Question

Rising nearly 1.68 km above Yosemite Valley and 2.68 km above sea level, Half Dome is a Yosemite icon and a great challenge to many hikers. (a) What would be the barometric pressure reading, in mmHg, at Half Dome? (b) Assume a hiker takes the same volume of air with each breath as he/she does while walking at sea level. Determine the ratio of the mass of oxygen inhaled for each breath at Half Dome compared to that at sea level.

Answer 1

Ans;- a) let P be the barometric pressure

P₀ be the atmospreic pressure = 760mmhg

m be the mass of air = 28.9 g/mol

g be acceleration due to gravity=9.8m/s²

R=universal gas constant =8.314J/mol/K

T=temperature=273K

Let be the height at half dome

Now

                =2.68-1.68

                =1km

                =1000m

Now,

= 760e^{(-28.9x9.8x1000/8.314x273)}

=760x1.4

   =1064mmhg

The barometric pressure at half dome is 1064mmhg

Ans b) Let m_s be the mass of oxygen at sea level

m_hd be the mass of oxygen at half dome

Now theoritically we know

density of air at sea level =1.225Kg/m³

Now height at half dome=1000m

density of at air at half dome = 1.202kg/m³

Now

mass = density/volume

since the volume remains same

we have

      

           =0.98/1