Gas Viscosity at High Pressure and High Temperature
Total Page:16
File Type:pdf, Size:1020Kb
Load more
Recommended publications
-
Solutes and Solution
Solutes and Solution The first rule of solubility is “likes dissolve likes” Polar or ionic substances are soluble in polar solvents Non-polar substances are soluble in non- polar solvents Solutes and Solution There must be a reason why a substance is soluble in a solvent: either the solution process lowers the overall enthalpy of the system (Hrxn < 0) Or the solution process increases the overall entropy of the system (Srxn > 0) Entropy is a measure of the amount of disorder in a system—entropy must increase for any spontaneous change 1 Solutes and Solution The forces that drive the dissolution of a solute usually involve both enthalpy and entropy terms Hsoln < 0 for most species The creation of a solution takes a more ordered system (solid phase or pure liquid phase) and makes more disordered system (solute molecules are more randomly distributed throughout the solution) Saturation and Equilibrium If we have enough solute available, a solution can become saturated—the point when no more solute may be accepted into the solvent Saturation indicates an equilibrium between the pure solute and solvent and the solution solute + solvent solution KC 2 Saturation and Equilibrium solute + solvent solution KC The magnitude of KC indicates how soluble a solute is in that particular solvent If KC is large, the solute is very soluble If KC is small, the solute is only slightly soluble Saturation and Equilibrium Examples: + - NaCl(s) + H2O(l) Na (aq) + Cl (aq) KC = 37.3 A saturated solution of NaCl has a [Na+] = 6.11 M and [Cl-] = -
VISCOSITY of a GAS -Dr S P Singh Department of Chemistry, a N College, Patna
Lecture Note on VISCOSITY OF A GAS -Dr S P Singh Department of Chemistry, A N College, Patna A sketchy summary of the main points Viscosity of gases, relation between mean free path and coefficient of viscosity, temperature and pressure dependence of viscosity, calculation of collision diameter from the coefficient of viscosity Viscosity is the property of a fluid which implies resistance to flow. Viscosity arises from jump of molecules from one layer to another in case of a gas. There is a transfer of momentum of molecules from faster layer to slower layer or vice-versa. Let us consider a gas having laminar flow over a horizontal surface OX with a velocity smaller than the thermal velocity of the molecule. The velocity of the gaseous layer in contact with the surface is zero which goes on increasing upon increasing the distance from OX towards OY (the direction perpendicular to OX) at a uniform rate . Suppose a layer ‘B’ of the gas is at a certain distance from the fixed surface OX having velocity ‘v’. Two layers ‘A’ and ‘C’ above and below are taken into consideration at a distance ‘l’ (mean free path of the gaseous molecules) so that the molecules moving vertically up and down can’t collide while moving between the two layers. Thus, the velocity of a gas in the layer ‘A’ ---------- (i) = + Likely, the velocity of the gas in the layer ‘C’ ---------- (ii) The gaseous molecules are moving in all directions due= to −thermal velocity; therefore, it may be supposed that of the gaseous molecules are moving along the three Cartesian coordinates each. -
Viscosity of Gases References
