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Really simple ideal gas question.

  1. Dec 14, 2007 #1
    You know what, I think I might be stupid. I tried this question for 20 minutes already and I am still stucked

    1. The problem statement, all variables and given/known data
    gas is confined in a tank at pressure of 10 atm and temp of 15C, if half of the gas is withdrawn and temp is raised to 65C, what's the new pressure?


    2. Relevant equations
    pV=nRT


    3. The attempt at a solution

    Don't even joke about this.

    Help would be welcomed.
     
  2. jcsd
  3. Dec 14, 2007 #2

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    P1V = nRT1 and P2V = (n/2)RT2. Does this help?
     
  4. Dec 14, 2007 #3
    i keep getting 21.6.

    does it reduce to T1/P1=T2/(2*P2)?

    answer is around 5.7
     
  5. Dec 14, 2007 #4

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    I get 5.87 atm. Divide 2nd eqn by 1st.
     
  6. Dec 14, 2007 #5
    can you show me the steps.

    i spent at least two hours on this already.

    After going through two years of calculus I can't even do this stupid problem.
     
  7. Dec 14, 2007 #6
    I'm not sure if this helps but if you use the pv = nRT formula, your pressure would have to be in kPa. Have you converted your value of atm from the given question?
     
  8. Dec 14, 2007 #7
    never mind. I've been doing it right since the beginning

    I just forgot the it's in K.

    I thought using the 0.08 for R can allow you to omit the C -> K conversion

    That's why i've been getting 21.6 all the time

    thank you anyways
     
  9. Dec 14, 2007 #8
    fyi, always use the units, even when bringing in a constant equation like that. In first year my chem prof use to take marks off for not including the unit conversions. This way you know that when the units cross out, its done correctly (units wise)
     
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