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Translational Kenetic Energy

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A spherical balloon is filled with helium atoms (Note 1 atm = 101.3 kPa).


    (i) What is the total translational KE of the atoms if the balloon has a diameter 13.0 cm at 37.0 °C and the pressure inside the balloon is 121.6 kPa?

    (ii) The above balloon was filled from a cylinder of volume 0.2 m3 containing helium gas of 190.0 atm and at the same temperature as in the balloon in (i). How many of the balloons in (i) can the cylinder inflate?

    3. The attempt at a solution

    (i) To find the translational kenetic energy I think I need to use the equation

    [tex]\frac{2}{3} nRT[/tex]

    But the problem is that the number of moles is not given to us. I tried finding it using the formula n=m/M, but that won't work because I know that the molar mass of helium is 4 g/mol but I don't know what the mass (m) is!

    T=37.0 °C + 273.15 = 310.15 K

    [tex]\frac{2}{3} n(8.314)(310.15)[/tex]

    I can't go any further.
    Any help is appreciated. :smile:
     
  2. jcsd
  3. Apr 8, 2010 #2

    ehild

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    You know the pressure and temperature of the gas. The balloon is a sphere, the diameter is given. Can you determine the volume?

    There is the universal gas law, which relates the pressure, volume, temperature and number of moles of the gas.

    ehild
     
  4. Apr 9, 2010 #3
    Thank you.... I get it. But there is still something wrong:

    [tex]V=\frac{4}{3} \pi r^3[/tex]
    [tex]V=\frac{4}{3} \pi (6.5)^3=1150.3[/tex]

    [tex]PV=nRT[/tex]
    [tex](121.6 kPa) \times 1150.3 = n (8.314) \times (310.15 K) [/tex]
    [tex]n= 54.24[/tex]

    [tex]\frac{2}{3}nRT= \frac{2}{3} (54.24) (8.314) (310.15)[/tex]
    [tex]=92829.13 J[/tex]

    But this is false because the correct answer must be 210.0 J. I tried different units for temprature and pressure etc but it didn't work... what's wrong?
     
  5. Apr 9, 2010 #4

    ehild

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    Use appropriate units: m^3, K, Pa.

    Also you used a wrong formula for the average translational energy. It is 1/2 RT per degrees of freedom, so 3/2 RT.

    ehild
     
  6. Apr 9, 2010 #5
    Oops that was a typo.

    But if I used Pa instead of KPa, the value for n will be huge:

    n=54245.3

    [tex]\frac{3}{2} nRT = \frac{3}{2} (54245.3)(8.314)(310.15)[/tex]

    The answer will be a VERY huge number. It's very far from 210.0 J! :(
    I used the right formula & the right units, why is my answer so wrong?
     
  7. Apr 9, 2010 #6

    ehild

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    Express all quantities in the basic units. kPa means 10^3 Pa, cm means 0.01 m. Use the units during your calculations.

    For example,

    [tex]
    V=\frac{4}{3} \pi (0.065 m)^3=1.1503 \cdot 10^{-3} m^3
    [/tex]

    ehild
     
  8. Apr 10, 2010 #7
    Thank you SO much!! I got it.

    Now, any hints on how to solve part (ii)? I absolutely have no clue how to approach this problem...
     
  9. Apr 10, 2010 #8

    ehild

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    Now you know the moles in the balloon. Calculate the moles in the cylinder.

    ehild
     
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