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elena55 [62]
3 years ago
7

An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volu

me of water,respectively. For a respiration rate of 40 breaths per minute, how many moles of water per minute are expelled from the body through the lungs?
Physics
1 answer:
LenaWriter [7]3 years ago
8 0

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

V_i = (4L)(0.5\%)

V_i = 0.02L

During exhalation, volume of water expelled is

V_e = (4L)(6.2\%)

V_e = 0.248L

During 40 breathes, total volume of water taken is

V_{it} = (40L)(0.02L) = 0.8L

During 40 breathes, total volume of water expelled out is

V_{et} = (40L)(0.248L) = 9.92L

Therefore resultant volume of water expelled out from the lung is

\Delta V = 9.92L-0.8L = 9.12

From the body through the lung we have that

n = \frac{PV}{RT}

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}

n = 0.373mol/min

Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min

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