Eötvös rule

[EndeavourCmdr]

I have a theory that I have been working on for a while, and I am trying to calculate for surface tension via the Eötvös rule.

I am wondering if any of the values of the density, molar mass or critical temperature of space its self are known and able to be expressed in terms of variables or equations?

Anyone have insight into this?

 

[statickid]

0.0

 

[jedidia]

not quite zero, but pretty close... it's also not constant. Can range from somewhere between 0.0001 ions /cm^3 up to a million molecules per cm^3, which is still a very high-quality vacuum. check out the wikipedia article on[ame="http://en.wikipedia.org/wiki/Interstellar_medium"] interstellar medium[/ame], or check some articles on the scientific google search page for a more indept treatment.

 

[EndeavourCmdr]

Okay then since my math skills aren't all THAT great, would someone please solve:

γV^2/3 = k(Tc - 6 K - T)

For the value of T where:

γ = 0
M = 0.001367 kg/mol
p = 3.078 * 10^-17 mol/m^3
V = M/p
k = 2.1×10^−7 J/(K·mol−2/3)
Tc = 33.67 K

V is the molar volume, Tc is the critical temperature of a liquid, and k is the Eötvös constant, and K simply means Kelvin.

It would be VERY much appreciated.

EDIT there was a major typo in k, now fixed.

 

[jedidia]

Not quite the math-wiz myself, but I'm sure someone will have a shot at it if you say what you want as l-value in the end...

 

[Notebook]

I'll have a go in Mathcad:
http://i89.photobucket.com/albums/k207/Notebook_04/EtvsRule_1_.jpg

Can't promise anything of course!

Solve it for T?

N.

 

[agentgonzo]

Temperature of deep-space is two-point-something kelvins. 2.7 from memory, but that may be slightly out.

 

[Notebook]

Put it in Maple 12, and got this for T:

http://i89.photobucket.com/albums/k207/Notebook_04/EotvosRule_1_.jpg

Haven't checked it yet, that could take some time...

Some info from Maple's Help:

Description
The function RootOf is a place holder for representing all the roots of an equation in one variable. In particular, it is the standard representation for Maple algebraic numbers, algebraic functions (see evala), and finite fields GF(p^k), p prime, k > 1 (see mod).
Maple automatically generates RootOfs to express the solutions to polynomial equations and systems of equations (see solve), eigenvalues of matrices (see LinearAlgebra:-Eigenvalues and rational function integrals (see int). Maple can apply diff, series, evalf, and simplify to RootOf expressions.
If x is not specified, then expr must be either a univariate expression or an expression in _Z. In this case, the RootOf represents the roots of expr with respect to its single variable or _Z, respectively. If the first argument is not an equation, then the equation expr = 0 is assumed.
The RootOf function checks the validity of its arguments and solves it for polynomials of degree one. The RootOf is expressed in a single-argument canonical form, obtained by making the argument primitive and expressing the RootOf in terms of the global variable _Z.

N.

 

[RisingFury]

0.0

Well, at least you can add +1 to your post count...

Not quite the math-wiz myself, but I'm sure someone will have a shot at it if you say what you want as l-value in the end...

And you got the urge to tell the whole forum you can't turn a basic equation around?

Okay then since my math skills aren't all THAT great, would someone please solve:

γV^2/3 = k(Tc - 6 K - T)

For the value of T where:

γ = 0
M = 0.001367 kg/mol
p = 3.078 * 10^-17 mol/m^3
V = M/p
k = 2.1×10^−7 J/(K·mol−2/3)
Tc = 33.67 K

V is the molar volume, Tc is the critical temperature of a liquid, and k is the Eötvös constant, and K simply means Kelvin.

It would be VERY much appreciated.

EDIT there was a major typo in k, now fixed.

[math]T=\frac{y}{k}*V^\frac{2}{3} -Tc - 6K[/math]
*Edited into LaTeX*

 

[Notebook]

Well, you've beaten me, and I never was much at algerbra!

N.

 

[fireballs619]

T = (y / k) * V^(2/3) - Tc - 6K

Just to clarify and make easier to read for others, this is

[math]T=\frac{y}{k}*V^\frac{2}{3} -Tc - 6K[/math]

I recommend using LaTeX, it makes math much more legible :thumbup:

 

[RisingFury]

Just to clarify and make easier to read for others, this is

[math]T=\frac{y}{k}*V^\frac{2}{3} -Tc - 6k[/math]

I recommend using LaTeX, it makes math much more legible :thumbup:

Just a minor nitpick...

[math]T=\frac{y}{k}*V^\frac{2}{3} -Tc - 6K[/math]

It's 6K not 6k, but thanks

 

[Notebook]

Still pondering this...

As y(gamma?) is set to zero, dosen't this turn into

T= -Tc-6K ?

Also if someone could walk me through the formula translation, that would be nice.
I always end up with

[math]
T=-\frac{y*V^\frac{2}{3}} k-Tc + 6K
[/math]

All input appreciated.

 

[Fizyk]

Now that I check it, I get:
[math]T = -\frac{\gamma V^{\frac{2}{3}}}{k} + T_c - 6K[/math]
Most of us are definitely wrong :lol: (Also, yeah, looks like it was gamma).

Start from:
[math]\gamma V^{\frac{2}{3}} = k(T_c - 6 K - T)[/math]

Divide by k:
[math]\frac{\gamma V^{\frac{2}{3}}}{k} = T_c - 6K - T[/math]

Subtract the left side from both sides and add T:
[math]T = T_c - 6K - \frac{\gamma V^{\frac{2}{3}}}{k}[/math]

BTW, we are orbinauts, we should know maths, and look how many of us are needed to transform a simple formula :facepalm:

 

[fireballs619]

We are given the values to evaluate this equation now.

γ = 0
M = 0.001367 kg/mol
p = 3.078 * 10^-17 mol/m^3
V = M/p
k = 2.1×10^−7 J/(K·mol−2/3)
Tc = 33.67 K

Are we doing this, or leaving it for EndeavourCmdr to solve?

 

[Notebook]

Thanks Fizyk, I'm working on it!

To fireballs619:

harder stuff is rationalising those units

Subtract the left side from both sides and add T:
Ah, a cunning plan, I'll talk to Baldrick/

N.

 

[Notebook]

Not having much succes with the units:
http://i89.photobucket.com/albums/k207/Notebook_04/EotvosRule_2_.jpg

Gettin a "Units miss-match" error message for the red text. Not sure if this because the last term goes to zero? or something else.

About as far as I can go with this.

N.