The atmospheric pressure decreases with height. experimental
it is determined that the rate at which the pressure p decreases in
function of the height h, is always proportional to p. Write the
differential equation that describes this model and solve general. Imagine
then p = 1 atm at sea level and p = 0, at 83 atm
5000 m altitude. Calculate what extent the pressure 0.70 atm
amounts.
I misread the question to say that pressure varies linearly with height. Of course, it actually says the rate of change of pressure with height varies linearly with pressure.
The question appeared twice, once in another language. When I got a Google translation of that it was clear that it referred to the pressure being 0.83 atmospheres at 5000 metres.
.
Thanks Alan,
This question is really confusing me.
My answer is also completely different from yours Alan so therefore it is bound to be wrong but maybe you can help clear up my confusion. I do not understand the wording of the question at all.
I don't get the 83atm 5000 m altitude - that is jsut for starters.
BUT
The atmospheric pressure decreases with height. Experimentally it is determined that the rate at which the pressure p decreases in function of the height h, is always proportional to p.
Write the differential equation that describes this model and solve general.
$$\\\frac{dP}{dh}=kP\\\\
P=e^{kh}\; +\; c\\\\
check\\
\frac{dP}{dh}=ke^{kh}=kP\qquad $That looks good$\\\\$$
Imagine then p = 1 atm at sea level and p = 0, at 83 atm 5000 m altitude.
When h=0 P=1 1=e^0+c SO c=1
$$P=e^{kh}+1$$
p=0 h = total confusion here p=0, 83atm, 5000m altitude.
It looks like there's a typo in the fifth line, ... and p=0, at 83 atm 5000 m altitude.
I'm choosing to think of it as ... and p = 0.83 atm at 5000 m altitude.
The d.e. will be
$$\displaystyle\frac{dp}{dh}=kp$$
where k is a constant.
Separating the variables and integrating,
$$\displaystyle \frac{dp}{p}=kdh,\text{ } \ln p=kh+C,$$
so
$$\displaystyle p=e^{kh+C}=e^{C}e^{kh}=Ae^{kh},\text{ A const}$$.
Given
$$p=1 \text{ when } h=0,\text{ so }\qquad A=1,\text{ and,}\\ p=0.83\text{ when }h=5000,\text{ so }\qquad 0.83=e^{5000k},\qquad k=\frac{1}{5000}\ln0.83.$$
$$\displaystyle \text{If }p=0.7,\quad 0.7=e^{kh},\quad kh=\ln0.7,\quad h=\frac{\ln0.7}{k}=5000\frac{\ln0.7}{\ln0.83}$$
$$\displaystyle h = 9571 m.$$
I misread the question to say that pressure varies linearly with height. Of course, it actually says the rate of change of pressure with height varies linearly with pressure.
The question appeared twice, once in another language. When I got a Google translation of that it was clear that it referred to the pressure being 0.83 atmospheres at 5000 metres.
.