Question

# You have a 26.39L sample of helium in a weather balloon. At ground level, the conditions...

You have a 26.39L sample of helium in a weather balloon. At ground level, the conditions are: T = 19.09°C, P = 1.019atm. You release the balloon and it rises to an altitude where the temperature has dropped to 5.05°C and the pressure is 0.7339atm. You also found that there was a small leak in your balloon and you have lost 11.247% of the helium gas that was originally in the balloon. What is the volume of the balloon? Report your answer in liters to 2 decimal places.

Ground conditions T= 19.09 deg.c =19.09+273.15 K=292.24 K P= 1.019 atm and V= 26.39 L

mole of helium at these conditions can be calculated from gas law which states that

n= number of moles= PV/RT , R= 0.08206 L.atm/mole.K

n= 1.019*26.39/ (0.08206*292.24)=1.121 moles

When the ballon is taken to an altitude, it lost 11.247% of helium die to leakage. So helium remaining in the balloon= (100-11.247)*1.121/100=0.9949 moles

At an altitude T= 5.05 deg. =5.05+273.15=278.2K P= 0.7339 n= 0.9949

Volume = nRT/P= 0.9949*0.08206*278.2/ 0.7339=30.94 L

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