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Homework Help: Ideal Gas Law and a gas bubble

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A gas bubble with a volume of 0.10 cm^3 is formed at the bottom of a 10.0 cm deep container of mercury. If the temperature is 27C at the bottom of the container and 37C at the top of the container, what is the volume of the bubble just beneath the surface of the mercury? assume that the surface is at atmospheric pressure.

    2. Relevant equations

    3. The attempt at a solution

    P2= 1.01x10^5Pa + (13.6x10^3kg/m3)(9.81m/s)(1.0x10^-3m)
    P2= 101130Pa

    P1V1/T1=P2V2/T2 > V2=P1V1T2/T1P2

    V2=(1.01x10^5Pa)(1.0X10^-3 m^3)(300K)/(310K)(101130Pa)

    V2=9.66x10^-4 m^3

    The answer in the back differs with mine and I have tried to redo the problem many times.
  2. jcsd
  3. Dec 2, 2008 #2
    Hi greyknight, welcome to PF :smile:

    Firstly, What do your labels 1 and 2 refer to? Which is bottom and which is top?


    What is 1.0x10^-3 m?

    What is 1.0x10^-3 m^3?
  4. Dec 2, 2008 #3
    Label 1 is the top and label 2 is the bottom.

    1.0x10^-3 m is the depth of the container. Originally 10.0 cm but I converted it to meters.

    1.0x10^-3 m^3 is the volume of the bubble on top converted from 0.10 cm^3.
  5. Dec 2, 2008 #4
    Well, in that case you are trying to find the volume V2 at the bottom, but the question asks you for the volume at the surface. Bubbles rise up! :smile:

    Check this conversion. 1x10^-3 m = 1/1000 m = 1mm.

    Check this conversion also. How do you convert cm^3 to m^3?
  6. Dec 2, 2008 #5
    10 cm = 0.1 m

    0.1 cubic centimeter = 1.0e-7 cubic meter

    I divided 0.10 by 100x100x100 or 0.10/100^(3)
  7. Dec 2, 2008 #6
    Wow, that was just a lot of sloppy math by me. Thanks, naresh. :)
  8. Dec 2, 2008 #7
    do you get the answer?
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