VISCOSITY OF GASES Marcia L. Huber and Allan H. Harvey The following table gives the viscosity of some common gases generally less than 2% . Uncertainties for the viscosities of gases in as a function of temperature . Unless otherwise noted, the viscosity this table are generally less than 3%; uncertainty information on values refer to a pressure of 100 kPa (1 bar) . The notation P = 0 specific fluids can be found in the references . Viscosity is given in indicates that the low-pressure limiting value is given . The dif- units of μPa s; note that 1 μPa s = 10–5 poise . Substances are listed ference between the viscosity at 100 kPa and the limiting value is in the modified Hill order (see Introduction) . Viscosity in μPa s 100 K 200 K 300 K 400 K 500 K 600 K Ref. Air 7 .1 13 .3 18 .5 23 .1 27 .1 30 .8 1 Ar Argon (P = 0) 8 .1 15 .9 22 .7 28 .6 33 .9 38 .8 2, 3*, 4* BF3 Boron trifluoride 12 .3 17 .1 21 .7 26 .1 30 .2 5 ClH Hydrogen chloride 14 .6 19 .7 24 .3 5 F6S Sulfur hexafluoride (P = 0) 15 .3 19 .7 23 .8 27 .6 6 H2 Normal hydrogen (P = 0) 4 .1 6 .8 8 .9 10 .9 12 .8 14 .5 3*, 7 D2 Deuterium (P = 0) 5 .9 9 .6 12 .6 15 .4 17 .9 20 .3 8 H2O Water (P = 0) 9 .8 13 .4 17 .3 21 .4 9 D2O Deuterium oxide (P = 0) 10 .2 13 .7 17 .8 22 .0 10 H2S Hydrogen sulfide 12 .5 16 .9 21 .2 25 .4 11 H3N Ammonia 10 .2 14 .0 17 .9 21 .7 12 He Helium (P = 0) 9 .6 15 .1 19 .9 24 .3 28 .3 32 .2 13 Kr Krypton (P = 0) 17 .4 25 .5 32 .9 39 .6 45 .8 14 NO Nitric oxide 13 .8 19 .2 23 .8 28 .0 31 .9 5 N2 Nitrogen 7 .0 12 .9 17 .9 22 .2 26 .1 29 .6 1, 15* N2O Nitrous -
Specific Latent Heat
SPECIFIC LATENT HEAT The specific latent heat of a substance tells us how much energy is required to change 1 kg from a solid to a liquid (specific latent heat of fusion) or from a liquid to a gas (specific latent heat of vaporisation). �����푦 (��) 퐸 ����������푐 ������� ℎ���� �� ������� �� = (��⁄��) = 푓 � ����� (��) �����푦 = ����������푐 ������� ℎ���� �� 퐸 = ��푓 × � ������� × ����� ����� 퐸 � = �� 푦 푓 ����� = ����������푐 ������� ℎ���� �� ������� WORKED EXAMPLE QUESTION 398 J of energy is needed to turn 500 g of liquid nitrogen into at gas at-196°C. Calculate the specific latent heat of vaporisation of nitrogen. ANSWER Step 1: Write down what you know, and E = 99500 J what you want to know. m = 500 g = 0.5 kg L = ? v Step 2: Use the triangle to decide how to 퐸 ��푣 = find the answer - the specific latent heat � of vaporisation. 99500 퐽 퐿 = 0.5 �� = 199 000 ��⁄�� Step 3: Use the figures given to work out 푣 the answer. The specific latent heat of vaporisation of nitrogen in 199 000 J/kg (199 kJ/kg) Questions 1. Calculate the specific latent heat of fusion if: a. 28 000 J is supplied to turn 2 kg of solid oxygen into a liquid at -219°C 14 000 J/kg or 14 kJ/kg b. 183 600 J is supplied to turn 3.4 kg of solid sulphur into a liquid at 115°C 54 000 J/kg or 54 kJ/kg c. 6600 J is supplied to turn 600g of solid mercury into a liquid at -39°C 11 000 J/kg or 11 kJ/kg d. -
Ideal Gas Law PV = Nrt R = Universal Gas Constant R = 0.08206 L-Atm R = 8.314 J Mol-K Mol-K Example
Ideal gas law PV = nRT R = universal gas constant R = 0.08206 L-atm R = 8.314 J mol-K mol-K Example: In the reaction of oxygen with carbon to produce carbon dioxide, how many liters of oxygen at 747 mm Hg and 21.0 °C are needed to react with 24.022 g of carbon? CHEM 102 Winter 2011 Practice: In the reaction of oxygen and hydrogen to make water, how many liters of hydrogen at exactly 0 °C and 1 atm are needed to react with 10.0 g of oxygen? 2 CHEM 102 Winter 2011 Combined gas law V decreases with pressure (V α 1/P). V increases with temperature (V α T). V increases with particles (V α n). Combining these ideas: V α n x T P For a given amount of gas (constant number of particles, ninitial = nafter) that changes in volume, temperature, or pressure: Vinitial α ninitial x Tinitial Pinitial Vafter α nafter x Tafter Pafter ninitial = n after Vinitial x Pinitial = Vafter x Pafter Tinitial Tafter 3 CHEM 102 Winter 2011 Example: A sample of gas is pressurized to 1.80 atm and volume of 90.0 cm3. What volume would the gas have if it were pressurized to 4.00 atm without a change in temperature? 4 CHEM 102 Winter 2011 Example: A 2547-mL sample of gas is at 421 °C. What is the new gas temperature if the gas is cooled at constant pressure until it has a volume of 1616 mL? 5 CHEM 102 Winter 2011 Practice: A balloon is filled with helium. -
