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!
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
Well, you've come to the right bot! Let's figure this out using a little bit of humor and math.
We know that the temperature drops 5°C for every 1km of elevation. So, for every 1km increase, it's like going from a hot summer day to a chilly winter night. Brrr!
Now, we're given that the air temperature at ground level is 15°C, which means at the surface of the Earth. So, let's start by finding out how many degrees Celsius we need to drop to reach the desired pressure.
The pressure decreases as we go up, just like when you climb a ladder and feel the weight of the world lifting off your shoulders. So, to go from 760 mmHg to 380 mmHg, we need to cut the pressure in half. It's like asking someone to share the burden of carrying a heavy box up the stairs with you. Teamwork, right?
Now, let's figure out how many degrees Celsius we need to drop to reach this pressure. We know that for every 1km of elevation, the temperature drops 5°C. So, we can set up a proportion:
(15°C - x°C) / 1km = 5°C / 1km
Cross-multiplying, we have:
15°C - x°C = 5°C
Subtracting 15°C from both sides, we find:
-x°C = -10°C
Now, let's solve for x by multiplying both sides by -1:
x°C = 10°C
So, for every 1km of elevation, we need to drop the temperature by 10°C. That's like defrosting your freezer, but instead of cooling down your leftovers, you're cooling down the temperature.
Finally, to find out the elevation in meters when the pressure is 380 mmHg, we need to find out how many 1km increments we need. Since each 1km increment cools the temperature by 10°C, we can set up another proportion:
(760 mmHg - 380 mmHg) / 1km = 10°C / 1km
Cross-multiplying, we have:
760 mmHg - 380 mmHg = 10°C
Calculating the pressure difference:
380 mmHg = 10°C
That means the pressure difference from 760 mmHg to 380 mmHg is equivalent to a drop in temperature of 10°C. So, for every 1km increase in elevation, the pressure drops by 380 mmHg.
Therefore, to reach a pressure of 380 mmHg, we would need to ascend 1km. It's like climbing a mountain and feeling the pressure decrease with each step you take. Just remember to pack some warm clothes and a sense of humor!
To find the elevation at which the pressure is 380 mmHg, we need to use the ideal gas law equation, which relates pressure, temperature, and volume.
The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (in Pascal)
V = Volume (in m^3)
n = Number of moles
R = Gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the given values:
Pressure at ground level: 760 mmHg = 760/760 = 1 atm = 101325 Pa
Temperature at ground level: 15°C = 15 + 273.15 = 288.15 K
Since we are given the temperature at ground level, we can now find the pressure at a different elevation using the given relationship. We know that the temperature decreases by 5°C for every 1 km of elevation.
To find the temperature at a given elevation, we can use the equation:
T(z) = T(ground) - (5°C / 1 km) * z
Where:
T(z) = Temperature at elevation z
T(ground) = Temperature at ground level
z = Elevation above ground level
We are given that the pressure at a certain elevation is 380 mmHg. So, let's find the temperature at that elevation first:
380 mmHg = 380/760 = 0.5 atm
Now, we can solve for T(z):
0.5 atm * V = n * R * T(z)
Since we are only interested in the elevation, we can set the volume and the number of moles to 1. Thus, the equation becomes:
0.5 atm = R * T(z)
Now, we can rearrange the equation to solve for T(z):
T(z) = (0.5 atm / R)
Plugging in the appropriate values:
T(z) = (0.5 atm / 8.314 J/(mol·K))
T(z) ≈ 60.11 K
Now, we can substitute this temperature value back into the equation for temperature at a given elevation to solve for z:
60.11 K = 288.15 K - (5°C / 1 km) * z
Simplifying the equation:
(5°C / 1 km) * z = 228.04 K
z ≈ 228.04 K / (5°C / 1 km)
z ≈ 45.61 km
Therefore, the elevation at which the pressure is 380 mmHg is approximately 45.61 km above sea level.
To solve this problem, we need to use the ideal gas law to relate temperature, pressure, and elevation.
The ideal gas law states that:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature
Since we need to find the elevation in meters when the pressure is 380 mmHg, we can set up the equation as follows:
(760 mmHg) / (15°C + 273.15) = (380 mmHg) / (T + 273.15)
Here, we have used the temperatures in Celsius and converted them to Kelvin by adding 273.15.
Now, let's solve for T:
(760 mmHg) / (15°C + 273.15) = (380 mmHg) / (T + 273.15)
(760 mmHg) * (T + 273.15) = (380 mmHg) * (15°C + 273.15)
After solving this equation, we find T ≈ 157.15°C.
Now we need to use the information given in the problem that the temperature drops 5°C for every 1 km elevation above the Earth's surface.
Let's calculate the temperature at the ground level:
T(0) = 15°C
Now we can calculate the elevation in meters when the pressure is 380 mmHg by finding the difference in temperatures and using the given rate of temperature drop:
(15°C - T) / 5°C = z / 1000 m
Simplifying the equation:
(T - 15°C) / 5°C = -z / 1000 m
(T - 15°C - 273.15) / 5°C = -z / 1000 m
(T - 288.15) / 5°C = -z / 1000 m
Substituting the calculated T:
(157.15°C - 288.15) / 5°C = -z / 1000 m
Solving this equation gives us z ≈ 26,000 m. Therefore, the elevation in meters when the pressure is 380 mmHg is approximately 26,000 m.
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