Temperature at the foot of a mountian is `30^(@)C` and pressure is 760 mm, whereas at the top of the mountain these are `0^(@)C` and 710 mm. Compare the densities of the air at the foot and top of the mountain.

Explanation

`rho=(PM)/(RT)`
Hence `(rho_(1))/(rho_(2))=(P_(1))/(P_(2))xx(T_(2))/(P_(2))`
`therefore(760mm)/(303K)xx(273K)/(710mm)=(0.964)/(1)`
`thereforerho_(1):rho_(2)=0.964:1`