On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept?

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I am playing around with data from exoplanets.org, and am interested in the plot of surface gravity to planet mass. I reproduced this plot after downloading their data and performed a non-linear regression model to fit the curve. Unfortunately,the covariance matrix of my fit has infinite values, so I am now trying out a linear fit on a log-log plot, shown below. My fit, for $y=ax+b$, is $a=0.9511$, $b=0.8631$.

I am now thinking about what I am plotting. I suspect there may not be anything interesting in $log frac{GM}{R^2}$ vs. $log M$, but, regardless, I am trying to understand if there is any meaning in the y-intercept.

What are possible explanations for the value $b$?

I think what you have established here is just that $ ho$ tends to increase with mass. The density of planets isn't constant.

Let $ ho = ho_0 (M/M_{earth})^{alpha}$, so that $M = (4/3)pi R^{3} ho_0 (M/M_{earth})^{alpha}$

Then $$g = frac{GM}{R^2} = frac{4pi G}{3} R ho$$

Replace $R$ with $(3M/4pi ho)^{1/3}$ so that $$ g = frac{4pi G}{3} left(frac{3M}{4pi ho} ight)^{1/3} ho$$ $$ g = left(frac{4pi}{3} ight)^{2/3} G M^{1/3} ho_0^{2/3} (M/M_{earth})^{2alpha/3}$$ $$ g = left(frac{4pi}{3} ight)^{2/3} GM_{earth}^{1/3} ho_0^{2/3} (M/M_{earth})^{(2alpha+1)/3}$$

So, bar a (highly possible) algebraic slip, if you plot $log g$ vs $log M$, the gradient is $(2alpha+1)/3$, which from your plot, gives $alpha simeq 0.92$ - i.e. the average planet density increases almost linearly with mass.

The intercept then is $$ b = log left[ left(frac{4pi}{3} ight)^{2/3} GM_{earth}^{1/3} ho_0^{2/3} ight],$$ which yields $ ho_0 simeq 3.5$ kg/m$^3$ (NB: I subtracted 2 from your $b$ to make it SI; giving a density of around 814 kg/m$^3$ at a Jupiter mass).

The fact that density is almost proportional to mass can be found from the same dataset. e.g. see below. Below 0.1 Jupiter masses, the relationship appears to break down, though in actual fact very few of the densities for such planets are very accurately measured (since it requires a radius from a transit), but it works well enough in the range you have plotted. The physics here is that gas giants are governed by a partially (electron) degenerate equation of state that results in them all having a similar radius from about a tenth of a Jupiter mass to about 50 Jupiter masses (albeit with considerable and largely unexplained scatter). Thus the density is proportional to mass. This relationship does not work for small rocky planets, where the radius decreases for smaller masses. Thus your nice line in log space does not continue to lower masses (see plot at bottom - NB: some of the low-mass points have huge uncertainties).

For planets of constant mean density you have: $$ M= ho imes 4pi r^3 $$ and the surface value of $g$ is: $$ g(r)=frac{GM}{r^2}=G imes ho imes 4 pi imes r $$ So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r o 0$ tells you the density. So for bodies of equal density $log(g(r)) o -infty$ as $r o 0$

In terms of mass: $$ g(M)={G(4pi ho)^{2/3}M^{1/3}} $$ So as $M o 0$ we have $g(M) o 0$ and again $log(g(M)) o -infty$ and the intercept of the $log-log$ plot as $M$ goes to zero alows you to calculate the mean density.

What is the geoid?

A depiction of the United States geoid. Areas in yellow and orange have a slightly stronger gravity field as a result of the Rocky Mountains.

Print Your Own Geoid

Please note that this model includes a wide, flat base that makes it easier to print on most home 3D printers. The base also serves as a stand to display the geoid model.

While we often think of the earth as a sphere, our planet is actually very bumpy and irregular.

The radius at the equator is larger than at the poles due to the long-term effects of the earth's rotation. And, at a smaller scale, there is topography&mdashmountains have more mass than a valley and thus the pull of gravity is regionally stronger near mountains.

All of these large and small variations to the size, shape, and mass distribution of the earth cause slight variations in the acceleration of gravity (or the "strength" of gravity's pull). These variations determine the shape of the planet's liquid environment.

If one were to remove the tides and currents from the ocean, it would settle onto a smoothly undulating shape (rising where gravity is high, sinking where gravity is low).

This irregular shape is called "the geoid," a surface which defines zero elevation. Using complex math and gravity readings on land, surveyors extend this imaginary line through the continents. This model is used to measure surface elevations with a high degree of accuracy.

Did you know?

Scientists at the National Geodetic Survey are working on a 15-year-long project to make accurate height measurement better, faster, and cheaper. This project is called the Gravity for the Redefinition of the American Vertical Datum, or GRAV-D for short. When it’s done in in 2022, you will be able to get accurate heights to within about an inch for most locations around the nation.

The Hooke’s Law Formula

The spring constant is a key part of Hooke’s law, so to understand the constant, you first need to know what Hooke’s law is and what it says. The good news it’s a simple law, describing a linear relationship and having the form of a basic straight-line equation. The formula for Hooke’s law specifically relates the change in extension of the spring, x, to the restoring force, F, generated in it:

The extra term, k, is the spring constant. The value of this constant depends on the qualities of the specific spring, and this can be directly derived from the properties of the spring if needed. However, in many cases – especially in introductory physics classes – you’ll simply be given a value for the spring constant so you can go ahead and solve the problem at hand. It’s also possible to directly calculate the spring constant using Hooke’s law, provided you know the extension and magnitude of the force.

Answers and Replies

Okay this is the last question of the lab and I don't get it
Basically its a lab where we find how amplitude, bob mass, and length affects the period T for each time.

