Calculation of specific weight,specific volume and density, determination of gas constant R

The below are some of the examples worked out to calculate specific weight, specific volume, specific gravity and density of substance(matter) at given conditions:
Problem 1:
At temperature of 100° F and 120 psi (pressure per square inch) absolute, calculation of above properties for methane (CH4):
solution: 
By using equation of state  =   PV= nRT  ( R= 96.3 ft/°R)
specific weight = P/RT = n/V = 120 × 144 / [96.3 (460+100)] = 0.320 lb/ft³  (note  ° R= 460 + ° F)

density = specific weight / gravity = 0.320/32.2 = 0.00994 slug/ft³

specific volume = 1/density = 1/0.00994 = 101 ft³/slug
_____________________________________________________________________________________

Problem 2:
For oil of volume 6 m³ and weighs 47 kN
solution:
specific weight = 47 kN / 6 m³ = 7.833 kN/m³

density =7833/9.81 = 798 kg/m³

specific gravity = specific weight of oil /specific weight of water = 7.833/9.79 =0.800

Note: specific gravity of liquid is the ratio of specific weight of particular liquid to specific of water at same conditions, and it specific gravity as no units it has only a numerical value.
specific gravity of substance is calculated as: = weight of substance / weight of equal volume of water
= specific weight of substance / specific weight of water
= density of substance / density of water
_____________________________________________________________________________________

Problem 3:
At 90° F and 300psi absolute, find gas constant and density for a gas which has 11.4 ft³/ lb.
solution:
Data given is the condition of gas i.e temperature and pressure and specific volume
R= P/(specific volume . T) = [(30.0 × 144)11.4]/460+90 = 89.5 ft/° R

density = specific weight / gravity = 1 / ( volume × gravity ) = 1 / (11.4 × 32.2) = 0.00272 slugs/ft³

Concentration Calculation and Methods of Determining Concentration of Compounds,Molarity,Molality,Normality with solved example

Concentration is the word used frequently in the calculation of material balance, energy balance, and analyzing the purity of the substance. In physics and chemistry, it is a regular word. Without gaining enough knowledge on concentration concepts you are not able to understand the depth concepts of environmental science, mass transfer and reaction engineering, etc.

Concentration is a term used to know the amount of matter present in the measurable volume, mostly followed by unit volume. We should know, that mass, weight and mole are three physical quantities that tell us about the matter content but in various phenomena let us discuss it.

MASS:  It is a physical matter that exists in the universe.

WEIGHT: It is the force exerted by that mass by the influence of gravity.

MOLE:  It is the number that represents the number of molecules present in available mass.

Consider a substance called "X" that occupies a unit volume, specifically a cube with dimensions:

Length (L) = 1 cm

Breadth (B) = 1 cm

Height (H) = 1 cm

This unit volume of substance "X" has a total volume of:

Volume (V) = L × B × H = 1 cm × 1 cm × 1 cm = 1 cubic centimeter (or 1 cm³)

Now, let's explore the properties and behavior of substance "X" within this defined unit volume.

Concentration thus can be calculated in terms of

  

	Concentration	Units
Mass	Mass/ volume	Kg/m3 or  g/cm³
Weight	Weight/ Volume	Kg/m3
Mole	Mole/volume	Kmole/m3

The density of the solid is the term used for the mass/volume ratio

Liquid density is molecular mass/ molar volume

Vapor density is given the ideal gas equation PV= nRT.

The Gram Atom: 

It is used to specify the amounts of chemical elements. It is defined as the mass in grams of an element that is equal numerically to its atomic weight.

• Gram atom of element =  weight in grams of the element / atomic weight of the element

Mole: 

 

A mole is defined as the amount of substance equal to molecular weight (formula weight)

The Gram Mole: 

 

It is used to specify the amounts of chemical compounds. It is defined as the mass in grams of the substance that is equal numerically to its molecular weight.

• The gram mole of compound = weight in grams of that compound/compound molecular weight

Mostly gram mole is commonly termed a mole in textbooks. If you see a mole consider it to be a gram mole.

