# Stuck on Ideal Gas Law

• February 27th 2013, 05:13 PM
dempsey1905
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?
• February 27th 2013, 06:11 PM
sakonpure6
Re: Stuck on Ideal Gas Law
• February 28th 2013, 03:41 AM
TheRedWizard
Re: Stuck on Ideal Gas Law
That looks like a homework question, in that case hope this site helps:

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• February 28th 2013, 09:45 AM
HallsofIvy
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 $\frac{V}{2.4}= \frac{219}{294}$.
• March 1st 2013, 07:46 AM
Irvingmiller
Re: Stuck on Ideal Gas Law
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• March 5th 2013, 01:06 AM
BabyShawn
Re: Stuck on Ideal Gas Law
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