Our finding was that length affects the period a lot. I did a bunch of graphs and tables and answered all the questions but don't know how I derive this

the question is "From your graph of T^2 vs. L
determine a value for g."

T here is the period in seconds and L is the length of the string in cm. Now I do have to graph and it looks pretty normal to me.

The slope is 0.03900 and the r is 0.9947
now I dont think this would help much at all
(unless its r-slope haha. )

I have to find g from this graph and calculate the percent of error
I know how to calculate the percent of error but I don't know how I can find g from this graph

the points are for x which is length

My first guess was i need to find an equation for this relationship and then
use the formula 4pi^2/g x L = T^2 but I don't know what to do?

What is the Gravity on Venus?

Venus is the virtual twin of Earth in many ways. Similar size, mass and density. But what is the gravity on Venus? According to our friends over at NASA, the answer is 8.87 m/s 2 . To translate that a little more, it is about 90% of the gravity here on Earth. A person who measures 100 kg when they leave home would tip the scales on the Venusian surface at 90 kg.

The surface gravity of Venus is not the only characteristic of the planet that nearly mirrors Earth. Venus has 86% of the volume that Earth has along with 82% of the mass. The planet’s density is nearly identical at 5.243 g/cm 3 .

In order to shed that ten kilos you would have to spend a couple of months in space. Once you arrived the real trouble would begin. Science has not been able to develop a spacesuit that could survive more than a few minutes in the harsh environment of Venus. To start with there is the 470C surface temperature. That is 9 times the temperatures in the hottest deserts here on Earth. The heat would not destroy your suit though. The atmosphere is 96% carbon dioxide and full of sulfuric acid clouds and droplets and ash from the volcanoes that dot the surface. The atmosphere is so thick that most meteors could not penetrate it, burning up before impact instead.

While there have been many large volcanoes here on Earth, there is no real comparison to the number, size, and extent of the volcanic activity on Venus. The Venusian surface is dominated by the more than 1,000 volcanoes or volcanic centers that are larger than 20 km. Lava flows are thought to have completely resurfaced the planet between 300 and 500 million years ago.

The reflective nature of the sulfuric acid in the atmosphere has made visual observation of the surface impossible. It was early in the 20th century, when astronomers were able to make spectroscopic, ultraviolet, and radar observations, before much was known about the planet. Surface features went undetected until radar observations were made in the 1970s.

Fifty years ago no one could have accurately told you much about Venus gravity. It was still a mystery at the beginning of the 20th century. In many ways it can be considered the Earth’s near twin, but the planet is still a host of mysteries that need to be solved. The Venus Express spacecraft has contributed a great deal of data. BepiColumbo and Akatsuki may be able to add a great deal more in 2014 and 2016, respectively. All we can do is wait and see.

We have written many articles about Venus for Universe Today. Here are some interesting facts about Venus, and here’s an article about the color of Venus.

We’ve also recorded an entire episode of Astronomy Cast all about Venus. Listen here, Episode 50: Venus.

Advances in Agronomy

4.1 Sheindorf–Rebhun–Sheintuch Model

The Sheindorf–Rebhun–Sheintuch (SRS) equation has been developed to describe competitive or multicomponent sorption where it is assumed that the single-component sorption follows the Freundlich equation ( Sheindorf et al., 1981 ). The derivation of SRS equation was based on the assumption of an exponential distribution of adsorption energies for each component. Specifically, the SRS model was developed to describe competitive equilibrium sorption for multicomponent systems where the sorption isotherms of single component follow the Freundlich equation. A general form of the SRS equation can be written as

where the subscripts i and j denote the metal components i and j, l is the total number of components, and α_i,j is the dimensionless competition coefficient for the adsorption of component i in the presence of component j. The parameters K_i and n_i are the Freundlich parameters representing a single-component system i as described in Eqn (1) above. By definition, α_i,j equals 1 when i = j. If there is no competition, i.e. α_i,j = 0 for all j ≠ i , Eqn (2) yields a single-species Freundlich equation for component i identical to Eqn (1) . The estimated α_Ni–Cd for Ni adsorption, in the presence of Cd, was larger than 1 for Windsor and Olivier soils, indicating noticeable decrease of Ni in the presence of Cd. In contrast, α_Ni–Cd for Ni adsorption on Webster soil was <1, which is indicative of small influence of competing Cd ions. These results are in agreement with the competitive sorption reported by Antoniadis and Tsadilas (2007) . Such small α_Ni–Cd implies that Ni adsorption in Webster soil was least affected in a competitive Ni–Cd system in comparison to the other two soils. Moreover, the estimated α_Cd–Ni for Cd adsorption was 0.61 for Windsor and 0.82 for Olivier, whereas the competitive coefficient of Cd/Ni was 4.00 for Webster Soil. Although the SRS equation may be regarded as a multicomponent model and does not imply certain reaction mechanisms, differences of competitive sorption between the neutral and the two acidic soils were illustrated based on the SRS models’ competitive selectivity parameters. In fact, Roy et al. (1986) suggested that the SRS parameters could be used to describe the degree of the competition under specific experimental conditions. Calculated results using the estimated α_Ni–Cd are given in Figs 5.8 and 9 and illustrate the capability of the SRS model in describing experimental data for competitive adsorption of Ni and Cd ( Liao and Selim, 2009 ). An F-test indicated that there was no statistical difference between our experimental results and SRS model calculations (at the 95% confidence level). Based on these calculations, the SRS model was capable of quantifying competitive adsorption for Ni and Cd. However, for both Ni and Cd, the SRS model deviated considerably from experimental data for high concentrations of the competing ions. This finding is consistent with the application of SRS made earlier by Gutierrez and Fuentes (1993) and illustrates the need for model improvement to better describe competitive adsorption of heavy metals over the entire range of concentrations.