The molecular weight of the compound is found from the atomic weights of elements involved in the formation of compounds. Summing all the atomic weights multiplied by their respective no of molecules or elements participates in the formation of the compound.

Example 1: Find the molecular weight of sodium hydroxide NaOH
                

 atomic weight of Na = 23

           atomic weight of O   = 16
           atomic weight of H   =  1
Now,

= 23 X 1

=16 X 1

=1 X 1

 one atom of each element is present in this compound so they multiplied with it, now add up

= 23 X 1 + 16 X 1 + 1 X 1

=  23 + 16 + 1 

= 40  

Ex 2: Find Molecular weights of H2SO4, Na2CO3, KMnO4

Remember the atomic weight of H= 1, S= 32, O = 16

molecular weight of H2SO4 = 2 X 1 + 1 X 32 + 4 X 16 = 98

Atomic weight of Na= 23, C = 12 , O = 16

molecular weight ofNa2CO3 = 2 X 23 + 1 X  12 + 3 X 16 = 106

Atomic weight of K = 39 , Mn = 55,  O = 16

molecular weight of KMnO4 = 1 X 39 + 1 X 55 + 4 X 16 = 158

Molarity, Molality and Normality are important terms used in the determination of concentration when concerned with chemical material balance or chemical reactions studies.

The equivalent weight of an element or compound: is defined as the ratio of the atomic weight or molecular weight to its valence. The valence of an element or a compound depends on the number of hydrogen ions (H+) accepted or hydroxyl ions (OH-) donated for each atomic weight or molecular weight.

Equivalent weight = Molecular weight/valence

Ex: (1)

Calculate the equivalent weight (EW) of the following ions or molecules:

Chemicals:                      K+      Mg2+        Na2CO3
Ionic/molecular weight:  39          24           106

For K+ valence is +1 so, 39/1 = 39

similarly,

Mg2+ valence is +2, so the equivalent weight is 24/2 = 12

Na2CO3 valence is +2, so the equivalent weight is 106/ 2 = 53

Equivalent weight does not have signs so if you find the valence of any sign, use the numerical value and molecular weight, divide the molecular weight with the numerical value of the valence.

Normality:

Normality Calculator

 

Equivalent Weight:

Volume (liters):

Concentration (g/L):

It is defined as the number of gram-equivalents of solute dissolved in one liter of solution. and designated by the symbol 'N'

Normality (N) =  Gram equivalent of solute / Volume of solution in a liter

where,
Gram equivalent  is the ratio of the weight of the compound to the equivalent weight of the compound

Molarity:

Molarity Calculator

 

Gram Moles of Solute:

Volume of Solution (liters):

It is defined as the number of gram moles of the solute dissolved in one liter of solution. It is designated by the symbol M.

Molarity (M) =  Gram moles of solute / Volume of solution in a liter

Molality:

Molality Calculator

 

Gram Moles of Solute:

Mass of Solvent (kg):

It is defined as gram moles of solute dissolved in one kilogram of solvent

Molality = gram moles of solute/mass of solvent in Kg

Note: to find the concentration of solute in g/l we can use the formula as
= Normality  X equivalent weight


Ex: (2)

Find the Equivalent weight of HCl, NaOH, H2SO4

HCl: 

molecular weight of HCl: 1 X 1 +  1 X 35.5 = 36.5

valence of HCl: 1

therefore equivalent weight of HCl = 36.5 / 1 = 36.5

NaOH:

molecular weight of NaOH: 1 X 23 + 1X 16 +  1 X 1 =40

valence of NaOH: 1

therefore equivalent weight of NaOH: 40 / 1 = 40

H2SO4:

molecular weight of  H2SO4: 2 X 1 + 1X 32 +  4 X 16 =98

the valence of  H2SO4: 2

therefore equivalent weight of  H2SO4: 98 / 2 = 49

Ex: (3)

