At the summit of Mt. Everest, the atmospheric pressure is 200 mmHg and the air density of 0.406 kg/m^3. Calculate the air temperature, in Celsius, given that the molar mass of air is 29.0 g/mol.

Assuming that the composition of the air does not change between sea level and the summit, calculate the percent decrease...

the atmospheric pressure is 200 mmHg and the air density of 0.406 kg/m^3. Calculate the air temperature, in Celsius, given that the molar mass of air is 29.0 g/mol.

D = m/V ... 406g/1000L = 0.406g/L

... I will use MM for molar mass

PV = nRT .... PV = (m/MM)RT

P(MM) = (m/V)RT <<<< sub m/V = D

P(MM) = DRT

(200/760)(29.0) = (0.406)(0.0821)T

T = 229K = -44 degC <<<< first answer

Assuming that the composition of the air does not change between sea level and the summit <<< If you assume this (composition does NOT CHANGE) then the percent oxygen will not decrease ... The question does not make sense.

Consider the air at the top of Mt. Everest :

Pressure, P = 200/760 atm

Density, d = 0.406 kg/m³ = 0.406 g/L

Molar mass, M = 29.0 g/mol

Gas constant, R = 0.08206 L atm / (mol K)

PV = nRT and n = m/M

Then, PV = (m/M)RT

Then, PM = (m/V)RT

Then, PM = dRT

Temperature, T = PM/dR = (200/760) × 29.0 / (0.406 × 0.08206) K = 229 K = -44°C

====

Percent decrease in oxygen gas from sea level to the top of Everest

= Percent decrease in air pressure from sea level to the top of Everest

= [(760 - 200) / 760] × 100(%)

= 73.7(%)

P = 200 mmHg

M = 29.0 g/mol

ρ = 0.406 kg/m³ = 0.406 g/L

R = 62.3636 L mmHg /K mol

T = PM/ ρR = 200*29.0/(0.406*62.3636) = 229.07 K = 229.07 - 273.15 °C = -44.08 °C

At sea level the density of air is generally considered at 1.225 Kg/m³

percentage decrease in air density at summit = (1.225 - 0.406)/1.225 = .6686 or 66.86%

density of a gas = P * molar mass / ( R * temperature)

Substitute and solve.

< Assuming that the composition of the air does not change between sea level and the summit >

Then the percent of oxygen is a constant if the composition does not change.

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