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Thermodynamics, ideal gas

  1. Apr 2, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A container contains 1g of [itex]O_2[/itex] at a pressure of 1 atm and temperature [itex]T_i=47°C[/itex]. At a certain time, due to an escape of gas, the pressure is 5/8 atm and the temperature decreases to 27°C.
    1)What is the volume of the container?
    2)How much mass of [itex]O_2[/itex] has escaped?


    2. Relevant equations
    PV=NRT.


    3. The attempt at a solution
    1g of [itex]O_2[/itex] is worth 1/32 mol of [itex]O_2[/itex]. Thus [itex]V=\frac{N_iRT_i}{P_i}[/itex]. I took care of converting degrees Celcius to kelvin, took 1atm =10⁵ pascal. This gave me [itex]V=8.31m^3[/itex].
    While [itex]N_f = \frac{P_fV}{RT_f} \approx 208.25 mol[/itex] which makes absolutely no sense. I should get less than 1/32, yet I get a number greater than 200. I don't know what I did wrong, I tried another approach leading to another inconsistency: [itex]N_f=1.5 N_i[/itex].
     
  2. jcsd
  3. Apr 2, 2012 #2

    I like Serena

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    Let's see...

    $$V_i={N_i R T_i \over P_i}={\frac 1 {32} \cdot 8.31 \cdot 320 \over 10^5}=8.31 \cdot 10^{-4} m^3$$
    How did you get ##V_i=8.31 m^3##?
     
  4. Apr 2, 2012 #3

    fluidistic

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    Hi! Thanks :biggrin: Misuse of calculator apparently. I must be extremely careful to put all parenthesis I think. I now get your result.
    Number of moles in the end: 0.005 or about 1/200. Thus about 0.16 g of [itex]O_2[/itex].
     
  5. Apr 2, 2012 #4

    I like Serena

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    Yeah, I thought as much.
    And this is a problem with numbers that are "nice", so you can easily do it without calculator! ;)
     
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