Question

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 volume 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?

Answer 1

The athlete expels 8.89 x 10^-2 moles of water per minute through the lungs.

Step-by-step explanation:

To calculate the number of moles of water per minute expelled by the athlete through the lungs, we need to determine the amount of water evaporated with each breath and then multiply it by the respiration rate. According to the information provided, an average breath is about 0.5 L, and each breath evaporates 4.0 x 10^-2 g of water. We can convert grams of water to moles by dividing by the molar mass of water (18.02 g/mol). So, the moles of water evaporated with each breath are (4.0 x 10^-2 g)/(18.02 g/mol) = 2.22 x 10^-3 mol/breath.

Next, we can calculate the number of breaths per minute multiplied by the moles of water evaporated per breath to find the moles of water expelled per minute. The respiration rate is given as 40 breaths per minute. Therefore, the moles of water expelled per minute are (2.22 x 10^-3 mol/breath) x 40 breaths/minute = 8.89 x 10^-2 mol/minute.

Answer 2

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 = (PV)/(RT)`

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

`n = ((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