Find the Equivalent weight of H3PO4, CaCl2, FeCL3, Al2(SO4)3, KMnO4

 H3PO4:

molecular weight of  H3PO4: 3 X 1 + 1X 31 +  4 X 16 =98

valence of H3PO4: 3

equivalent weight of  H2SO4: 98 / 3 = 32.67

CaCl2 :

molecular weight of CaCl2 : 1 X 40 + 2 X 35.5 = 111

valence of CaCl2 :2

equivalent weight of  CaCl2 : 111 / 2 = 55.5

FeCl3 :

molecular weight of FeCl3 : 1 X 56 + 3 X 35.5 = 162.5

valence of FeCl3: 3

equivalent weight of  FeCl3 : 162.5 / 3 = 54.17

Al2(SO4)3 :

molecular weight of Al2(SO4)3 : 2 X 27 + 3 X 32 + 2X 16 =342

valence of Al2(SO4)3: 6

equivalent weight of  Al2(SO4)3 : 342 / 6 = 57

KMnO4 :

molecular weight of KMnO4 : 1 X 39 + 1 X 55 + 4X 16 =158

valence of KMnO4: 5

equivalent weight of  KMnO4 : 158 / 5 = 31.6

Ex: (4)

Find the normality, molarity and molality of the caustic solution containing 20% NaOH by weight, the density of the solution is 1.196 kg/l

Basis: 100 kg of solution 
so, the solution contains 20 kg of NaOH (20X 100/100) and 80 kg of water(solvent)
The density of solution given as 1.196 kg/l

Now find the volume of the solution by using mass and density as: volume = mass/density
= 100/ 1.196 = 83.62 liters


Moles of NaOH in solution = weight of NaOH in the solution / molecular weight of NaOH
                                            = 20/40 = 0.5 kmol = 500 mol (also called gram mole or g mole)


Molarity =  gram mole of NaOH / volume of solution in a liter
              =  500 / 83.52
              =  5.98 mol/ liter

For NaOH the valence = 1
Equivalent weight = molecular weight/valence
                           =  40 / 1 = 40 
Gram equivalent weight = weight of NaOH / equivalent weight
                                    =  20000 gram / 40   (since we have 20 kg converted to gram it becomes 20X1000 g)
                                    =    500


Normality =  gram equivalent of NaOH / volume of solution in a liter
                 = 500 / 83.52
                 = 5.98


Molality= gram moles of NaOH /  kg of solvent
               = 500 / 80 
               = 6.25 mol/kg


Ex: (5)

To prepare 1 N of the Na2CO3 solution, how many grams of Na2CO3 must be dissolved in 1.0 L of water?

Ans:
By using the Normality formula = gram equivalent of solute/volume of solution in a liter
Given:
1.0-liter solution, 1 N of Na2CO3
We know the molecular weight Na2CO3 = 106, valence = 2
Gram equivalent = weight of Na2CO3 / equivalent weight of Na2CO3

1N = weight of Na2CO3/ 53/1 liter solution
weight of Na2CO3 = 53 g

Specific gravity is used for indirect measurement of concentrations of aqueous solutions. which is the ratio of the density of solution at T1 (temperature)to that of the density of water at T2 (temperature)and the specific gravity of a gas is the ratio of the average molecular weight of the gas to the molecular weight of air at the same temperature.

In water analysis, the impurities are in traces and their concentration is expressed in mg/liter or Parts per million.

  

An Introduction About Diffusion Concept in Mass Transfer

Diffusion is a phenomenon that deals with the microscopic movements of the molecules in all the phases, learning and studying the concept of diffusion will help to understand mass transfer operations, chemical reaction engineering and transportation concepts. The kinetic Theory of gases supports the diffusion occurrence in a system,

Diffusion is understood as a process of movement of particles from a higher concentration section to a lower concentration section of the system. Diffusion of the component is caused by this concentration difference which is called the concentration gradient. For example, when a bottle containing perfume is opened the smell is sensed as the molecules diffuse into the air due to the concentration difference, so till the equilibrium is attained these molecules will move all over the room if an exit fan is present at the top of the room and push out the air with some constant flow rate and the with the same flow rate the molecule will transfer into the system from the bottle.

