Answer:
Δh_air=714m
Explanation:
Given data
[tex]P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}[/tex]
Solution
ΔP=P₁-P₂
=(ΔhHg)×pHg×g
=(Δh_air)× p_air ×g
Then
Δh_air=(pHg+ΔhHg)÷p_air
[tex]=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m[/tex]
Δh_air=714m