Question

(a) if the partial pressure of oxygen in air at 1.0 atm is 160mmHg, what is the partial pressure on the summit of Mt. Whitney, where atmospheric pressure is 440mmHg? Assume the % of oxygen is the same. (b) how many moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm? (c) How many moles of oxygen are in 1.00L of air on the summit of Mt. Whitney? (d) Comapre the vaules of part (b) and (c) and use that to explain why it is harder to breathe at highter altitudes.

Answer 1

a,) The partial pressure of oxygen in air at 1.0 atm is 16mmHg

convert into pressure unit by

P = ρ*g*h

ρ= density of mercury =13593 Kgm^-3, g = garvity = 9.81 ms^-2 and h = height of mercury in colmn.

partial pressure of oxygen is

when h= 160mmHg = 0.160m

so partial pressure of oxygen is Po = 13593 Kgm^-3 *9.81 ms^-2*0.160m =21335.5728Kgm^-1s^-2= 21335.5728 Pa

1atm = 101325Pa

So percentage of oxyegn partail presuure is = oxygen pressure * 100/atmosphere pressure

So percentage of oxyegn partail presuure is = 21335.5728 Pa*100/101325Pa= 21.0565%

So percentage of oxyegn partail presuure is = 21.0565%

The pressure on the summit of Mt. Whitney, where atmospheric pressure is 440mmHg

now the percentage of oxyegn partail presuure is = 21.0565% of 440mmHg

now the percentage of oxyegn partail presuure is = {21.0565 * 440}/100 mmHg= 92.649mmHg

Hence the percentage of oxyegn partail presuure at the summit of Mt. Whitney is = = 92.649mmHg

b.

Moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm

Note we assuming all gases are perfect gas

We know that at standard temperature(0^oC) and pressure(1atm) the volume per mole of perfect gas is = 22.414L/mol

so  22.414L perfect gas contain = 1mole

so 1L contain perfect gas amount in mole is = 1mole*1L/  22.414L = 0.04096mol

Moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm = 0.04096mol*21.0565%

So Moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm = 0.04096mol*21.0565/100 =0.0086248mol

So Moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm =0.0086248mol

First to find the atmospheric pressure is 440mmHg in bar

by using

P = ρ*g*h

ρ= density of mercury =13593 Kgm^-3, g = garvity = 9.81 ms^-2 and h = height of mercury in colmn.

partial pressure of oxygen is

when h= 440mmHg = 0.440m Hg

soatmospheric pressure of is P = 13593 Kgm^-3 *9.81 ms^-2*0.440m =58672.8252 Pa

1atm = 101325Pa

So %pressure at  summit of Mt. Whitney of air is   = 58672.8252 Pa*100/101325Pa = 57.9056%

So %pressure at  summit of Mt. Whitney of air is = 57.9056%

so less pressure amount is 100%-57.9056% = 42.0944%

Hence there gases have less presure so they more volume42.0944%of 1atn volume

thus we know in air at 1 STP = 22.414L

At  summit of Mt. Whitney of have pressure volume is = 22.414L*42.0944%+22.414L= 31.849L

so

31.849L contain = 1 mol

than 1L contain anount of mole   summit of Mt. Whitney = 1mol*1L/31.849L = 0.0313 mole

Than amount of oxygen at summit of Mt. Whitney =  0.0313 mole * *21.0565% = 0.00661mol

So  amount of oxygen at summit of Mt. Whitney = 0.00661mol

d.)

Comapre the vaules of part (b) and (c)

part (b )Moles of oxygen are in 1.00L of air at standard atmospheric pressure of 1 atm =0.0086248mol

part (c)Moles of oxygen are in 1.00L at summit of Mt. Whitney = 0.00661mol

from the value of part (b) and Part (c) we see as the height increase the pressure amount is decrease hence amount of oxygen pressure is also decreases so its concentration in one volume is decreases so oxygen amount decreases so  it is harder to breathe at higher altitudes.