The motion of air masses through atmosphere can be approximated as adiabatic and

Question

The motion of air masses through atmosphere can be approximated as adiabatic and reversible. Air can be treated as an ideal gas with average molar mass 29 g mol-1 and average heat capacity 29 J K-1 mol-1.

b. Suppose the average atmospheric pressure near the earth's surface is P0 and the temperature is T0. The air is displaced upward until the temperature is T and the pressure is P. Determine the relationship between P and T. (Hint: consider the process as occuring in two steps, equate the sum of the two entropy changes to (delta S total=0))

c. In the lower atmosphere, the dependence of pressure on height, h, above the Earth's surface can be approximated as ln(P/P0)= -Mgh/RT

Where M is the molar mass, g the acceleration due to gravity, and R the gas constant. If the air temperature near the equator is 38 degrees C (roughly 100degrees F) calculate the air temperature at the summit of Mount Kilimanjaro, roughly 5.9km above sea level.

Explanation / Answer

The process can be divided into two steps. One constant pressure process and the second one constat temperature process.

In the case of constant pressure process, the pressure is kept constant at P0, while the temperature is changed from T0 to T.

entropy change = CP*ln(T/TO).

Now the process is Carried out at constant temperature T while changing the pressrue from Po to P.

The entropy change = -R ln (P/P0)

Total entropy change = CP*ln(T/O) - Rln *(P/Po) =0

ln (T/To)= (R/CP)*ln (P/Po)=

T/TO= (P/PO) R/CP = (P/PO) (CP-CV)/CP

but CP/CV= Y

hence T/TO= (P/PO) (Y-1)/Y

which is the equation for adiabatic process.

2.

Given ln(P/PO)= -Mgh/RTO

at h=0, P= P0

at h= h, ln (P/PO)= -Mgh/RT (1)

but T/To= (P/PO) Y-1/Y

P/PO   =(T/TO) Y/(Y-1)

ln(P/PO)= (Y-1)/Y *ln (T/TO)

Eq.1 becomes, (Y-1)/Y*ln (T/TO)= -Mgh/RTo

Y= CP/CV= 29/(29-8.314) =1.4, TO= 38 deg.C= 38+273= 311K, h= 50m, g =9.8 m/sec2 and M =29/1000 =0.029, R= 8.314 J/mole.K

{(1.4-1)/1.4}*ln (T/311)= -0.029*9.8*5.9*1000/(8.314*311)

Ln(T/311)= -2.3, T/311 = 0.103, T= 311*0.103= 32.14 deg.K= 32.14-273= -240.86deg.c

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