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3 years ago

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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?

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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c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

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At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

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c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

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