The temperature at the foot of a mountain is `30^(@)C` and pressure is `760 mm Hg`, whereas at the top of the densities of air at the foot and top of the mountain.

Explanation

Let `P_(1)=760 mm`, `T_(1)=30^(@)C=303 K`, `P_(2)=710 mm`, `T_(2)=273 K`
We have
`rho=(PM)/(RT)`
where `rho` is density, `M` is molar mass, and `R` is gas constant
Let the density at the top be `(rho_(2))`
`rho_(2)=(P_(2)M)/(RT_(2))`
Let the density at the foot be `(rho_(1))`
`rho_(1)=(P_(1)M)/(RT_(1))`
Hence
`(rho_(1))/(rho_(2))=(P_(1))/(T_(1))xx(T_(2))/(T_(1))=(760)/(303)xx(273)/(710)=(0.964)/(1)`
`:. rho_(1) : rho_(2)=0.964 : 1`