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Homework Help: Boyle's Law Question

  1. May 17, 2010 #1
    Hi, I was working through a past exam paper when I came across a Boyle's Law question in which the answer to the second part confused me. I will write the question, part one and its solution and the second part. Here is the question :

    A rigid cylinder contains [tex]0.08 m^3[/tex] of helium gas at a pressure of 750 kPa. Gas is released from the cylinder to fill party balloons. During the filling process, the temperature remains constant. When filled, each balloon holds [tex]0.020 m^3[/tex] of helium gas at a pressure of 125 kPa.

    (a) Calculate the total volume of the helium gas when it is at a pressure of 125 kPa.

    Solution :

    Using [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

    [tex](7.5*10^5)(0.08) = (1.25*10^5)V_{2}[/tex]

    [tex]V_{2} = 0.48 m^3[/tex]

    (b) Determine the maximum number of balloons which can be fully inflated by releasing gas from the cylinder.

    My answer to this question was as follows :

    [tex]Number of balloons = \frac{V_{2}}{0.020} = \frac{0.48}{0.020} = 24 balloons[/tex]

    But when I looked at the answer sheet this is what was written:

    [tex]Volume of gas available = 0.48 - 0.08 = 0.40 m^3[/tex]

    [tex]Number of balloons = \frac{0.40}{0.020} = 20 balloons[/tex]

    My confusion is in the part where the volume of 0.08 cubic metres is taken off the total volume obtained from a pressure of 125 kPa. I know that the 0.08 is the same volume in the cylinder but why is this taken away from it, if the pressure is changed isn't the gas the new volume of 0.48 cubic metres?

    I would really appreciate any helpful advice given, thank you in advance.
     
  2. jcsd
  3. May 17, 2010 #2

    phyzguy

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    The new volume of the gas is 0.48, as you calculated, but 0.08 of it is still in the cylinder, hence not available to fill balloons. Once the pressure in the cylinder is the same as the balloons, these is no pressure difference to force the remaining gas out the cylinder into the balloons. If you were to pump the remaining gas out of the cylinder, you could fill 4 more balloons, but you would need a pump to do that.
     
  4. May 17, 2010 #3
    Ah, I think I understand so as you fill each balloon the volume is decreasing inside the cylinder and hence the pressure is too, but how am I supposed to know when the pressure inside the cylinder reaches 125 kPa, how do I know it does that when I have filled 20 balloons?
     
  5. May 17, 2010 #4

    phyzguy

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    That's what you calculated. When the pressure has decreased to 125 kPa, the volume has increased to 0.48 m^3. Of that 0.48 m^3, 0.08 m^3 is still inside the cylinder, and 0.40 m^3 is in the 20 balloons that you filled.
     
  6. May 17, 2010 #5
    So, no matter what pressure the gas inside the cylinder is reduced to, I will still be left with the 0.08 m3 that was in there at the beginning? For example if I was to escape it into balloons that filled the same volume but at a pressure of 100 kPa :


    [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

    [tex]\frac{(7.5*10^5)(0.08)}{1*10^5} = V_{2} = 0.6 m^3[/tex]

    So now, to work out the number of balloons I can fill all I have to do is remember that no matter what, 0.08 m3 of the gas will always remain in the cylinder:

    Volume of gas available = 0.6 - 0.08 = 0.52 m3

    Number of balloons = [tex]\frac{0.52}{0.02}[/tex] = 26 balloons

    is this correct?
     
  7. May 17, 2010 #6

    phyzguy

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    Yes, correct. The volume of the cylinder is the same no matter what the pressure of the gas in it is.
     
  8. May 17, 2010 #7
    Thanks for your help and patience :smile:
     
  9. May 17, 2010 #8

    phyzguy

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    Glad I could help.
     
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