Chapter 3 3.4-2 the Compressibility Factor Equation of State
Chapter 3 3.4-2 The Compressibility Factor Equation of State The dimensionless compressibility factor, Z, for a gaseous species is defined as the ratio pv Z = (3.4-1) RT If the gas behaves ideally Z = 1. The extent to which Z differs from 1 is a measure of the extent to which the gas is behaving nonideally. The compressibility can be determined from experimental data where Z is plotted versus a dimensionless reduced pressure pR and reduced temperature TR, defined as pR = p/pc and TR = T/Tc In these expressions, pc and Tc denote the critical pressure and temperature, respectively. A generalized compressibility chart of the form Z = f(pR, TR) is shown in Figure 3.4-1 for 10 different gases. The solid lines represent the best curves fitted to the data. Figure 3.4-1 Generalized compressibility chart for various gases10. It can be seen from Figure 3.4-1 that the value of Z tends to unity for all temperatures as pressure approach zero and Z also approaches unity for all pressure at very high temperature. If the p, v, and T data are available in table format or computer software then you should not use the generalized compressibility chart to evaluate p, v, and T since using Z is just another approximation to the real data. 10 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 112 3-19 Example 3.4-2 ---------------------------------------------------------------------------------- A closed, rigid tank filled with water vapor, initially at 20 MPa, 520oC, is cooled until its temperature reaches 400oC. -
Ideal Gasses Is Known As the Ideal Gas Law
ESCI 341 – Atmospheric Thermodynamics Lesson 4 –Ideal Gases References: An Introduction to Atmospheric Thermodynamics, Tsonis Introduction to Theoretical Meteorology, Hess Physical Chemistry (4th edition), Levine Thermodynamics and an Introduction to Thermostatistics, Callen IDEAL GASES An ideal gas is a gas with the following properties: There are no intermolecular forces, except during collisions. All collisions are elastic. The individual gas molecules have no volume (they behave like point masses). The equation of state for ideal gasses is known as the ideal gas law. The ideal gas law was discovered empirically, but can also be derived theoretically. The form we are most familiar with, pV nRT . Ideal Gas Law (1) R has a value of 8.3145 J-mol1-K1, and n is the number of moles (not molecules). A true ideal gas would be monatomic, meaning each molecule is comprised of a single atom. Real gasses in the atmosphere, such as O2 and N2, are diatomic, and some gasses such as CO2 and O3 are triatomic. Real atmospheric gasses have rotational and vibrational kinetic energy, in addition to translational kinetic energy. Even though the gasses that make up the atmosphere aren’t monatomic, they still closely obey the ideal gas law at the pressures and temperatures encountered in the atmosphere, so we can still use the ideal gas law. FORM OF IDEAL GAS LAW MOST USED BY METEOROLOGISTS In meteorology we use a modified form of the ideal gas law. We first divide (1) by volume to get n p RT . V we then multiply the RHS top and bottom by the molecular weight of the gas, M, to get Mn R p T . -
Thermal Properties of Petroleum Products
UNITED STATES DEPARTMENT OF COMMERCE BUREAU OF STANDARDS THERMAL PROPERTIES OF PETROLEUM PRODUCTS MISCELLANEOUS PUBLICATION OF THE BUREAU OF STANDARDS, No. 97 UNITED STATES DEPARTMENT OF COMMERCE R. P. LAMONT, Secretary BUREAU OF STANDARDS GEORGE K. BURGESS, Director MISCELLANEOUS PUBLICATION No. 97 THERMAL PROPERTIES OF PETROLEUM PRODUCTS NOVEMBER 9, 1929 UNITED STATES GOVERNMENT PRINTING OFFICE WASHINGTON : 1929 F<ir isale by tfttf^uperintendent of Dotmrtients, Washington, D. C. - - - Price IS cants THERMAL PROPERTIES OF PETROLEUM PRODUCTS By C. S. Cragoe ABSTRACT Various thermal properties of petroleum products are given in numerous tables which embody