Given:

$P_s=\text{760 mm of Hg (Barometric pressure at sea level)}$

$P_h=\text{725 mm of Hg (Barometric pressure on a mountain)}$

$ρ_a=1.19kg/m^3 \text{(density of air assumed constant with change in elevation)}$

To find:

h (Elevation of the mountain)

Sol:

Now, we know that the difference in barometric (atmospheric) pressure at sea level and that on the mountain is due the elevation difference i.e. a pressure equivalent to the extra height of air column equal to the elevation of the mountain acts at the sea level as compared to on the mountain.

$∴P_s=P_h+ρ_a gh$

$∴ρ_a gh=P_s-P_h$

Now,

$\begin{aligned} \text{1 mm of Hg} &= \text{0.001 m of Hg} \\ &=0.001\times13600\times9.81 \\ &=133.416 N/m^2 \text{or pascal (Since P=hρg) and ρ(density) of Mercury (Hg)} \\ &=13600 kg/m^3 \end{aligned}$

$\begin{aligned} ∴ ρ_a gh &=(760-725)\times0.001\times13600\times g \\ ∴ρ_a h&=(760-725)\times13.600 \\ ∴h&=(760-725)x13.600/ρ_a \\ ∴h&=(760-725)\times13600/1.19 \\ ∴h&=400 m \end{aligned}$

Thus, the elevation of the mountain from the sea level is 400mtrs.