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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?
Wiki User
∙ 14y ago
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
Wiki User
∙ 11y ago
To find the volume of the balloon at the new pressure and temperature, we can use the combined gas law equation: (P1V1) / (T1) = (P2V2) / (T2). Plug in the known values and solve for V2. The temperature is given in Celsius, so remember to convert it to Kelvin by adding 273. The volume of the balloon at the new location will be around 4.92 x 10^4 L.
Wiki User
∙ 14y agoWhat 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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