the results of a critical study of the data in the literature, together with unpublished data obtained at the Bureau of Standards. The tables contain what appear to be the most reliable values at present available. The experimental basis for each table, and the agreement of the tabulated values with experimental results, are given. Accompanying each table is a statement regarding the esti- mated accuracy of the data and a practical example of the use of the data. The tables have been prepared in forms convenient for use in engineering. CONTENTS Page I. Introduction 1 II. Fundamental units and constants 2 III. Thermal expansion t 4 1. Thermal expansion of petroleum asphalts and fluxes 6 2. Thermal expansion of volatile petroleum liquids 8 3. Thermal expansion of gasoline-benzol mixtures 10 IV. Heats of combustion : 14 1. Heats of combustion of crude oils, fuel oils, and kerosenes 16 2. Heats of combustion of volatile petroleum products 18 3. Heats of combustion of gasoline-benzol mixtures 20 V. -
Mol of Solute Molality ( ) = Kg Solvent M
Molality • Molality (m) is the number of moles of solute per kilogram of solvent. mol of solute Molality (m ) = kg solvent Copyright © Houghton Mifflin Company. All rights reserved. 11 | 51 Sample Problem Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water. mol of solute m = kg solvent 1 m ol KF ()13.5g 58.1 = 0.250 kg = 0.929 m Copyright © Houghton Mifflin Company. All rights reserved. 11 | 52 Mole Fraction • Mole fraction (χ) is the ratio of the number of moles of a substance over the total number of moles of substances in solution. number of moles of i χ = i total number of moles ni = nT Copyright © Houghton Mifflin Company. All rights reserved. 11 | 53 Sample Problem -Conversions between units- • ex) What is the molality of a 0.200 M aluminum nitrate solution (d = 1.012g/mL)? – Work with 1 liter of solution. mass = 1012 g – mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ; – mass water = 1012 g -43 g = 969 g 0.200mol Molality==0.206 mol / kg 0.969kg Copyright © Houghton Mifflin Company. All rights reserved. 11 | 54 Sample Problem Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water. mol of NaCl χ=NaCl mol of NaCl + mol H2 O 1 mol NaCl ()10.0g 58.5g NaCl = 1 m ol NaCl 1 m ol H2 O ()10.0g+ () 100.g H2 O 58.5g NaCl 18.0g H2 O = 0.0299 Copyright © Houghton Mifflin Company. -
Changes in State and Latent Heat
Physical State/Latent Heat Changes in State and Latent Heat Physical States of Water Latent Heat Physical States of Water The three physical states of matter that we normally encounter are solid, liquid, and gas. Water can exist in all three physical states at ordinary temperatures on the Earth's surface. When water is in the vapor state, as a gas, the water molecules are not bonded to each other. They float around as single molecules. When water is in the liquid state, some of the molecules bond to each other with hydrogen bonds. The bonds break and re-form continually. When water is in the solid state, as ice, the molecules are bonded to each other in a solid crystalline structure. This structure is six- sided, with each molecule of water connected to four others with hydrogen bonds. Because of the way the crystal is arranged, there is actually more empty space between the molecules than there is in liquid water, so ice is less dense. That is why ice floats. Latent Heat Each time water changes physical state, energy is involved. In the vapor state, the water molecules are very energetic. The molecules are not bonded with each other, but move around as single molecules. Water vapor is invisible to us, but we can feel its effect to some extent, and water vapor in the atmosphere is a very important http://daphne.palomar.edu/jthorngren/latent.htm (1 of 4) [4/9/04 5:30:18 PM] Physical State/Latent Heat factor in weather and climate. In the liquid state, the individual molecules have less energy, and some bonds form, break, then re-form. -
