Asked by Eurt
The temperature of earth atmosphere drops 5°C for every 1km elevation above the earth surface. If the air temperature at ground level is 15°C and 760 mmHg, determine the elevation in m when the pressure is 380 mmHg. Assume the air is an ideal gas with molecular weight of 29 g/mol.
I know that we need to find T(z) but I don't know. Help please!
I know that we need to find T(z) but I don't know. Help please!
Answers
Answered by
Edmond Baking
Find a relation between T and P
D
dT/dZ= -5 DEG C/1000 m
Integrating: setting limits of T=288.15 at z=0 yields an equation
T=-0.005z + 288.15 (eqn 1)
Sub for T in eqn 1 and integrating yields
(Integral) dP/P (from Pb to Pa) = -(g*MW/R) (integral) dZ/(288.15-0.005z)
Since za=0
ln(Pa/Pb) = -(g*MW/R) * 288.15/(288.15-0.005z)
Sub the values and using shift solve
The answer would be 5550 m ~~ 5.55km
D
dT/dZ= -5 DEG C/1000 m
Integrating: setting limits of T=288.15 at z=0 yields an equation
T=-0.005z + 288.15 (eqn 1)
Sub for T in eqn 1 and integrating yields
(Integral) dP/P (from Pb to Pa) = -(g*MW/R) (integral) dZ/(288.15-0.005z)
Since za=0
ln(Pa/Pb) = -(g*MW/R) * 288.15/(288.15-0.005z)
Sub the values and using shift solve
The answer would be 5550 m ~~ 5.55km
Answered by
Anonymous
P2/P1 = (To-Bh/To)^(gMW/RB)
where:
P2=380mmHg
P1= 760mmHg
To=15+273.15 =288.15°K
B=5°C/1000m
g=9.81
MW=29kg/kmol
R=8314 Pa×m³/kmol×K
Then you can shift solve on your calculator. Thanks me later.
h = 5,551.1031meters
where:
P2=380mmHg
P1= 760mmHg
To=15+273.15 =288.15°K
B=5°C/1000m
g=9.81
MW=29kg/kmol
R=8314 Pa×m³/kmol×K
Then you can shift solve on your calculator. Thanks me later.
h = 5,551.1031meters
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