Figure 5.8 . Competitive adsorption isotherms for Ni in the presence of different concentrations of Cd. Solid curves are SRS model calculations.

Figure 5.9 . Competitive adsorption isotherms for Cd in the presence of different concentrations of Ni. Solid curves are SRS model calculations.

The suitability of the multicomponent SRS equation for describing the competitive adsorption isotherms of trace elements on soil and soil minerals have been investigated by several researchers. A general procedure of applying the SRS equation is first to obtain the Freundlich distribution coefficient K_F and reaction exponent b or n by fitting the single-component isotherms to Freundlich equation, followed by estimating the competition coefficients α_i,j through fitting the experimental isotherms of binary and ternary mixtures to the SRS equation ( Roy et al., 1986 ). Although the SRS equation does not imply specific reaction mechanisms, the competition coefficients α_i,j in the equation can be used to evaluate the relative selectivity of the sorbent to the heavy metal species. It is demonstrated that SRS equation with competition coefficients estimated through nonlinear least-squares optimization successfully described the experimental competitive adsorption isotherms of Ni and Cd on three different soils ( Liao and Selim, 2009 ). Gutierrez and Fuentes (1993) employed the SRS equation to represent the competitive adsorption of Sr, Cs and Co in Ca-montmorillonite suspensions. They found that the SRS competition coefficients α_i,j obtained from experiment data of binary mixtures successfully predicted the competitive adsorption of ternary mixture Sr–Cs–Co. Similarly, Bibak (1997) found that values of SRS competitive coefficients obtained from binary sorption experiments successfully predicted sorption data of the ternary solute mixture Cu–Ni–Zn. The SRS equation was successfully used to describe for competitive sorption of Cd, Ni, and Zn on a clay soil by Antoniadis and Tsadilas (2007) . In addition, SRS equation was also used by Wu et al. (2002) in representing the competitive adsorption of molybdate, sulfate, selenate, and selenite on γ-Al₂O₃ surface where relative affinity coefficient was used instead of competitive coefficients. The relative affinity coefficients were calculated as the ratios of the proton coefficients of competing anions. The simulation result showed that the sorption affinity of anions on γ-Al₂O₃ surface decreased in the order of MoO₄ 2− > SeO₃ 2− > SeO₄ 2− > SO₄ 2− .

On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept? - Astronomy

The center of gravity is a geometric property of any object. The center of gravity is the average location of the weight of an object. We can completely describe the motion of any object through space in terms of the translation of the center of gravity of the object from one place to another, and the rotation of the object about its center of gravity if it is free to rotate. If the object is confined to rotate about some other point, like a hinge, we can still describe its motion. In flight, both airplanes and rockets rotate about their centers of gravity. A kite, on the other hand, rotates about the bridle point. But the trim of a kite still depends on the location of the center of gravity relative to the bridle point, because for every object the weight always acts through the center of gravity.

Determining the center of gravity is very important for any flying object. How do engineers determine the location of the center of gravity for an aircraft which they are designing?

In general, determining the center of gravity (cg) is a complicated procedure because the mass (and weight) may not be uniformly distributed throughout the object. The general case requires the use of calculus which we will discuss at the bottom of this page. If the mass is uniformly distributed, the problem is greatly simplified. If the object has a line (or plane) of symmetry, the cg lies on the line of symmetry. For a solid block of uniform material, the center of gravity is simply at the average location of the physical dimensions. (For a rectangular block, 50 X 20 X 10, the center of gravity is at the point (25,10, 5) ). For a triangle of height h, the cg is at h/3, and for a semi-circle of radius r, the cg is at (4*r/(3*pi)) where pi is ratio of the circumference of the circle to the diameter. There are tables of the location of the center of gravity for many simple shapes in math and science books. The tables were generated by using the equation from calculus shown on the slide.

For a general shaped object, there is a simple mechanical way to determine the center of gravity:

If we just balance the object using a string or an edge, the point at which the object is balanced is the center of gravity. (Just like balancing a pencil on your finger!)
Another, more complicated way, is a two step method shown on the slide. In Step 1, you hang the object from any point and you drop a weighted string from the same point. Draw a line on the object along the string. For Step 2, repeat the procedure from another point on the object You now have two lines drawn on the object which intersect. The center of gravity is the point where the lines intersect. This procedure works well for irregularly shaped objects that are hard to balance.

If the mass of the object is not uniformly distributed, we must use calculus to determine center of gravity. We will use the symbol S dw to denote the integration of a continuous function with respect to weight. Then the center of gravity can be determined from:

where x is the distance from a reference line, dw is an increment of weight, and W is the total weight of the object. To evaluate the right side, we have to determine how the weight varies geometrically. From the weight equation, we know that:

where m is the mass of the object, and g is the gravitational constant. In turn, the mass m of any object is equal to the density, rho, of the object times the volume, V:

We can combine the last two equations:

If we have a functional form for the mass distribution, we can solve the equation for the center of gravity:

cg * W = g * SSS x * rho(x,y,z) dx dy dz

where SSS indicates a triple integral over dx. dy. and dz. If we don't know the functional form of the mass distribution, we can numerically integrate the equation using a spreadsheet. Divide the distance into a number of small volume segments and determining the average value of the weight/volume (density times gravity) over that small segment. Taking the sum of the average value of the weight/volume times the distance times the volume segment divided by the weight will produce the center of gravity.

You can view a short movie of "Orville and Wilbur Wright" explaining how the center of gravity affected the flight of their aircraft. The movie file can be saved to your computer and viewed as a Podcast on your podcast player.

