A, 20 - 3 (3000/1000)
=20 - 9
= 11 deg C
B. if it really is "adiabatic",no heat lost or gained by gas during drop in pressure.
for an ideal gas with g =1.4
Po Vo^g = P V^g
To * Vo^(g-1) = T * V^(g-1)
for atmosphere the pressure is approximately
P = Po e^-(M G/R T)y
at 3000 meters using T =293:
M = mol mass of air = 28.8 *10^-3 kg/mol
G = 9.8 m/s^2 gravity
R = 8.315 J/mol deg K gas constant
so
ln (P/1 atm) = -.347
P = .706 atm
so what is the volume ratio V/Vo of a mol for example ?
(V/Vo)^1.4 = =Po/P = 1/.706 = 1.42
1.4 ln (V/Vo) = 1.416
ln V/Vo = 1.01 expands
V/Vo = 2.75 expansion ratio
so what is T actually?
To * Vo^(g-1) = T * V^(g-1)
T = To (Vo/V)^(g-1) with g = 1.4
T = 293 (.364)^.4
ln (T/293) = .4 ln (.364)
ln (T/293) = -.4245
T = 293(.654)
T = 192 K
= -81 C
CHECK MY ARITHMETIC !!!
Given a lapse rate of 3 degrees C/1000 meters and a temp of 20 degrees C at 0 meters:
A.What is the air sitting at 3,000 meters elevation?___degrees C
B.If a sample of air were lifted from 0 meters to 3,000 meters elevation, what would its new temp be?_____C(Assume no condensation is occuring.)
I don't undertsnad how to figure this out at all....
3 answers
so what is the volume ratio V/Vo of a mol for example ?
(V/Vo)^1.4 = =Po/P = 1/.706 = 1.42
1.4 ln (V/Vo) = .3507
ln V/Vo = .25046 expands
V/Vo = 1.28 expansion ratio
Vo/V = .778
so what is T actually?
To * Vo^(g-1) = T * V^(g-1)
T = To (Vo/V)^(g-1) with g = 1.4
T = 293 (.778)^.4
ln (T/293) = .4 ln (.778)
ln (T/293) = -.1004
T = 293(.904)
T = 265 K
= -8 C
CHECK MY ARITHMETIC !!!
(V/Vo)^1.4 = =Po/P = 1/.706 = 1.42
1.4 ln (V/Vo) = .3507
ln V/Vo = .25046 expands
V/Vo = 1.28 expansion ratio
Vo/V = .778
so what is T actually?
To * Vo^(g-1) = T * V^(g-1)
T = To (Vo/V)^(g-1) with g = 1.4
T = 293 (.778)^.4
ln (T/293) = .4 ln (.778)
ln (T/293) = -.1004
T = 293(.904)
T = 265 K
= -8 C
CHECK MY ARITHMETIC !!!
The way I would do it is as follows:
You gave a lapse rate of 3ºC per 1000m
Air cools the higher you get (to a certain point, but that doesn't interest us for this example)
If the air at 0 is 20ºC and it cools by 3º every 1000m - then the air at 3000m
is 9ºC cooler - namely 11ºC
20 - 9 = 11ºC
________________________________________
Problem B; you didn't give a lapse rate for this example, so I am assuming that you are supposed to use the correct dry adiabatic lapse rate which is:
10ºC for 1000m (or 5.5ºF per 1000')
Thus, assuming that the temp. at 0 is still 20ºC the air at 3000m would be 30ºC cooler (3x10)- namely -10ºC
20 - 30 = -10ºC
At least, that's the way I would do it.
(Actually, the normal lapse rate is 3.5ºF per 1000' or 6.4ºC for 1000m)
The wet adiabatic rate is 3.3ºF per 1000' or 6ºC per 1000m
FALLING air always WARMS at the DRY adiabatic rate
You gave a lapse rate of 3ºC per 1000m
Air cools the higher you get (to a certain point, but that doesn't interest us for this example)
If the air at 0 is 20ºC and it cools by 3º every 1000m - then the air at 3000m
is 9ºC cooler - namely 11ºC
20 - 9 = 11ºC
________________________________________
Problem B; you didn't give a lapse rate for this example, so I am assuming that you are supposed to use the correct dry adiabatic lapse rate which is:
10ºC for 1000m (or 5.5ºF per 1000')
Thus, assuming that the temp. at 0 is still 20ºC the air at 3000m would be 30ºC cooler (3x10)- namely -10ºC
20 - 30 = -10ºC
At least, that's the way I would do it.
(Actually, the normal lapse rate is 3.5ºF per 1000' or 6.4ºC for 1000m)
The wet adiabatic rate is 3.3ºF per 1000' or 6ºC per 1000m
FALLING air always WARMS at the DRY adiabatic rate
0 answers