The density of air on the Earth decreases almost linearly with height from 1.22kgm–3 at sea level
to 0.74 kgm–3 at an altitude of 5000m.
Atmospheric pressure at the Earth’s surface on a particular day is 100000Pa. The value of g
between the Earth’s surface and an altitude of 5000 m can be considered to have a constant
value of 9.7ms–2.
What will be the atmospheric pressure at an altitude of 5000m?
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60656Pa, rounded to units place.
Explanation:The relationship between the density ρ and air pressure p is given as
ρ1ρ2=p1p2
where it is assumed that temperature does not alter with altitude.
∴ If ρs is density of air at sea level and ρ5000 the value of density at an altitude of 5000m, and similarly for air pressure, we have the equation
p5000=ps×ρ5000ρs
Inserting given values we obtain
p5000=105×0.741.22Pa
⇒p5000=60656Pa, rounded to units place.
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reddz:
But according to the mark scheme the answer is supposed to be 52 000 Pa... because I got the exact same answer...
To find the pressure at an altitude of 5000 m,
You need to find the average density =[(0.74+1.22)/2], then multiply this value by 5000 and then by 9.7. This is from the formula: Pressure= density x acceleration of free fall x height. You should get 47530 Pa which is the pressure due to the air below the 5000m high point down to sea level. Your final answer must 100000Pa- 467530 Pa to get the pressure at 5000m altitude which gives approximately 52000 Pa. Hope this helps. :)
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