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9000 / 101325 =0.977 atm

n = 2.4 x 10^2 x 0.977 / 0.08206 x 273 K=10.5 moles He

mass He = 4.0 x 10.5 =42 g

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11y ago
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14y ago

What is it you are trying to find: the temperature outside the balloon's new location? or could it be the volume the gas now occupies? Also, for clarification, is the inital volume 3.35104 L?

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Q: A weather balloon is filled with helium that occupies a volume of 3.35 104 L at 0.995 ATM and 32.0C After it is released it rises to a location where the pressure is 0.720 ATM and the temperature?
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