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Homework Help: Ideal Gas Law Problem

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data
    You have a balloon (stretchy) filled with one liter of He at 1 atm and 298 Kalvin. The balloon is suddenly placed into a room at .5 atm and 298 Kalvin. What is the temperature of the gas inside the balloon a little while after this happens?


    2. Relevant equations
    P*V=n*R*T


    3. The attempt at a solution
    So first, I calculated the number of moles of He in the balloon, using the ideal gas law as follows...
    (1 atm)*(1 L)=n*.0821*(298 K)
    n=.0409 mol

    Then, I said that...
    P1*V1=P2*V2
    (1atm)*(1L)=(.5atm)*(V2)
    therefore, V2 = 2L

    From there, (using ideal gas law again...)
    (.5atm)*(2L)=.0409 mol(.0821)*(T)
    T=298K
    but this answer didn't make sense to me...shouldn't temperature in the balloon change?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 8, 2010 #2
    This is only true if the temperature doesn't change! Your work is correct, though; the temperature should stay the same.
     
  4. Jan 8, 2010 #3
    So...I was thinking at first that I could use P1*V1/T1=P2*V2/T2 so I would get
    (1atm)*(1L)/(298K)=(.5atm)*(V2)/T2...but then I have two unknowns?

    Then I tried that eqn with
    (.5atm)*V2=.0409mol*.0821*T2

    But these two equations are not indpendent? Help?
     
  5. Jan 9, 2010 #4
    You're confusing yourself. The "temperature of the gas inside the balloon a little while after this happens" should be the same as the temperature of the room, or 298K.
     
  6. Jan 9, 2010 #5

    Borek

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    Staff: Mentor

    For me wording is ambiguous. "A little while after" doesn't mean "after thermal equilibrium has been achieved". It can mean whatever you want. Say, 5 minutes. How is it related to thermal equilibrium? No idea.

    --
    methods
     
  7. Jan 9, 2010 #6
    okay...so if I try and calculate the temp in the balloon "shortly after the action is performed" (exact wording from the problem), it shouldn't be 298K ? I also find this problem ambiguous, because I have no idea whether or not thermal equilibrium is reached, which would happen eventually...
     
  8. Jan 9, 2010 #7
    Okay...i just realized that in the problem it states that the balloon is insulated...which means it shouldn't be 298 K (right?)...so what if i try an energy balance
    I know that change in Q=0, change in W=0, and assume changes in kinetic and potential energy both equal zero, which means the change in entropy = 0
    so, H = U + deltaP*V

    H= (.5atm)*(1L)

    but i don't remember how to calculate delta H, isn't it in a table or something?
     
  9. Jan 9, 2010 #8

    vela

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    Why do you think the gas in the balloon did no work when it expanded against the 0.5 atm outside?
     
  10. Jan 10, 2010 #9
    any other ideas on how to approach this problem? I feel like I'm heading in the wrong direction with this
     
  11. Jan 10, 2010 #10

    vela

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    You said the balloon is insulated, so the gas expands adiabatically. For adiabatic expansions, you have [itex]PV^\gamma=\texttt{constant}[/itex] where [itex]\gamma=c_p/c_v[/itex].
     
  12. Jan 10, 2010 #11
    Thanks a ton! I had forgotted about that equation...
     
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