no rush please (practice for conversion factors) 1. A helium balloon (4.00 g/mol
ID: 854134 • Letter: N
Question
no rush please
(practice for conversion factors) 1. A helium balloon (4.00 g/mol) used for long distance flying occupies a volume of 1.2 x 10^7 L under ambient conditions and altitude. Assume that ambient conditions are 25 degree C and 1 atm of pressure and these do not change (1 mole occupies 24.5 L). a) What is the number of commercial gas cylinders used to fill this balloon (7.65 m^3 per cylinder)? i. Write how you break down this problem in steps; ii. List all the conversion factors used to cancel out the units; iii. Combine all the conversion factors into a 1-line equation; iv. What is the appropriate number of significant figures in the answer? b) What is the mass of helium in this balloon? (break it down similarly to above) c) What is the density of He gas, in kg/m^3? (break it down similarly to above)Explanation / Answer
a)
volume of commercial gas cylinder = 7.65m^3 = 7.6*10^3 lt (bcz 1m^3 = 1000lt)
Helium baloon volume = 1.2*10^7 lt
Number of commercial cylinders required = 1.2*10^7 lt/7.6*10^3
= 1578.95 cylinders or 1579 cylinders
b)
24.5 lt Ocuupies ...........1 mole
1.2*10^7 lt .................?
= 489795.92 moles
wt = no.of moles *molwt
= 489795.918*4
= 1959183.67 gm
= 1959.184 Kg
c)
density = Mass/Volume
= 1959.184 Kg/1.2*10^4 m^3
= 0.163 Kg/m^3
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