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Removing a gass: Pressure ?

  1. Jul 27, 2017 #1
    1. The problem statement, all variables and given/known data

    In a container we have a gas, at P = 11.0 atm & T = 20.0 C. We remove 2/3rds of the gas and increase the temperature to T = 75.0 C.What's the pressure of the gas that remains?

    2. Relevant equations

    PV = nRT

    3. The attempt at a solution

    Normally this would be a regular "put the numbers in the formula" exercise, but I'm having trouble understanding just what "remove 2/3rds of the gas" means. If the gas was, say, a block of wood, I'd be removing 2/3ds of its mass. With that logic, I figured it was the n (moles) that were decreased, and that the volume V was kept the same (same container). But I didn't get the wanted result from this, so I was wrong.

    Then I figured maybe the change happens so fast that the gas doesn't manage to spread and have the same Volume again, so both n & V should be n' = n/3 & V' = V/3. That didn't work either. So I'm kinda stuck on what that phrase means.

    Any help is appreciated!
     
  2. jcsd
  3. Jul 27, 2017 #2
    I think you're original method is on the right track, but you didn't include your final answer in your post, so I'm going to have to guess that what I did here is different.

    Pretty much baked into the definition of the gas is the property that it expands to fill any container, so I wouldn't worry about changing the volume of the gas. Besides what if I took molecules randomly out of the gas instead of scooping out the top [itex]2/3[/itex]? That's also a viable interpretation of "remove [itex]2/3[/itex]."

    The first thing we need to do is solve for [itex]n_iR/V[/itex] in terms of the initial quantities. Doing so gives [itex]P_i/T_i[/itex]. Now we need to solve for [itex]n_iR/V[/itex] in terms of the final quantities, where [itex]n_f=n_i/3[/itex]: [tex]P_fV=n_fRT_f=\frac{n_iRT_f}{3}[/tex] so [itex]n_iR/V=3P_f/T_f[/itex] Now we equate the two representations of [itex]n_iR/V[/itex] and solve for [itex]P_f[/itex]: [tex]\frac{P_i}{T_i}=\frac{3P_f}{T_f}[/tex] so [itex]P_f=P_iT_f/3T_i[/itex]. At this point, feel free to plug in the numerical quantities and count your sig figs. Right now it looks like if you plugged everything in as is, you'd get a pressure in [itex]\text{atm}[/itex], so no need to change units unless you really want to.
     
  4. Jul 27, 2017 #3
    That's exactly what I did, but I missed the fact that I had to convert the temperatures from Celcius to Kelvin, so my result didn't match with the book's. That said:

    Pi = 11.0 atm
    Tf = 75.0 + 273.15 = 348.15 K
    Ti = 25.0 + 273.15 = 298.15 K

    Pf = Pi*Tf/*Ti = 11.0atm * 348.15 K/3*298.15 K = 4.28 atm

    Thanks a ton for the help!
     
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