Adhesion of atoms, ions, bimolecules or molecules of gas, liquid or dissolved solids to a surface is called adsorption. This process creates a film of the adsorbate –the molecules or atoms being accumulated, on the surface of the adsorbent.

Examples:

Activated charcoal adsorbs gases like CO₂ , SO₂, Cl₂ etc.
Pt or Ni metal kept in contact with a gas adsorbs the gas - Hydrogenation of oils.
Animal charcoal, when added to acetic acid solution and shaken vigorously, adsorbs acetic acid.
Molasses is decolourised by activated charcoal.

The molecules of gases or liquids or the solutes in solutions adher to the surface of the solids. In adsorption process, two substances are involved. One is the solid or the liquid on which adsorption occurs and it is called adsorbent. The second is the adsorbate, which is the gas or liquid or the solute from a solution which gets adsorbed on the surface.

Adsorbent: The substance on whose surface the adsorption occurs is known as adsorbent.

Adsorbate: The substance whose molecules get adsorbed on the surface of the adsorbent ( i.e. solid or liquid ) is known as adsorbate.

Adsorption is different from absorption. In absorption, the molecules of a substance are uniformly distributed in the bulk of the other, whereas in adsorption molecules of one substance are present in higher concentration on the surface of the other substance.

Types of adsorption:

Depending upon the nature of forces existing between adsorbate molecules and adsorbent, the adsorption can be classified into two types:

1. Physical adsorption (physisorption): If the force of attraction existing between adsorbate and adsorbent are Vander Waal’s forces, the adsorption is called physical adsorption. It is also known as Vander Waal’s adsorption. In physical adsorption the force of attraction between the adsorbate and adsorbent are very weak, therefore this type of adsorption can be easily reversed by heating or by decreasing the pressure.

2. Chemical adsorption (chemisorption): If the force of attraction existing between adsorbate and adsorbent are almost same strength as chemical bonds, the adsorption is called chemical adsorption. It is also known as Langmuir adsorption. In chemisorption the force of attraction is very strong, therefore adsorption cannot be easily reversed.

Comparison between Physisorption and Chemisorption

Low heat of adsorption usually in the range of 20-40 kJ mol -1

Force of attraction are Van der Waal's forces

It usually takes place at low temperature and decreases with increasing temperature

It is reversible

It is related to the ease of liquefaction of the gas

It is not very specific

It forms multi-molecular layers

It does not require any activation energy

Factors affecting adsorption:

The extent of adsorption depends upon the following factors:

Nature of adsorbate and adsorbent.
The surface area of adsorbent.
Activation of adsorbent.
Experimental conditions. E.g., temperature, pressure, etc.

Adsorption Isotherm:

Adsorption process is usually studied through graphs known as adsorption isotherm. That is the amount of adsorbate on the adsorbent as a function if its pressure or concentration at constant temperature .The quantity adsorbed is nearly always normalized by the mass of the adsorbent to allow comparison of different materials.

Basic Adsorption Isotherm:

From the above we can predict that after saturation pressure P_s, adsorption does not occur anymore, that is there are limited numbers of vacancies on the surface of the adsorbent. At high pressure a stage is reached when all the sites are occupied and further increase in pressure does not cause any difference in adsorption process. At high pressure, Adsorption is independent of pressure.

Type of Adsorption Isotherm:

Five different types of adsorption isotherm and their characteristics are explained below.

Type I Adsorption Isotherm:

Type I Adsorption Isotherm

The above graph depicts Monolayer adsorption.
This graph can be easily explained using Langmuir Adsorption Isotherm.
If BET equation, when P/P₀<<1 and c>>1, then it leads to monolayer formation and Type I Adsorption Isotherm is obtained.
Examples of Type-I adsorption are Adsorption of Nitrogen (N₂) or Hydrogen (H) on charcoal at temperature near to -1800°C.

Type II Adsorption Isotherm:

Type II Adsorption Isotherm

Type II Adsorption Isotherm shows large deviation from Langmuir model of adsorption.
The intermediate flat region in the isotherm corresponds to monolayer formation.
In BET equation, value of C has to be very large in comparison to 1.
Examples of Type-II adsorption are Nitrogen (N₂ (g)) adsorbed at -1950°C on Iron (Fe) catalyst and Nitrogen (N₂ (g)) adsorbed at -1950°C on silica gel.

Type III Adsorption Isotherm:

Type III Adsorption Isotherm

Type III Adsorption Isotherm also shows large deviation from Langmuir model.
In BET equation value if C <<< 1 Type III Adsorption Isotherm obtained.
This isotherm explains the formation of multilayer.
There is no flattish portion in the curve which indicates that monolayer formation is missing.
Examples of Type III Adsorption Isotherm are Bromine (Br₂) at 790°C on silica gel or Iodine (I₂) at 790°C on silica gel.

Type IV Adsorption Isotherm:

Type IV Adsorption Isotherm

At lower pressure region of graph is quite similar to Type II. This explains formation of monolayer followed by multilayer.
The intermediate flat region in the isotherm corresponds to monolayer formation.
The saturation level reaches at a pressure below the saturation vapor pressure. This can be explained on the basis of a possibility of gases getting condensed in the tiny capillary pores of adsorbent at pressure below the saturation pressure (P_S) of the gas.
Examples of Type IV Adsorption Isotherm are of adsorption of Benzene on Iron Oxide (Fe₂O₃) at 500°C and adsorption of Benzene on silica gel at 500°C.

Type V Adsorption Isotherm:

Type V Adsorption Isotherm

Explanation of Type V graph is similar to Type IV.
Example of Type V Adsorption Isotherm is adsorption of Water (vapors) at 1000°C on charcoal.
Type IV and V shows phenomenon of capillary condensation of gas.