This concept of molecule behavior is used in separation operation and purification operations such as distillation, extraction, absorption, reverse osmosis, etc. Diffusion can also occur by pressure gradient even with temperature gradient or even with an external force field which acts as the activity gradient. Mostly diffusion concept is developed by the physical properties of the molecules. For a system, at equilibrium, no net diffusion occurs. which is explained as when a molecule diffuses from one phase to another phase in a system containing two phases through an interface, then at the same time another molecule will diffuse from the opposite direction which the net diffusion is set to zero due to the cancellation of the opposite direction, as said this occurs at equilibrium only.

Classification: 

1. Molecular diffusion 

2. Eddy or Turbulent diffusion

Importance of Molecular diffusion:

Due to the thermal energy in a molecule, it gains the ability to move through another set of molecules in a system is known to be thermal diffusion. So in the same way molecules will tend to travel due to the driving force of concentration difference. To understand this let us discuss an example which we involve regularly during the rush hour.

Let us consider a road that you want to cross from one side, say ‘A’ at the same time there are a group of people waiting to cross the same from the opposite side, say ‘B’. Now when the traffic is halted by the traffic signal, all the sudden you start to move toward the zebra crossing and diffuse into the mass of people who are in a hurry as you, things to be considered are:

You cannot move without colliding (facing each other) with anyone
You cannot move in a straight line, you have to take some zig-zig turns into the gaps between the people,
Finally, you will reach point ‘B’ let us compare to some physical phenomena

• You will move at some rate i.e. speed
• The distance you covered is less than the displacement
• Opposite people also move with some speed or rate
• Your body weight also will affect the speed at which you travel, the heavier you are the slower you will move

Finally, we have some options from above that could be compared to the molecular level,

You will be replaced by the one molecule of the solute in solution ‘A’(which contains a high concentration of solute A) and another mass of people will take the position of the solvent molecules present in solution ‘B’, as you and opposite people cross the road in the same way solute molecule will cross the interface and move into the solution B, at the same time solvent molecules (of solution B which are on the opposite side) will cross the interface and move towards the solution A.

The molecule will have the rate you have so it also covers a distance in zig-zig mode colliding with other molecules, the distance traveled per second will be the rate of diffusion, when a certain fixed area is considered along the path which the molecule moves then the rate become as the flux which is moles/ (area X time) and this rate depends on the molecular weight also and the number of molecule present which we say as concentration, the rate will exist till the molecule reaches the destination, that is the space in solution B to be occupied sufficiently and saturated with molecules of B, this state is called equilibrium.

Diffusion Calculator

Input Parameters

- Diffusion Coefficient (D): The diffusion coefficient of the substance (m^2/s)

- Length (L): The length of the system (m)

- Time (t): The time of diffusion (s)

- Initial Concentration (C0): The initial concentration of the substance (mol/m^3)

- Boundary Concentration (Cb): The concentration of the substance at the boundary (mol/m^3)

Select Calculation Type

- Concentration Profile: Calculate the concentration of the substance at different points in the system

- Diffusion Flux: Calculate the diffusion flux of the substance

- Diffusion Coefficient: Calculate the diffusion coefficient of the substance

 

Calculator

Diffusion Coefficient (D):

Length (L):

Time (t):

Initial Concentration (C0):

Boundary Concentration (Cb):

Select Calculation Type: Concentration Profile Diffusion Flux Diffusion Coefficient

Diffusivities (cm²/s) of gases at standard atmospheric pressure, (101.325 KPa), T in Kelvin

  Gas mixture system diffusivity is used for mass transfer calculations and to design the unit operation which involves in handling mass transfer like distillation, absorption etc.

