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Thread: Stuck on Ideal Gas Law

  1. #1
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    Stuck on Ideal Gas Law

    A weather balloon carries instruments that
    measure temperature, pressure, and humidity as
    it rises through the atmosphere. Suppose such a
    balloon has a volume of 1.2 m 3 at sea level where
    the pressure is 1 atm and the temperature is
    20 ° C. When the balloon is at an altitude of 11 km
    (36,000 ft) the pressure is down to 0.5 atm and the
    temperature is about − 55 ° C.
    What is the volume of the balloon then?
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  2. #2
    Member sakonpure6's Avatar
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    Re: Stuck on Ideal Gas Law

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  3. #3
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    Re: Stuck on Ideal Gas Law

    That looks like a homework question, in that case hope this site helps:

    online tutoring at tutorsville.net
    Last edited by TheRedWizard; February 28th 2013 at 02:44 AM.
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  4. #4
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    Re: Stuck on Ideal Gas Law

    Ideal gas law: PV= NRT. where "P" is the pressure, "V" is the volume, "N" is the number of molecules, "R" is a constant, and "T" is the temperature in degrees Kelvin. 20 degrees C is about 20+ 274= 294 degrees K and -55 C is about -55+ 274= 219 degrees K.

    You are told that "such a balloon has a volume of 1.2 m 3 at sea level where the pressure is 1 atm and the temperature is 20 ° C."
    So (1)(1.2)= (NR)(294).

    You are also told that "When the balloon is at an altitude of 11 km (36,000 ft) the pressure is down to 0.5 atm and the temperature is about − 55 ° C." and asked to find the volume.
    So (0.5)V= (NR)(219).

    Divide that equation by the first: \frac{(0.5)V}{(1)(1.2)}= \frac{(NR)(219)}{(NR)(294)}
    which reduces to \frac{V}{2.4}= \frac{219}{294}.
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    Re: Stuck on Ideal Gas Law

    Nice.It is very helpful post for me.
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    Re: Stuck on Ideal Gas Law

    Unexpected ending in doctor season 1 stimulates me to buy more Video.
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