Freundlich Adsorption Isotherm:

In 1909, Freundlich expressed an empirical equation for representing the isothermal variation of adsorption of a quantity of gas adsorbed by unit mass of solid adsorbent with pressure. This equation is known as Freundlich Adsorption Isotherm or Freundlich Adsorption equation or simply Freundlich Isotherm.

x/m = adsorption per gram of adsorbent which is obtained be dividing the amount of adsorbate (x) by the weight of the adsorbent (m).

P is Pressure, k and n are constants whose values depend upon adsorbent and gas at particular temperature .

Though Freundlich Isotherm correctly established the relationship of adsorption with pressure at lower values, it failed to predict value of adsorption at higher pressure. This relation is called as the freundlich adsorption isotherm. As see the following diagram. The value of x/m is increasing with increase in p but as n>1 it does not increase suddenly. This curve is also called the freundlich isotherm curve.

Taking the logarithms of a first equation.

Hence, if a graph of log x/m is plotted against log p, it will be a straight line in the following diagram.

From this the value of slope equal to 1/n and the value of intercept equal to log k can be obtained. Over and above, it the graph of log x/m against log p comes out to be a straight line, it can be assured that the freundlich adsorption isotherm is satisfied for this system.

Langmuir Adsorption Isotherm:

In 1916, Irving Langmuir published a new model isotherm for gases adsorbed to solids, which retained his name. It is a semi-empirical isotherm derived from a proposed kinetic mechanism. This isotherm was based on different assumptions one of which is that dynamic equilibrium exists between adsorbed gaseous molecules and the free gaseous molecules.

It is based on four assumptions:

The surface of the adsorbent is uniform, that is, all the adsorption sites are equivalent.
Adsorbed molecules do not interact.
All adsorption occurs through the same mechanism.
At the maximum adsorption, only a monolayer is formed: molecules of adsorbate do not deposit on other, already adsorbed, molecules of adsorbate, only on the free surface of the adsorbent.

Langmuir suggested that adsorption takes place through this mechanism:

A(g) = unadsorbed gaseous molecule

B(s) = unoccupied metal surface

AB = Adsorbed gaseous molecule.

The direct and inverse rate constants are k and k -1

Based on his theory, Langmuir derived an equation which explained the relationship between the number of active sites of the surface undergoing adsorption and pressure. This equation is called Langmuir Equation.

θ= the number of sites of the surface which are covered with gaseous molecule,

K =is the equilibrium constant for distribution of adsorbate between the surface and the gas phase .

The basic limitation of Langmuir adsorption equation is that it is valid at low pressure only.

At lower pressure, KP is so small, that factor (1+KP) in denominator can almost be ignored. So Langmuir equation reduces to

At high pressure KP is so large, that factor (1+KP) in denominator is nearly equal to KP. So Langmuir equation reduces to

Adsorbents:

The material upon whose surface the adsorption takes place is called an adsorbent .Activated carbon is used as an adsorbent

Adsorbents are used usually in the form of spherical pellets, rods, moldings, or monoliths with hydrodynamic diameters between 0.5 and 10 mm.
They must have high abrasion resistance, high thermal stability and small pore diameters, which results in higher exposed surface area and hence high surface capacity for adsorption.
The adsorbents must also have a distinct pore structure which enables fast transport of the gaseous vapors.

Most industrial adsorbents fall into one of three classes:

Oxygen-containing compounds - Are typically hydrophilic and polar, including materials such as silica gel and zeolites.
Carbon-based compounds - Are typically hydrophobic and non-polar, including materials such as activated carbon and graphite.
Polymer-based compounds - Are polar or non-polar functional groups in a porous polymer matrix.

Activated carbon is used for adsorption of organic substances and non-polar adsorbates and it is also usually used for waste gas (and waste water) treatment. It is the most widely used adsorbent since most of its chemical (eg. surface groups) and physical properties (eg. pore size distribution and surface area) can be tuned according to what is needed. Its usefulness also derives from its large micropore (and sometimes mesopore) volume and the resulting high surface area.

Mechanism of Adsorption Using Adsorbent:

Applications of adsorption:

The principle of adsorption is employed,

in heterogeneous catalysis.
in gas masks where activated charcoal adsorbs poisonous gases.
in the refining of petroleum and decolouring cane juice.
in creating vacuum by adsorbing gases on activated charcoal.
in chromatography to separate the constituents' of a mixture.
to control humidity by the adsorption of moisture on silica gel.
in certain titrations to determinate the end point using an adsorbent as indicator (Example: Flouroscein).

Procedure of adsorption:

500 ml of 0.5N oxalic acid solution is prepared. Five well cleaned, dried, reaction bottles (250 ml) are taken and are labeled. About 2 g of the activated animal charcoal are accurately weighed and transferred carefully into each of the bottles. By means of a burette 50, 40, 30, 20 and 10 ml of 0.5N oxalic acid are added followed by 0, 10, 20, 30 and 40 ml of distilled water so that the total volume (50 ml) remains constant in each bottle. These bottles were shaken thoroughly nearly for an hour by means of a mechanical shaker and they are set aside in a trough containing water to reach equilibrium.

The supernatant liquid of each of the bottles are filtered through a small dry filter paper and the filtrate is collected in properly labeled conical flasks.(The initial 5 ml or 10 ml of the filtrate is rejected) 10 ml of the filtrate is pipetted out into a clean conical flask. It is titrated against standardized KMnO4 solution until a pink colour appears. The titration is repeated to get concordant values. From the titre values, the concentration of oxalic acid remaining and hence the amount of oxalic acid adsorbed are calculated.

