A vessel will be used to store radon gas. So to find out how fast it would leak, nitrogen gas was placed in the vessel. After 48 hours, a certain amount of nitrogen leaked out. How long would it take for the same amount of radon to leak from this vessel?
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Gas leaking out of a container, as described in this problem, is labeled as effusion. We know that the speed at which a gas effuses depends on its molar mass. The relationship between the rates of effusion can be expressed as:
(Rate of Effusion Gas A / Rate of Effusion Gas B) = SqRt (Molar Mass Gas B / Molar Mass Gas A)
The molar mas of Nitrogen is 28g/mol (Nitrogen is a diatomic gas)
The molar mass of Radon is 222g/mol
Plug it into the equation:
SqRt( 222g/mol / 28.0g/mol)
= 2.8
This means that Radon gas will effuse 2.8 times slower than Nitrogen gas. If it took 48hrs for an amount of Nitrogen to effuse, it would take the same amount of Radon gas 135hrs.
Given what we discussed previously about larger molar masses resulting in slower effusion, the answer looks to be reasonable.
(Rate of Effusion Gas A / Rate of Effusion Gas B) = SqRt (Molar Mass Gas B / Molar Mass Gas A)
The molar mas of Nitrogen is 28g/mol (Nitrogen is a diatomic gas)
The molar mass of Radon is 222g/mol
Plug it into the equation:
SqRt( 222g/mol / 28.0g/mol)
= 2.8
This means that Radon gas will effuse 2.8 times slower than Nitrogen gas. If it took 48hrs for an amount of Nitrogen to effuse, it would take the same amount of Radon gas 135hrs.
Given what we discussed previously about larger molar masses resulting in slower effusion, the answer looks to be reasonable.