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Perfect gas equation

  1. Nov 2, 2006 #1
    Hi, I'm not sure how to do the following. Could someone help me out?

    Q. At sea level, where the barometric pressure is 101.3 kN/m^2 and the temperature is 20 degrees celcius, a balloon is filled with 2.5 kg of hydrogen also at 20 degrees celcius.

    (i) Find the volume of the balloon at sea level.

    (ii) If the balloon rises quickly to an altitude of 2000m where the barometric pressure is 80 kN/m^2 and the temperature is 2.0 degrees celcius find its volume.

    (iii) If the balloon stays at 2000 m altitude until its temperature stabilises at 2 degrees celcius find its volume.

    The pressure difference between the hydrogen inside the balloon and the air outside may be neglected.

    My working:

    (i) I just used the perfect gas equation as follows.

    [tex]
    V_{H_2 } = \frac{{m_{H_2 } R_{H_2 } T}}{P}
    [/tex]

    Is that right?

    (ii) For this part I used:

    [tex]
    \frac{{P_1 V_1 }}{{T_1 }} = \frac{{P_2 V_2 }}{{T_2 }}
    [/tex]

    where the values of symbols with 1 (one) as the subscript are the same as those given or used in part (i) while P_2 = 80 kN/m^2 and T_2 = 295 K.

    Solve for V_2 to find the volume of the balloon?

    (iii) For this part I essentially just used [tex]\frac{{P_1 V_1 }}{{T_1 }} = \frac{{P_2 V_2 }}{{T_2 }}[/tex] again. However, this time wouldn't P_1 = P_2 = 80 kN/m^2 since the barometric pressure hasn't changed from part (ii)? Also, the temperature "stabilises at 2 degrees celcius" which is the same temperature as in part (ii). But this seems to imply that the required volume is just the same as that found in part (ii).

    I'm not really sure what to do here. Can someone check my working and point me in the right direction? Thanks.
     
    Last edited: Nov 2, 2006
  2. jcsd
  3. Nov 2, 2006 #2

    Andrew Mason

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    Since the balloon rises quickly, the temperature of the balloon does not have time to change to the surrounding temperature. In other words, there is no heat flow into or out of the balloon, so the expansion is adiabatic.

    Use the adiabatic condition:

    [tex]PV^\gamma = K[/tex]

    to determine the final volume. For air, being mostly diatomic, [itex]\gamma = 1.4[/itex].

    AM
     
  4. Nov 2, 2006 #3
    Thanks for explaining part (ii). I can accept that the final volume can be found from one of the 'adiabatic versions' of the gas equation but I would like to know why the equation I originally used in my opening post, is not valid in this case. I've read through my notes but I still don't know when the equation I originally used is applicable.
     
  5. Nov 2, 2006 #4

    NateTG

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    Quick or insulated things generally follw the adiabatic condition - no heat exchange between the gas and the outside world.
     
  6. Nov 3, 2006 #5

    Andrew Mason

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    The short answer is that you do not know what the temperature of the balloon is so you can't use PV=nRT to find the volume.

    As the balloon rises quickly and the ambient pressure decreases, the gas in the balloon expands and does work on the atmosphere. This causes the hydrogen gas to lose internal energy (since there is no heat being added, the energy required to perform that work must come from internal energy of the gas) so the gas cools. The relationship between Pressure and Volume is given by the adiabatic condition. Since P = nRT/V, the adiabatic condition can also be expressed as:

    [tex]TV^\gamma/V = TV^{(\gamma -1)} = K/nR = \text{Constant}[/tex]

    AM
     
  7. Nov 3, 2006 #6
    I think I needed to properly identify the system to which I'm applying the equations, but when I tried this question I used information about the surroundings which don't apply to the hygrogen in the balloon. Thanks for clearing that up.
     
    Last edited: Nov 3, 2006
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