The rate at which barometric pressure decreases with...

The rate at which barometric pressure decreases with altitude is proportional to the barometric pressure at that altitude. If the barometric pressure is measured in inches of mercury, and the altitude in feet, then the constant of proportionality is $3.7 \cdot 10^{-5} .$ The barometric pressure at sea level is 29.92 inches of mercury. (a) Calculate the barometric pressure at the top of Mount Whitney, 14,500 feet (the highest mountain in the US outside Alaska), and at the top of Mount Everest, 29,000 feet (the highest mountain in the world). (b) People cannot easily survive at a pressure below 15 inches of mercury. What is the highest altitude to which people can safely go?

Transcript

00:01 In this question, it is given that the rate at which the barometric pressure decreases with an altitude is proportional to the pressure at that altitude.

00:11 It means d .p by d .h is proportional to pressure p, which means if we remove this proportionality symbol, the constant given can be written as minus 3 .7 into 10 to the power minus 5 multiplied by p.

00:31 So, this expression can be further reduced into the variable separable form, pp upon p equals to minus 3 .7 into 10 to the power minus 5 d h.

00:46 Now if you integrate on the both on both sides we get log p equals to minus 3 .7 into 10 to the power minus 5 h plus log c if we keep log c on the left side we get this expression which further can be written as log of p upon c equals to minus 3 .7 into 10 to the power minus 5 h which means p upon c can be written as e to the power minus 3 .7 into 10 to the power minus 5 into h which implies p equals to c e to the power minus 3 .7 into 10 to the power minus 5 h so this is the expression for the pressure in terms of altitude as well as the constant c here constant c simply represents the initial pressure which is at the c level we can take this constant as the pressure at the c level or p0 and rewrite the above expression as p equals to p .0 e to the power minus 3 .7 into 10 to the power minus 5 h.

02:12 After substituting the value of p .0 mentioned in the question, the expression becomes p equals to 29 .92 e to the power minus 3 .7 into 10 to the power minus 5h.

02:27 Let this expression as equation number 1.

02:30 Now we will substitute for the part 1.

02:33 It is required that we need to calculate the pressure at height 14 ,500 feet.

02:44 Therefore, after the substitution of value h in the expression 1, we get the barometric pressure as 29 .92 into e to the power minus 3 .7 into 10 to the power minus 5 multiplied by 14 ,500...

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