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4. A tank contains `26.0\ kg` of `O_2` gas at a gauge pressure of `8.70\ atm`. If the oxygen is replaced by helium, how many kilograms of the latter will be needed to produce a gauge pressure of `7.00\ atm`?
4. A tank contains 26.0\ kg of O_2 gas at a gauge pressure of 8.70\ atm. If the oxygen is replaced by helium, how many kilograms of the latter will be needed to produce a gauge pressure of 7.00\ atm? `PV=nRT`
`m_(He) =2.68\ kg`
Given:
`m_(O_2) = 26.0 kg`
`P_1 = 8.70\ atm + 1.00\ atm = 9.70\ atm`
`P_2 = 7.00\ atm + 1.00\ atm = 8.00\ atm`
Known
atomic mass `am_(O_2) = 32\ u=32.0times10^(-3) kg//mol`
atomic mass `am_(He) = 4\ u = 4.00times10^(-3) kg//mol`
Equation
`PV=nRT`
Solution:
Assume oxygen and helium are ideal gases. under same pressure, same temperature and same volume, number of mol will be same.
Since `T` an `V` keep same.
`P/n = (RT)/V =` constant
For oxygen:
`P_1/n_1 = P_2/n_2`
`n_2 = n_1P_2/P_1 = ( 26.0 kg)/(32.0times10^(-3) kg//mol)(8.00\ atm)/(9.70\ atm)`
`n_2 = 6.70 times 10^2\ mol`
At same condition, helium will have the same number of mol
`m_(He) = (6.70 times 10^2\ mol)(4.00times10^(-3) kg//mol)`
`m_(He) =2.68\ kg` |