700N (rough estimate of the weight of the equipment and passenger). Calculate the the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is 0 degrees celcius and that air is an ideal gas under these conditions. What factors limit the maximum altitude attainable by this method for a given load? (neglect variables like wind).

    answer:37 degrees cecius.
    ok so I drew FBD and the sum of forces includes the buoyant force in this case upward and that equals the two downward forces of the air in the balloon and the weight of passenger. However, I don't know what air to assume (helium?) inside the balloon and I do not know the weight of the passenger. So I'mthinking of somehow solving to get the weight of the air inside balloon which is like rho of the gas times the volume (1800m^3) to get its mass and then use this mass to apply to gas law PV=nRT which I can rework into T=Pv/nR. I also did a lot of other scratch work that doesn't lead me to the correct answer so I'm getting a little confused how to begin. any ideas on this problem will be greatly appreciated.
     
    Last edited: Nov 19, 2005
  10. Nov 19, 2005 #9

    lightgrav

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    it is a HOT AIR balloon, so inside the balloon is hot air.

    If the air inside was 0 C (=273 K) there would be enough lift to hold up the cold air only, not the equipment etc.
    Force upward, by the air outside, must equal the weight of this much cold air.
    You want to have less air inside, at high T (to keep same V and same P)
    so as to hold up equipment.
     
  11. Nov 20, 2005 #10
    What about the temperature how to figure? how much less air to keep P and V constant?
     
  12. Nov 20, 2005 #11

    lightgrav

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    ditch as much air as you replace (mass-wise) with equipment.
    raise the T to compensate for delta n
     
  13. Nov 20, 2005 #12

    Astronuc

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    This is a bouyancy problem. The air inside must be less dense, but the pressure must equal or slightly exceed the outside air. In order to maintain pressure at a lower density, the temperature must be increased.

    With PV = nRt => P = (n/V)RT or P = [itex]\rho[/itex]RT

    Then with P the same, one has [itex]\rho[/itex]RTc = ([itex]\rho\,-\,\Delta\rho[/itex])RTh, where Tc and Th are the cold and hot temperatures.

    The bouyant force, FB = [itex]\Delta\rho[/itex]Vg and the force of the air at density [itex]\rho[/itex] is just [itex]\rho[/itex]Vg.

    Take the density of air at 0°C as 1.28 kg/m3.
     
  14. Nov 20, 2005 #13
    I see the reasoning for this it makes me understand physics a little better and lead me to how I should begin the problem. Thanks.
     
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