Help needed with gas laws chemistry problems!!?

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(%)

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.

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%