Equilibrium height of an air parcel

[Raihan amin]

Homework Statement

An air parcel is investigated to study the weather.It rises up and rests at an equilibrium height in the atmosphere,where its weight is exactly balanced by the upward buoyant force. We shall assume
ideal gas law to hold for all processes and neglect the mass of the air parcel. If the air parcel is displaced vertically, it is often found to oscillate about the equilibrium position. The frequency of this oscillation is called “Brunt-Vais¨al¨a” frequency. Now suppose the air parcel is displaced upwards adiabatically from the equilibrium
position. Let M(b) and M(a)be the molar mass of the air parcel and the atmosphere respectively.and C(b) and C(a) be the molar heat capacity at constant volume for the air parcel and the atmosphere respectively .
Find the equilibrium height of the air parcel.
Hints:First find the lapse rate for both the air parcel and the atmosphere. Then find out the acceleration of the air parcel.

2. Homework Equations :

I first find the lapse rate ,Γ=dT/dz for both of them. And it is
Γ(a)=-M(a)g/C(a) and
Γ(b)=-M(a)gT(b)/C(b)T(a)
And z''=g((M(a)T(b)/M(b)T(a))-1)


But i couldn't find out equilibrium height equating z" to zero. Please help me to find this.
(The original problem (1st one) is posted here)

 

[Raihan amin]

Here i don't know whether M(a) and M(b) are variable or not

 

[PeterDonis]

@Raihan amin please do not post images of your attempt at a solution. Use the PF LaTeX feature to post equations directly. You can find help on how to do that here:

https://www.physicsforums.com/help/latexhelp/

 

[Bilal Rajab Abbasi]

@Rahan amin Make simultaneous equations and solve. Find the variable you were asking about.

Mass always is a constant wheras Weight varies with height. A variable here would specifically be the temperature of the test substances you are using.
Also equate the equations like make two equations where the unknown is a subject and write them according to L.H.S and R.H.S...
This will make other variables known to you.Last you just have to put the values and the answer is there. You might want to make a graph for the equilibrium as well. Max Height can therefore be found using extrapolation of graph.

 

[Raihan amin]

Also equate the equations like make two equations where the unknown is a subject and write them according to L.H.S and R.H.S...
This will make other variables known to you.

Could you please explain symbolically ?

 

[Bilal Rajab Abbasi]

For example;
You can simplify Γ(a)=-M(a)g/C(a) and
Γ(b)=-M(a)gT(b)/C(b)T(a)
And z''=g((M(a)T(b)/M(b)T(a))-1)
this equation by using Linear Algebra...
Replace the symbol M with "x" and z with Y
i.e, y"=g((x(a)T(b)...
If you are using differentiation to find the answer then it is very simple...
You can choose for any value and make an equation...
e.g take 1,2,3 for M and input in the equation
Y=g*(x(1)t(1))
for the function x(a) and t(a), there must be a value
if the Molar Mass M is 28.97 g/mol
input the value and you get z

But instead of using Time Period, use 1/T=F which is the frequency.
You used two equations for T, just use one single value for frequency, i.e(from the source)..which will be constant.
When the weight is balanced by the force, there is no net force meaning no acceleration.
F=ma
F=resultant force
a=acceleration
F=0
a=0
Thanks..!

 

[Raihan amin]

To be frank, still i haven't got anything.would you please do some calculations and show us?