In order to test the validity of Freundlich adsorption isotherm plot log (x/m) against log Ce. The slopes and intercepts of the plot will give 1/n and log k respectively and hence n and K can be calculated.

Validity of Langmuir adsorption equation can be tested by plotting Ce/(x/m) Vs Ce. A linear plot obtained shoe the applicability of the isotherm. Calculate the constants "a" and "b" from the slope and intercept on the ordinate axis.

Answer Desk

Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, . the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2pisqrt<>>>$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $mu_mathrm$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.

Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.

The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.

Mass,Weight and, Density

I Words: Most people hardly think that there is a difference between "weight" and "mass" and it wasn't until we started our exploration of space that is was possible for the average person to experience, even indirectly, what it must mean to be "weightless". Everyone has been confused over the difference between "weight" and "density". Which of us hasn't fallen for the old riddle: What weighs more, a pound of lead or a pound of feathers? Now that Astronauts regularly seem to be demonstrating "weightlessness", ( yet we know they still have all of the matter or mass they had when they were on earth,) more of us are beginning appreciate the weight - mass confusion. Like many confusing concepts, after they are finally understood, they are still difficult to explain to others who don't understand. We hope we can explain the difference between mass, weight and density so clearly that you will have no trouble explaining the difference to your students.

Mass: This concept is so basic that, like length and time, it is really impossible to define. Isaac Newton called mass the quantity of matter . We can talk all around it but we will finally have to admit that our words fail. Some say mass is the amount of matter in something (and hope that no one asks: What is matter?). Others say mass is the measure of an object's inertia (which assumes we understand the elusive property of inertia). To add to the confusion, mass is related to an object's inertia but it also is related to how hard objects are attracted to the earth. Better minds than ours have been confused over the meaning of the concept "mass" and even today, better minds than ours contemplate what mass really means. Our way of giving up on the impossible task of defining mass is to say: mass is the measure of the amount of "stuff" in something. This definition is properly confusing and you can work on the meaning of "stuff"! In the metric system mass is measured in kilograms and grams and these will be the units we will most often use. (In the United States today, almost no one knows what the unit of mass is called--it's not the pound. The pound is a unit of weight--more about weight in the next paragraph. The more mass something has, the harder it is to move or, the more sluggish it is. The correct US unit of mass is called the "slug"--short for sluggishness-- but, as we said, almost no one uses this unit today.)

Weight: If you can finally accept the concept mass even if we have been unable to define it, weight is easy: The weight of a mass is the force that the earth pulls on the mass. We hope you have a feeling for what force means (and we will discuss it later). The entire idea of weight can be understood as the force of gravity on something. Usually we spend most of our time on Earth so our weight is the force that the earth pulls on us. If we get further away from the earth, the force the earth pulls on us is less and we weigh less. If you lived on Mars, the above definition would probably change to: "The weight of a mass is the force that Mars pulls on the mass." The whole idea of weight is related to the force of gravity (and we hate just to use the word "gravity" since it can bring up even more confusion). It would be correct to say, no matter where you might be in the universe that "the weight of a mass is the force of gravity on the mass." In the metric system force is measured in newtons hence weight is also measured in newtons. You will learn later that on the surface of the earth, a mass of 1 kilogram weighs 9.8 newtons. (You will probably never learn anywhere that on the surface of the earth, one slug weighs 32.2 pounds--don't worry about it, very few people know this!) The pound is the US unit of force hence the US unit of weight is also the pound. We will use newtons for the unit of force (and weight) almost always in the discussions that follow.

Density: There are two kinds of density, "weight density" and "mass density". We will only use mass density and when we say: "density", we will mean "mass density". Density is mass per volume. Lead is dense, Styrofoam is not. The metric system was designed so that water will have a density of one gram per cubic centimeter or 1000 kilograms per cubic meter. Lead is about 10 times as dense as water and Styrofoam is about one tenth as dense as water.

II Purpose of the Activity:

The purpose of this activity is to investigate the meaning of mass, weight and density by looking at how each might be measured.

III Materials required for the Activity:

At least one box of #1 (small) paper clips, 20 (or more) long thin rubber bands (#19 will work--they are 1/16" thick and 3+" long), drinking straws, a fine tipped marking pen (Sharpie), scotch tape, 40 (or more) 1oz or 2oz plastic portion cups (Dixie sells them in boxes of 800 for less than $10--see if your school cafeteria has them), lots of pennies (to use as "weights"), light string, 20 (or more) specially drilled wooden rulers or cut sections of wooden molding, about a pound or two of each of the following as available: sand, rice, sawdust, fine crushed Styrofoam (place Styrofoam packing or cups in a plastic bag and pound--yes, this can make a mess), lead shot (can this be used with children? Lead shot can be obtained at a gun and ammo shop), any other finely ground solid material which can be used to illustrate a variety of different densities. (We have avoided liquids since they seem to make a bigger mess.)

IV What the teacher must do in advance of the activity:

We feel the most difficult preparation item for this activity will be making the drilled ruler or preparing sections of wood properly drilled. (Is it possible to assign tasks like this to parents who have better shop facilities at home?) These will become the balance beams for our mass balance. The actual dimensions of the balance beam is not too critical and a wooden ruler is about right--the best are old ones that have lost the metal strip they often have along one edge. What follows is a description and illustration of how the balance beam should be constructed:

The center hole should be exactly in the center so that when the beam is supported by a nail through this center hole, it will spin around freely. The two end holes should be in line with the center hole and equidistant on each side. Two other holes should be drilled on a line directly above the center hole. The hole size is not critical, about 1/8 of an inch is fine. (We sincerely hope it will not be too difficult for you to accomplish this. We fooled around with metal hangers, etc. but nothing simple seemed to work as well as a carefully prepared stick.)

