Question 1178345
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**.