A high altitude balloon contains 6.0 g of helium in 104L at 240 K. Assuming ideal gas behavior , how many grams of helium would have to be added to increase the pressure to 4.0 × 10^-3 atm
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Given - weight of helium - 6.0 g Volume - 104L
Temperature - 240 K
Pressure - 4.0 × 10^-3 atm
Find - Grams of helium required.
Solution - Using ideal gas law equation - PV = nRT
In this equation, p refers to pressure, v refers to volume, n is number of moles, R is gas constant and T is temperature.
Keeping the values in equation-
4*10^-3*104 = n*0.082*240
number of moles = 2.03 moles
Weight of helium required = number of moles*molecular weight
Weight of helium required = 2.03*4 = 8.12gm
6g of helium is already present, therefore,
Weight = 8.12 - 6 = 2.12 g
Hence, weight of helium required to increase the pressure is 2.12 g.
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