It will be necessary to poke holes in the portion cups. Should you do this in advance or can your students do it? A small nail works well for this purpose. Probably time could be saved in class if the necessary strings were cut to length in advance (and perhaps even tied to the cups).

Building the "Weight Scale" requires some careful cutting of a straw that can be done with a good pair of scissors or a sharp knife. We think kids can do all of it but it will take time. You should build one prototype Weight Scale in advance so you can work out the details of construction and decide how much of the cutting should be done in advance.

Teaching outline and Presentation suggestions:

(Once again we remind you that we really don't know the best way to teach these concepts to young students. The following are suggestions for construction and use of the equipment and how we envisioned one might use this stuff but only you can know the best way to present this material to your students.)

Measuring mass: Mass is usually measured with a balance. The idea is to compare the unknown object with the mass of a known amount. Illustrated below is the device we will use to measure mass and we will call it "the Mass Balance".

Since everyone seems to have lots of pennies and all pennies are about the same mass, we will use the penny as our standard of mass. (It turns out that the average penny has a mass of about 2.6 grams and you can convert to grams if you wish but for now, we will simply determine mass in "pennies".) The mass measurement is accomplished simply by placing the unknown object in one cup of the Mass Balance and finding out how many pennies placed on the other side it takes to achieve balance. You should first check the Mass Balance with nothing in either cup to see if it is properly "zeroed". You should notice that the balance is most sensitive when the upper paper clip is in the center hole (in fact it is really too sensitive here) and it will be less sensitive when you use the higher holes. Slight errors in the zero reading can be corrected by using shorter or longer string sections on the appropriate side. Make sure all paper clips rotate freely in the drilled holes. The balance will not work properly if the paper clips hang up. The Mass Balance can be loaded with the cups on the table and pulling upward slightly on the support paper clip will test the balance condition. We suggest that students begin by matching pennies on the left with pennies on the right (and they should discover that all pennies aren't really the same--this is real!) After the students become familiar with the use of the balance, we suggest that nearly equal volumes of the assorted materials (sand, rice, metal shot, Styrofoam) be measured. If you are using the 1 oz Dixie portion cups, it is possible to draw a line on the cup 1.4 cm above the bottom and it will represent 10 cubic centimeters and a line 2.3 cm above the bottom of the cup will represent 20 cubic centimeters (or milliliters).

A very important question to consider now is: If you used this Mass Balance on the moon or on Mars, would the same amount of material on one side require the same number of pennies on the other side to balance it as it did on Earth? Naturally there is no easy way for us to perform such an experiment but, having your students think about this should help them to start understanding the difference between weight and mass. Mass or as Newton would say, the quantity of matter in an object, does not change when you change your location in space but, as we will see shortly, weight does change.

Measuring weight: Using a carefully segmented straw, a bent paper clip, a rubber band, some string, a small cup, a 3X5 card and some scotch tape we will construct the "Weight Scale" shown below:

( Since we had difficulty in joining the segment of rubber band to the string, we decided to show you a "carrick bend" which works quite well for this situation. A simple slip knot works well on the bottom where the rubber band attaches to the bent paper clip bail. A later construction detail diagram will give you a better illustration of the Weight Scale.)

The students will calibrate this Weight Scale with pennies and mark the 3 X 5 card with a marking pen during the calibration exercise. Attaching the rubber band to the bail of the cup is easily accomplished with a slip knot but attaching the string to the rubber band is a slight problem--a suggested knot is shown with the illustration. The whole idea is to have the zero of the scale at the bottom of the card using the string-rubber band junction as the pointer. With about 25 pennies in the cup, the rubber band will stretch to about the top of the card. (You hold the scale with the string which has been passed through a small piece of straw taped to the card.) The students will carefully load the cup with pennies and mark the card at about 5 penny intervals.

A more detailed construction of the Weight Scale is shown "below." (Again, we suggest that you construct one in advance so you can evaluate how difficult it is to build.) Note that in the construction diagram "below", we show how the straw should be sectioned so that it can be attached to the 3 X 5 card. (In the final scale, naturally, the string and rubber band fit inside of the straw sections.) It is important that the lower length of the straw be made just long enough to extend from the top of the paper clip "bail" to the bottom of the card with the rubber band sticking out the top. You must be able to tie the rubber band to the string and have the junction of the two be on the lower end of the 3 X 5 with nothing in the cup.

This is the "below" referred to in the above paragraph. We decided it would take a large image to show the necessary details so, if you have the time, click here for Weight Scale Construction Details.

After the students have calibrated the Weight Scale, it might be fun to have them see if they can guess how many pennies have been loaded into their cup by another student. From this they will learn how to read between the marks they have placed on their cards (this is called interpolation) and they will also learn that the scale is really not too accurate. However, all instruments are less than perfect at some level and this crude scale should help them to realize this fact. (We think it is nice that the scale is quite inexpensive and students who wish can construct one at home.)

This Weight Scale can also be used to measure some of the other materials that were measured with the Mass Balance. Hopefully they will find that they will get pretty close to the same answer in "pennies" for the mass as measured on the Mass Balance and the weight as measured on the Weight Scale. (So you ask--what is the difference between weight and mass?) Now comes the key question to ask the class: If you took the Mass Balance and your calibrated Weight Scale to the Moon, do you think they would give the same measurement as on Earth? Remember, you always balance the unknown object against several pennies with the Mass Balance but you just let the unknown object pull down against the calibrated rubber band on the Weight Scale. We hope that this thought experiment will help the students see that the Mass Balance will measure the same no matter where you locate it in space but the Weight Scale, which measures how hard gravity pulls down on the object, will give a smaller reading on the moon. (This is confusing stuff and most college students will have difficulty understanding it. Perhaps if your kids start thinking about it early enough, they may come to a better understanding of the difference between weight and mass when they are older.)

