Suppose that the earth's atmosphere is an ideal gas with molecular weight μand that the gravitat

Suppose that the earth's atmosphere is an ideal gas with...

Key Concepts

Ideal Gas Law

The ideal gas law provides a fundamental relationship between pressure, temperature, and density of a gas. It allows us to relate macroscopic quantities like pressure and temperature with microscopic properties, which is crucial for deriving equations that describe atmospheric behavior.

Hydrostatic Equilibrium

Hydrostatic equilibrium refers to the balance between the upward pressure gradient force and the downward gravitational force in a fluid. This concept is key in atmospheric physics, where it leads to the differential equation that relates the rate of change of pressure with altitude.

Adiabatic Processes

An adiabatic process is one in which no heat is exchanged with the surroundings. In the context of the atmosphere, when air expands or compresses adiabatically, its pressure and temperature are interrelated through specific thermodynamic relationships governed by the adiabatic index.

Adiabatic Lapse Rate

The adiabatic lapse rate describes the rate at which temperature changes with altitude in an air parcel undergoing an adiabatic process. It emerges from the interplay of adiabatic expansion (or compression) and the hydrostatic balance, quantifying the vertical temperature gradient.

Isothermal Atmosphere

An isothermal atmosphere is one where the temperature remains constant with height. This assumption simplifies the integration of the pressure gradient equation, leading to an exponential decrease in pressure with altitude known as the barometric formula.

Exponential Atmosphere (Barometric Formula)

The exponential atmosphere, or barometric formula, results from applying the hydrostatic equilibrium to an isothermal environment. It characterizes the pressure variation with altitude as an exponential function, incorporating parameters like temperature, gravitational acceleration, and molecular weight.

Transcript

00:01 This problem's solution lies with these three equations denoted in red are the red arrows.

00:07 The 1 minus h over 142 ,442 to the 5 .255 -876 power times the reference c -level atmospheric pressure.

00:19 In a standard condition, it would be 14 .7, and that would be calculating atmospheric pressure mathematically and pounds per square inch, which can also be converted into kilopascal's metric units.

00:34 Next, the second arrow is the air density of pounds per cubic foot formulation, and that's based on taking the molar mass of air, 29 grams per mole, times the atmospheric pressure in psi, which we calculated in the first equation, and then dividing that into 10 .73 times the 10 .73 is the ideal, gas law constant for air times the absolute temperature in rankins, which is, correlates to the fahrenheit scale only.

01:06 So that's the fahrenheit degrees plus 460.

01:09 Then he can compute the pounds per cubic foot air density.

01:15 And so the conversions are listed below that, below that formula for converting from rankins to, from fahrenheit to rankins and celsius to fahrenheit.

01:27 And then and then of course the conversion factor between kilograms per cubic meter and pounds per cubic foot air densities.

01:36 And then the last formula is the approximation of temperature based on altitude change.

01:44 And there is a standard reference point, but it's also predicated on weather fluctuations in interference pattern.

01:52 So it's really hard to actually equate a formulation for exact temperature changes with elevation above sea level, but it's approximately a negative 3 .5 degree fahrenheit fluctuation or change per 1 ,000 feet of elevation above sea level outside of weather fluctuations.

02:15 So we'll see them that the temperature change is a standard condition fluctuation based on that metric.

02:22 So everything else is basically converting back and forth from us and metric unit.

02:34 But we have to be very careful about the reference atmospheric pressure, and this question's specifications list 101 kilopascals instead of 101 .35.

02:48 And also this formula here to compute psi is cap.

02:54 Between mean c level and 36 ,086 feet.

02:59 Just a reference point above that is a different formulation.

03:04 So the problem specifies 101 .1 .0 kilopascals.

03:12 So if we convert that, it actually becomes 14 .65 psi instead of 14 .7.

03:20 So that's something to be aware of.

03:23 And so if we go out to 50 % or half of the reference c -level atmospheric pressure of 14 .65 psi, that would equate to 7 .325 psi.

03:39 So we have to find an altitude for which this 50 % atmospheric pressure metric, we have to find that specific altitude.

03:51 So there is a formulation as listed that can equate the altitude for which the atmospheric pressure will drop by 50%.

04:00 So we simply can input the value and get also the metric conversion of 50 .5 kilopascals is 7 .325 psi atmospheric pressure.

04:16 So if we actually combine those first two equations through substitution, for example, we start with the air density formula and we substitute the atmospheric pressure psi formula to combine the air pressure and temperature as functions of altitude.

04:37 So technically, part a would be answered by this combined equation here.

04:44 So it's the molar mass of air times the atmospheric pressure psi formula with a reference sea level atmospheric pressure of 14 .65 rather than 14 .7.

04:57 And then of course divided into the ideal gas constant for air times the absolute temperature and rankins equating air density.

05:05 So that would be the air pressure and temperature as functions of altitude describing the.

05:16 Mathematical relationship between these two and the functions as functions of altitude.

05:23 So from that point on, we can start with a reference c level temperature of 20 degrees celsius, converting that to fahrenheit, 68 degrees fahrenheit, and then converting it once again to rankin's absolute temperature, 528 degrees rankins.

05:39 And then after that, we equate the 50 % atmospheric pressure point.

05:45 Of 7 .35 psi to equal the psi equation, the atmospheric pressure and psi equation.

05:57 So right here would be that descriptor.

06:00 And then we simply solve for h.

06:02 So, of course, we could simplify by dividing both sides by 14 .65.

06:07 This then reduces it down to 0 .5 is equal to quantity of one minus the ratio of the feed elevation over 145 -442 to the 5 .25 -8776 power.

06:23 And so we really want to simplify the exponent out of it so we can get the inverse or the the inverse root of the exponent.

06:32 And so we can cancel that out.

06:33 So when we apply it to 0 .5, we get 0 .876 is equal to a pretty easy algebraic expression, 1 minus the ratio of the feet altitude over 145 -4 -14.

06:46 And then it's simply just solving for the h.

06:50 So we equate 18 ,000 35 feet or 5 ,498 .5 meters above the reference sea level point is when the atmospheric pressure will drop by 50%...

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