Safety Advice. Cryogenic Liquefied Gases
Safety advice. Cryogenic liquefied gases. Properties Cryogenic Liquefied Gases are also known as Refrigerated Liquefied Gases or Deeply Refrigerated Gases and are commonly called Cryogenic Liquids. Cryogenic Gases are cryogenic liquids that have been vaporised and may still be at a low temperature. Cryogenic liquids are used for their low temperature properties or to allow larger quantities to be stored or transported. They are extremely cold, with boiling points below -150°C (-238°F). Carbon dioxide and Nitrous oxide, which both have higher boiling points, are sometimes included in this category. In the table you may find some data related to the most common Cryogenic Gases. Helium Hydrogen Nitrogen Argon Oxygen LNG Nitrous Carbon Oxide Dioxide Chemical symbol He H2 N2 Ar O2 CH4 N2O CO2 Boiling point at 1013 mbar [°C] -269 -253 -196 -186 -183 -161 -88.5 -78.5** Density of the liquid at 1013 mbar [kg/l] 0.124 0.071 0.808 1.40 1.142 0.42 1.2225 1.1806 3 Density of the gas at 15°C, 1013 mbar [kg/m ] 0.169 0.085 1.18 1.69 1.35 0.68 3.16 1.87 Relative density (air=1) at 15°C, 1013 mbar * 0.14 0.07 0.95 1.38 1.09 0.60 1.40 1.52 Gas quantity vaporized from 1 litre liquid [l] 748 844 691 835 853 630 662 845 Flammability range n.a. 4%–75% n.a. n.a. n.a. 4.4%–15% n.a. n.a. Notes: *All the above gases are heavier than air at their boiling point; **Sublimation point (where it exists as a solid) Linde AG Gases Division, Carl-von-Linde-Strasse 25, 85716 Unterschleissheim, Germany Phone +49.89.31001-0, [email protected], www.linde-gas.com 0113 – SA04 LCS0113 Disclaimer: The Linde Group has no control whatsoever as regards performance or non-performance, misinterpretation, proper or improper use of any information or suggestions contained in this instruction by any person or entity and The Linde Group expressly disclaims any liability in connection thereto. -
Molar Gas Constant
Molar gas constant The molar gas constant (also known as the universal or ideal gas constant and designated by the symbol R ) is an important physical constant which appears in many of the fundamental equations inphysics, chemistry and engineering, such as the ideal gas law, other equations of state and the Nernst equation. R is equivalent to the Boltzmann constant (designated kB) times Avogadro's constant (designated NA) and thus: R = kBNA. Currently the most accurate value as published by the National Institute of Standards and Technology (NIST) is:[1] R = 8.3144621 J · K–1 · mol–1 A number of values of R in other units are provided in the adjacent table. The gas constant occurs in the ideal gas law as follows: PV = nRT or as: PVm = RT where: P is the absolute pressure of the gas T is the absolute temperature of the gas V is the volume the gas occupies n is the number of moles of gas Vm is the molar volume of the gas The specific gas constant The gas constant of a specific gas, as differentiated from the above universal molar gas constant which applies for any ideal gas, is designated by the symbol Rs and is equal to the molar gas constant divided by the molecular mass (M ) of the gas: Rs = R ÷ M The specific gas constant for an ideal gas may also be obtained from the following thermodynamicsrelationship:[2] Rs = cp – cv where cp and cv are the gas's specific heats at constant pressure and constant volume respectively. Some example values of the specific gas constant are: –1 –1 –1 Ammonia (molecular mass of 17.032 g · mol ) : Rs = 0.4882 J · K · g –1 –1 –1 Hydrogen (molecular mass of 2.016 g · mol ) : Rs = 4.1242 J · K · g –1 –1 –1 Methane (molecular mass of 16.043 g · mol ) : Rs = 0.5183 J · K · g Unfortunately, many authors in the technical literature sometimes use R as the specific gas constant without designating it as such or stating that it is the specific gas constant.