Using Pressure and Temperature differences to find height

AI Thread Summary
The discussion focuses on calculating the height of a mountain using temperature and pressure readings from a barometer at different elevations. The temperature lapse rate is utilized to establish a linear relationship between temperature and height, leading to the formulation of temperature equations. However, the assumption of constant air density is identified as a significant error, as it affects pressure calculations. A more accurate approach involves integrating the pressure gradient equation, which accounts for varying density with height. The conversation emphasizes the need for proper integration to resolve the discrepancies in the height calculation.
M_Abubakr
Messages
8
Reaction score
0

Homework Statement


At the top of a mountain the temperature is -5 degree C and a mercury barometer reads 566 mm, whereas the reading at the foot of the mountain is 749 mm.

Homework Equations


Assuming a temperature lapse rate of 0.0065 K/m and R = 287 J/kg K, calculate the height of the mountain.

The Attempt at a Solution


The lapse rate tells me how much temperature increases or decreases with height. So I can say that Temperature is a function of height T(d). And I'm assuming it to be linear function so I used the equation of a line which is y=mx+c.
where
y=T(d)
m=0.0065K/m
x=d
and c is the y intercept which is the temperature when height is 0 so the equation will be

T(d)= -0.0065d+c

I've taken slope as negative because temperature decreases with change in height which is obvious.

The pressure at the peak will be (from the mercury barometer reading)
P1=ρgh1
P1=13560x9.81x566x10^-3
P1= 75291.3576m

The pressure at the foot of the mountain is: (also from the mercury barometer reading)
P2=ρgh2
P2=13560x9.81x749x10^-3
P2= 99634.6764m

From the formula PV=mRT
we get P=ρRT
I assumed that the density of air will remain constant with change in height.
so what I did was I made rho subject.
ρ=P/RT
so at peak
ρ=P1/RT1
and at the foot
ρ=P2/RT2
ρ=ρ
P1/RT1=P2/RT2
Cancelling R on both sides will give us
P1/T1=P2/T2
making T2 subject will give us the temperature at the foot of the mountain.

T2=P2T1/P1
T2=(99634.67640)*(273-5)/(75291.3576)
T2=354.65
Which is the temperature at the foot of the mountain. Its also the y intercept c of the linear function T(d)= -0.0065d+c because it tells us the temperature when height is zero.

so the equation we'll end up with will be

T(d)= -0.0065d+354.65
To find the height of the peak from the foot of the mountain we'll put T(d) = 273-5 = 268K

268 = -0.0065d+354.65
after making d subject and solving further we'll get d=13330.76923m

Can anyone tell me what I did wrong in my attempt to solve this question? I know this is not the correct answer.
 
Physics news on Phys.org
If the temperature at height d is -5C, and the lapse rate is 0.0065 C/m, in terms of d, what is the temperature at the ground? In terms of d, what is the absolute temperature as a function of elevation z and height d?

Assuming the density of the air is constant was not a good idea. For arbitrary locations, $$\frac{dP}{dz}=-\rho g=\frac{P}{RT}g$$You need to integrate this between z = 0 and z = d, and match the pressures at the foot and at z = d.
 
  • Like
Likes Dr Dr news and scottdave
Good suggestion with the z=0 being sea level and p and T being the sea level standard values.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top