when does air rise up and why?
Answers
Answer:
because it is lighter than every thing and
you know what atoms of air has so much space that gravity also applies very less
thank you
Warm air rises because it is less dense than cold air.
Back in ancient Greece, Archimedes theorized that when an object is immersed in a fluid, it is buoyed upwards by a force equal to the weight of the fluid displaced by the object. Thus, less dense objects or fluids float atop denser fluids due to Archimedes principle.
An analogy is useful in understanding the buoyant force. For example, if you drop a cork into a bowl of water, it floats because the density of cork ( 0.25 gm/cm3 ) is less than the density of water ( 1 gm/cm3 ). Similarly, oil will separate from water and float on top of the water because the density of oil is approximately ( 0.92 gm/cm3 ), which is less than the density of water. Thus, the density of a parcel of warm air will rise above a parcel of colder air because it is less dense.
The Math for a Cork Floating in a Bowl of Water
To prove that a cork will float on a layer of water we equate the downward gravitational force acting on the cork to the upwards buoyant force exerted by the water on the cork.
The gravitational force acting on the cork is simply its mass, mcork times the acceleration of gravity, g . The buoyant force acting on the cork is equal to the volume of the water that is displaced by the cork Vdisp times the density of water, ρw and the acceleration of gravity, g .
mcork×g=Vdisp×ρw×g.
We know the values of all the parameters except the displacement volume, so let’s solve for it.
Vdisp=ρw×gmcork×g=ρwmcork .
For a one-gram cork, the displacement volume, Vdisp is 1 cm3 , so the volume of a one-gram of cork is equal to its mass, mcork , divided by its density, ρcork .
Vcork=mcork/ρcork .
Thus, the volume of a one gram cork is 4 cm3 .
Since Vcork>Vdisp , not all of the corks volume is displaced by the water. In fact, a quarter of the cork’s volume ( 1 cm3 ) is displaced by the water, so it is partially submerged, and three-quarters ( 3 cm3 ) of the cork has to float above the surface of the water in order for the buoyant force to be in equilibrium with the gravitational force.
By the way, if the cork were denser than water, then the displacement volume would be greater than the volume of the cork, and the cork would sink.
The Math for a Parcel of Warm Air Floating above a Layer of Cold Air
The gravitational force acting on a parcel of warm air is simply its mass, mwarm times the acceleration of gravity, g . The buoyant force acting on the parcel of warm air is equal to the volume of the cold air that it displaces Vdisp times the density of cold air, ρc and the acceleration of gravity, g .
mwarm×g=Vdisp×ρc×g.
We know the values of all the parameters except the displacement volume, so let’s solve for it.
Vdisp=ρc×gmwarm×g=ρcmwarm.
Let us compare the Vdisp of the cold air displaced to volume of the warm air Vwarm . Since PV=nRT , we can solve for the temperatures T′s , ie.,
Twarm=PVwarm/nR , and
Tcold=PVcold/nR
Since Twarm>Tcold , that is the temperature of warm air is greater than the temperature of cold air, then
PVwarm/nR > PVcold/nR , leading to
Vwarm>Vcold .
This result shows us that the volume of cold air is the displacement volume, and thus the parcel of warm air must rise until it either reaches an altitude that is the same temperature as the surrounding ambient air, or gets trapped by an inversion layer, at which point it stops rising and just “floats” atop the cooler air, just like oil “floats” above the vinegar in a salad dressing bottle.