Measuring Density: Since density is mass per volume, the most straight forward way of measuring the density of something is to measure its mass, then measure its volume and divide the mass by the volume. We could do exactly that in this activity but at this point we have no good way to measure volume. If you have a graduated cylinder (they aren't expensive but most elementary schools don't have them) you could use it with some water to mark the small "portion cups" at specific volumes. (We have already suggested that the small 1 oz cups will hold 10 cubic centimeters when filled to a point 1.4 cm above the bottom and it will hold 20 cubic centimeters when filled to a point 2.3 centimeters above the bottom.) Rather than actually measuring the density, we feel it will be sufficient for the students to appreciate that the same volume can be a large mass or a small mass depending upon the material involved. Our plan is to have the same volume of several different materials and measure their mass with the Mass Balance. Hopefully this exercise will help the students to begin to see the relationship between mass, volume and density.

(The next page begins the "Student Activity Sheet". We suggest that these be reproduced in sufficient numbers for the entire class. Whether the students work individually or in groups is best decided by you, however, some of the exercises need at least two people to hold and use the apparatus.)

Mass, Weight and Density--how matter is measured, how it interacts with other matter and how it fills space.

Which is heavier, a pound of feathers or a pound of lead? If you have never heard this old trick question before--think about it. Now try this one: which takes up more space, a pound of feathers or a pound of lead? Finally, think about this one: which weighs more 100 pennies on the earth or 100 pennies on the moon? The answer to each of these questions requires that you understand the difference between mass, weight and density.

You will measure the mass of objects by comparing them to the mass of pennies with a thing we will call a "Mass Balance". Although mass is usually measured in kilograms or grams, we will measure mass in "pennies". The Mass Balance is shown below.

This balance measures mass in "penny" units.

First test the Mass Balance to see if it is "zeroed". When you lift it by the center paper clip, it should stay fairly level. When the clip is in the top hole, it will balance easily. If you put the clip in the bottom hole, it probably will be too sensitive to balance at all. Test the Mass Balance by placing 5 pennies in both cups and gently lift it off the table--it should balance. Have one student secretly place a number of pennies in one cup and see if you can figure out how many pennies there are in the cup by matching them with pennies in the other cup. (You will find that not all pennies are exactly the same.) After you learn how to use the Mass Balance, you will be given several different materials to measure. Always measure the same volume of the given materials, that is, always fill the material to the same level on the cup on the left and find its mass by placing pennies in the cup on the right. Record your data in a table like the one below:

Name of material being measured

Mass of material in "pennies"

(name first materal meaured here)

(record its mass in pennies here)

Here is an important question to think about: If you took your Mass Balance to the Moon and repeated this experiment, would you get the same result? (Have your teacher discuss with you what mass means--this is quite confusing to many people.)

You will construct and calibrate a Weight Scale. This scale works by measuring how far a certain number of pennies are able to stretch a piece of rubber band. The scale is illustrated below:

Measuring weight in units of pennies.

After you have constructed the Weight Scale (your teacher will give you instructions) you will calibrate it using pennies. While one student carefully holds the Weight Scale by the string the other student will make a line on the card. First mark on the card where the knot touches the card with nothing in the cup (this will be the zero line on the scale). Now add 5 pennies to the cup and carefully mark where the knot touches the card. (Practice with pencil first until you learn how to do this.) Carefully "calibrate" your scale by adding 5 pennies at a time and marking where the knot touches the card. The numbers on your scale will be 0, 5, 10, 15, 20, 25 you probably will not be able to get any more. (Don't expect the numbers to be evenly spaced--rubber bands don't work that way.) If you have calibrated your scale well, you should be able to tell how many pennies someone has placed in the cup without actually counting them. As with the mass balance, you should weigh some of the materials your teacher provides. Always fill the cup with the same volume of material. Make a table like the following:

Name of material being weighed

Weight of material in "pennies"

(give the name of first material weighed here)

(give the weight of the material in pennies here)

Although your measurements are not perfect (no measurements ever are) you can use your calibrated Weight Scale to find the weight of assorted things and you could even use it to count pennies. Now comes a very important question: If you took your calibrated Weight Scale to the Moon, would it work the same way it did on Earth? If you filled it with the same amount of pennies, would the rubber band stretch to the same mark? Have your teacher discuss this with you and see if you can understand the difference between weight and mass.

Density tells us how much stuff has been packed into a certain amount of space. Lead is very dense, Styrofoam is not very dense at all. Density can be measured in grams per cubic centimeter. In our experiments, density could be measured in "pennies per cup". In your experiments with the Mass Balance, you always measured the mass of the same volume of material. Use the data you took to list the materials you measured in order of density with the most dense at the top of the list down to the least dense at the bottom of the list. You don't have to calculate the density in "pennies per cup" since you always measured the same volume of material--just look at your data to make the list.

After you have made your list of the densities of the materials, think about the following important question: How do you think the density of a substance would change if you measured it on the Moon rather than on the Earth?

Additional questions to think about:

1. You have an object and you want to know if it will float in water. To answer the question: "will it float?" do you need to know the objects mass, weight or density?

2. A student's mass on Earth is 50 kilograms. If this student went to the Moon, would her mass be more, less, or the same?

3. A student's weight on Earth is 100 pounds. If this student went to the Moon, would he weigh more, less, or the same?

4. A block of wood easily floats on water when on the Earth. If the same block of wood were taken to the Moon, would it float on water? (Really give this some careful thought, most people can not answer this question.)