s.no	System	200	273.15	293.15	373.15	473.15	573.15	673.15
1	Ar-CH4	-	-	-	0.306	0.467	0.657	0.876
2	Ar-CO		0.168	0.187	0.290	0.439	0.615	0.815
3	Ar-CO2		0.129	0.078	0.235	0.365	0.517	0.689
4	Ar-H2		0.698	0.794	1.228	1.876	2.634	3.496
5	Ar-He	0.381	0.645	0.726	1.088	1.617	2.226	2.911
6	Ar-Kr	0.064	0.117	0.134	0.210	0.323	0.456	0.605
7	Ar-N		0.168	0.190	0.290	0.439	0.615	0.815
8	Ar-Ne	0.160	0.277	0.313	0.475	0.710	0.979	1.283
9	Ar-O2		0.166	0.189	0.285	0.430	0.600	0.793
10	Ar-SF6	-	-	-	0.128	0.202	0.290	0.389
11	Ar-Xe	0.052	0.095	0.108	0.171	0.264	0.374	0.498
12	CH4-H2	-	-	0.782	1.084	1.648	2.311	3.070
13	CH4-He	-	-	0.723	0.992	1.502	2.101	2.784
14	CH4-N2	-	-	0.220	0.317	0.480	0.671	0.890
15	CH4-O2	-	-	0.210	0.341	0.523	0.736	0.978
16	CH4-SF6	-	-		0.167	0.257	0.363	0.482
17	CO-CO2	-	-	0.162	0.250	0.38
18	CO-H2	0.408	0.686	0.772	1.162	1.743	2.423	3.196
19	CO-He	0.365	0.619	0.698	1.052	1.577	2.188	2.882
20	CO-Kr		0.131	0.581	0.227	0.346	0.485	0.645
21	CO-N2	0.133	0.208	0.231	0.336	0.491	0.673	0.878
22	CO-O2	-	-	0.202	0.307	0.462	0.643	0.849
23	CO-SF6	-	-	-	0.144	0.226	0.323	0.432
24	CO2-C3H8	-	-	0.084	0.133	0.209	--	-
25	CO2-H2	0.315	0.552	0.412	0.964	1.470	2.066	2.745
26	CO2-H2O	-	-	0.162	0.292	0.496	0.741	1.021
27	CO2-He	0.300	0.513	0.400	0.878	1.321	-	-
28	CO2-N2	-	-	0.160	0.253	0.392	0.553	0.733
29	CO2-N2O	0.055	0.099	0.113	0.177	0.276	-
30	CO2-Ne	0.131	0.227	0.199	0.395	0.603	0.847	-

s.no	System	200	273.15	293.15	373.15	473.15	573.15	673.15
31	CO2-O2			0.159	0.248	0.38	0.535	0.710
32	CO2-SF6				0.099	0.155
33	D2-H2	0.631	1.079	1.219	1.846	2.778	3.866	5.103
34	H2-He	0.775	1.320	1.490	2.255	3.394	4.726	6.242
35	H2-Kr	0.340	0.601	0.682	1.053	1.607	2.258	2.999
36	H2-N2	0.408	0.686	0.772	1.162	1.743	2.423	3.196
37	H2-Ne	0.572	0.982	0.317	1.684	2.541	3.541	4.677
38	H2-O2		0.692	0.756	1.188	1.792	2.497	.299
39	H2-SF6			0.208	0.649	0.998	1.400	1.851
40	H2-Xe		0.513	0.122	0.890	1.349	1.885	2.493
41	H2O-N2			0.242	0.399
42	H2O-O2			0.244	0.403	0.645	0.882	1.147
43	He-Kr	0.330	0.559	0.629	0.942	1.404	1.942	2.550
44	He-N2	0.365	0.619	0.698	1.052	1.577	2.188	2.882
45	He-Ne	0.563	0.948	1.066	1.592	2.362	3.254	4.262
46	He-O2			0.641	0.697	1.092	1.640	2.276
47	He-SF6				1.109	0.592	0.871	1.190
48	He-Xe	0.282	0.478	0.538	0.807	1.201	1.655	2.168
49	Kr-N2		0.131	0.149	0.227	0.346	0.485	0.645
50	Kr-Ne	0.131	0.228	0.258	0.392	0.587	0.812	1.063
51	Kr-Xe	0.035	0.064	0.073	0.116	0.181	0.257	0.344
52	N2-Ne			0.258	0.483	0.731	1.021	1.351
53	N2-O2			0.202	0.307	0.462	0.643	0.849
54	N2-SF6				0.148	0.231	0.328	0.436
55	N2-Xe		0.107	0.123	0.188	0.287	0.404	0.539
56	Ne-Xe	0.111	0.193	0.219	0.332	0.498	0.688	0.901
57	O2-SF6			0.097	0.154	0.238	0.334	0.441