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2. A tire is filled with air at `15°C` to a gauge pressure of `220\ kPa`. If the tire reaches a temperature of `38°C`, what fraction of the original air must be removed if the original pressure of `220\ kPa` is to be maintained?
2. A tire is filled with air at 15°C to a gauge pressure of 220\ kPa. If the tire reaches a temperature of 38°C, what fraction of the original air must be removed if the original pressure of 220\ kPa is to be maintained? `PV = nRT`
`7.4%` must be removed
Given:
`T_1 = 273 + 15 = 288\ K`
`T_2 = 273 + 38 = 311\ K`
Equations:
`PV = nRT`
Solution:
Since `T` and `V` keeps the same.
`nT = (PV)/R = `constant
`n_1T_1 = n_2T_2`
`n_2/n_1 = T_1/T_2`
`n_2/n_1 = (288\ K)/(311\ K)`
`n_2/n_1 = 0.926`
`1-0.926 = 0.074 = 7.4%` must be removed |