C the variation in atmospheric pressure p with elevation h...

(c) The variation in atmospheric pressure P with elevation h above sea level can be expressed as: P = P0 * e^(-M * g * h / (R * T)) where P0 is the pressure at sea level, e is the base of the natural logarithm, M is the molar mass of air, g is the acceleration due to gravity, R is the ideal gas constant, T is the temperature, and h is the elevation above sea level. To calculate the atmospheric pressure at the top of Clingmans Dome, we can substitute the given values into the equation: P0 = 1 atm T = 20°C = 293.15 K M = 28.8 * 10^(-3) kg/mol h = 2025 m Plugging these values into the equation, we get: P = 1 atm * e^(-28.8 * 10^(-3) kg/mol * 9.8 m/s^2 * 2025 m / (8.314 J/(mol*K) * 293.15 K)) Simplifying the equation, we get: P = 1 atm * e^(-0.081 * 2025) Calculating the exponential term, we get: P = 1 atm * e^(-164.025) Finally, we can calculate the atmospheric pressure at the top of Clingmans Dome by evaluating the exponential term and multiplying it by 1 atm: P ≈ 1 atm * 1.5 * 10^(-71) atm Therefore, the atmospheric pressure at the top of Clingmans Dome is approximately 1.5 * 10^(-71) atm.

Transcript

00:01 All right, in this problem, we are asked to calculate different pressure values given partial pressure of oxygen and vapor pressures.

00:12 A big key point here in the problem statement, they give us this very important formula to solve these problems.

00:19 I say that the pressure, the partial pressure of oxygen, is equal to the percentage of oxygen in the atmosphere converted to a decimal, which would be 0 .209, times the pressure of the atmosphere, so pressure atm, minus the vapor pressure in our lungs.

00:41 And that vapor pressure is specifically the saturated water vapor pressure, just that's omnipresent in our lungs.

00:48 But anyway, in part a, they ask us to find if, given that p, the vapor pressure, is equal to 6300 pascales, they want us to find partial pressure of oxygen, and they give us that the pressure of the atmospheres is equal to 101 kpa.

01:13 So really, this problem is just a plug -in chug, really, because they want us to find the pressure of the oxygen, right? this is what we're after in that first problem.

01:23 And they give us both of these things.

01:25 The only thing we've got to watch out for is the fact that this guy here is in pascal's, and this guy here is in kilo -pascals.

01:32 So that means we got to convert it.

01:33 So we'll convert that guy.

01:35 We can do it with dimensional analysis.

01:38 Although if you can do it in your head, you probably don't even need to do this step.

01:40 We know that for every one kilo -pascal, there is a thousand pascal.

01:46 So that means the pressure of the atmosphere is 101 ,000 pascales.

01:52 All right.

01:53 And then it's just a matter of plugging it in.

01:56 So that means the pressure of the oxygen is equal, to 0 .209 times the pressure of the atmosphere, which is 101, oops, 101 ,000 minus the pressure, the vapor pressure.

02:15 So that's the 6300.

02:18 And you plug that into your calculator there, or do it in your head, you know, if you're that impressive, then you would get 19 ,792 .3 .3 pascal's, all right, but we always got to worry about our significant figures and so on, so forth.

02:38 So we look at our data that we're given, we see that this value here has three significant figures.

02:44 This data has two, so that means our answer can only have two.

02:47 So we look at this nine to decide if it rounds up or stays the same.

02:53 And since seven is bigger than five, it would go ahead and round up...

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