Calculate the pressure and density of air at an elevation of 4500 m above sea level if the atmospheric pressure and density at sea level are 101.35 kPa and 1.2255 kg/m.
Answers
Answer:
Here is your solution
Explanation:
ANSWER
(a)
Consider an atmospheric layer of thickness dy at height y and cross section area A at static equilibrium.
Mass of the layer = density×volume=Number of atoms per unit volume×volume×mass of an atom
M=ρAdy=mANdy
⟹ρ=mN.......(i)
For equilibrium, upward force = downward force + weight
PA−(P+dP)A=mNAdy g
where m:mass of an atom,N:Number of atoms per unit volume
dP=−mANgdy
Substituting from (i),
dP=−Aρgdy..........(ii)
From gas law,
P=NkT
P=ρkT/m
dP=
m
kT
dρ..........(iii)
From (ii) and (iii),
m
kT
dρ=−Aρgdy
ρ
dρ
=−αdy where α is a constant
∫
ρ
o
ρ
ρ
dρ
=∫
0
y
−αdy
lnρ−lnρ
o
=−αy
ln
ρ
o
ρ
=−αy
ρ=ρ
o
e
−αy
...........(iv)
Putting y=y
o
,ρ=ρ
o
α=1/y
o
............(v)
From (iv) and (v),
ρ=ρ
o
e
−y/y
o
Hence proved.
(b) Density ρ= Mass / Volume
ρ= (Mass of the payload + Mass of helium) / Volume
=(m+Vρ
He
)/V
=(400+1425×0.18)/1425
=0.46kgm
−3
From part (a)
ρ=ρ
0
e
−y/y
0
log
e
(ρ/ρ
0
)=−y/y
0
∴y=−8000×log
e
(0.46/1.25)
=−8000×(−1)
=8000m=8km
Hence, the balloon will rise to a height of 8 km.
If u r satisfied mark me as brainliest