Question 1178345: The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature, the pressure is 103.5 kPa at sea level and 89 kPa at h = 1,000 m. (Round your answers to one decimal place.)
(a) What is the pressure at an altitude of 3500 m?
kPa
(b) What is the pressure at the top of a mountain that is 6251 m high?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Set up the Differential Equation**
The problem states that the rate of change of pressure (P) with respect to altitude (h) is proportional to P. This can be written as:
* dP/dh = kP
where k is the proportionality constant.
**2. Solve the Differential Equation**
This is a separable differential equation. We can solve it as follows:
* dP/P = k dh
* ∫(dP/P) = ∫(k dh)
* ln|P| = kh + C
* P = Ae^(kh)
where A = e^C.
**3. Use the Given Information to Find A and k**
We are given:
* P = 103.5 kPa at h = 0 m (sea level)
* P = 89 kPa at h = 1000 m
Using the first condition:
* 103.5 = Ae^(k * 0)
* A = 103.5
So, the equation becomes:
* P = 103.5e^(kh)
Using the second condition:
* 89 = 103.5e^(1000k)
* 89 / 103.5 = e^(1000k)
* ln(89 / 103.5) = 1000k
* k = ln(89 / 103.5) / 1000
* k ≈ -0.0001484
Therefore, the equation is:
* P = 103.5e^(-0.0001484h)
**4. Calculate the Pressure at 3500 m**
* P = 103.5e^(-0.0001484 * 3500)
* P ≈ 103.5e^(-0.5194)
* P ≈ 103.5 * 0.5947
* P ≈ 61.55 kPa
Rounded to one decimal place:
* P ≈ 61.6 kPa
**5. Calculate the Pressure at 6251 m**
* P = 103.5e^(-0.0001484 * 6251)
* P ≈ 103.5e^(-0.9276)
* P ≈ 103.5 * 0.3954
* P ≈ 40.92 kPa
Rounded to one decimal place:
* P ≈ 40.9 kPa
**Answers**
(a) The pressure at an altitude of 3500 m is approximately **61.6 kPa**.
(b) The pressure at the top of a mountain that is 6251 m high is approximately **40.9 kPa**.
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