W 15 “Isotherms of Real Gases”
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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 -
Muddiest Point – Entropy and Reversible I Am Confused About Entropy and How It Is Different in a Reversible Versus Irreversible Case
1 Muddiest Point { Entropy and Reversible I am confused about entropy and how it is different in a reversible versus irreversible case. Note: Some of the discussion below follows from the previous muddiest points comment on the general idea of a reversible and an irreversible process. You may wish to have a look at that comment before reading this one. Let's talk about entropy first, and then we will consider how \reversible" gets involved. Generally we divide the universe into two parts, a system (what we are studying) and the surrounding (everything else). In the end the total change in the entropy will be the sum of the change in both, dStotal = dSsystem + dSsurrounding: This total change of entropy has only two possibilities: Either there is no spontaneous change (equilibrium) and dStotal = 0, or there is a spontaneous change because we are not at equilibrium, and dStotal > 0. Of course the entropy change of each piece, system or surroundings, can be positive or negative. However, the second law says the sum must be zero or positive. Let's start by thinking about the entropy change in the system and then we will add the entropy change in the surroundings. Entropy change in the system: When you consider the change in entropy for a process you should first consider whether or not you are looking at an isolated system. Start with an isolated system. An isolated system is not able to exchange energy with anything else (the surroundings) via heat or work. Think of surrounding the system with a perfect, rigid insulating blanket. -
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. -
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. -
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. -
Isothermal Process It Is the Process in Which Other Physical Quantities Might Change but the Temperature of the System Remains Or Is Forced to Remain Constant
Sajit Chandra Shakya Department of Physics Kathmandu Don Bosco College New Baneshwor, Kathmandu Isothermal process It is the process in which other physical quantities might change but the temperature of the system remains or is forced to remain constant. For example, the constant temperature of human body. Under constant temperature, the volume of a gas system is inversely proportional to the pressure applied, the phenomena being called Boyle's Law, written in symbols as 1 V ∝ P 1 Or, V = KB× where KB is a constant quantity. P Or, PV = KB …………….. (i) This means whatever be the values of volume and pressure, their product will be constant. So, P1V1 = KB, P2V2 = KB, P3V3 = KB, etc, Or, P1V1 = P2V2 = P3V3, etc. The requirements for an isothermal process are as follows: 1. The process should be carried very slowly so that there is an ample time for compensation of heat in case of any loss or addition. 2. The boundaries of the system should be highly conducting so that there is a path for heat to flow into or flow away from a closed space in case of any energy loss or oversupply. 3. The boundaries should be made very thin because the resistance of the substance for heat conduction will be less for thin boundaries. Since in an isothermal change, the temperature remains constant, the internal energy also does not change, i.e. dU = 0. So if dQ amount of heat is given to a system which undergoes isothermal change, the relation for the first law of thermodynamics would be dQ = dU + dW Or, dQ = 0 + PdV Or, dQ = PdV This means all the heat supplied will be utilized for performing external work and consequently its value will be very high compared to other processes. -
The First Law of Thermodynamics Continued Pre-Reading: §19.5 Where We Are
Lecture 7 The first law of thermodynamics continued Pre-reading: §19.5 Where we are The pressure p, volume V, and temperature T are related by an equation of state. For an ideal gas, pV = nRT = NkT For an ideal gas, the temperature T is is a direct measure of the average kinetic energy of its 3 3 molecules: KE = nRT = NkT tr 2 2 2 3kT 3RT and vrms = (v )av = = r m r M p Where we are We define the internal energy of a system: UKEPE=+∑∑ interaction Random chaotic between atoms motion & molecules For an ideal gas, f UNkT= 2 i.e. the internal energy depends only on its temperature Where we are By considering adding heat to a fixed volume of an ideal gas, we showed f f Q = Nk∆T = nR∆T 2 2 and so, from the definition of heat capacity Q = nC∆T f we have that C = R for any ideal gas. V 2 Change in internal energy: ∆U = nCV ∆T Heat capacity of an ideal gas Now consider adding heat to an ideal gas at constant pressure. By definition, Q = nCp∆T and W = p∆V = nR∆T So from ∆U = Q W − we get nCV ∆T = nCp∆T nR∆T − or Cp = CV + R It takes greater heat input to raise the temperature of a gas a given amount at constant pressure than constant volume YF §19.4 Ratio of heat capacities Look at the ratio of these heat capacities: we have f C = R V 2 and f + 2 C = C + R = R p V 2 so C p γ = > 1 CV 3 For a monatomic gas, CV = R 3 5 2 so Cp = R + R = R 2 2 C 5 R 5 and γ = p = 2 = =1.67 C 3 R 3 YF §19.4 V 2 Problem An ideal gas is enclosed in a cylinder which has a movable piston. -
Real Gases – As Opposed to a Perfect Or Ideal Gas – Exhibit Properties That Cannot Be Explained Entirely Using the Ideal Gas Law
Basic principle II Second class Dr. Arkan Jasim Hadi 1. Real gas Real gases – as opposed to a perfect or ideal gas – exhibit properties that cannot be explained entirely using the ideal gas law. To understand the behavior of real gases, the following must be taken into account: compressibility effects; variable specific heat capacity; van der Waals forces; non-equilibrium thermodynamic effects; Issues with molecular dissociation and elementary reactions with variable composition. Critical state and Reduced conditions Critical point: The point at highest temp. (Tc) and Pressure (Pc) at which a pure chemical species can exist in vapour/liquid equilibrium. The point critical is the point at which the liquid and vapour phases are not distinguishable; because of the liquid and vapour having same properties. Reduced properties of a fluid are a set of state variables normalized by the fluid's state properties at its critical point. These dimensionless thermodynamic coordinates, taken together with a substance's compressibility factor, provide the basis for the simplest form of the theorem of corresponding states The reduced pressure is defined as its actual pressure divided by its critical pressure : The reduced temperature of a fluid is its actual temperature, divided by its critical temperature: The reduced specific volume ") of a fluid is computed from the ideal gas law at the substance's critical pressure and temperature: This property is useful when the specific volume and either temperature or pressure are known, in which case the missing third property can be computed directly. 1 Basic principle II Second class Dr. Arkan Jasim Hadi In Kay's method, pseudocritical values for mixtures of gases are calculated on the assumption that each component in the mixture contributes to the pseudocritical value in the same proportion as the mol fraction of that component in the gas. -
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. -
Ch 19. the First Law of Thermodynamics
Ch 19. The First Law of Thermodynamics Liu UCD Phy9B 07 1 19-1. Thermodynamic Systems Thermodynamic system: A system that can interact (and exchange energy) with its surroundings Thermodynamic process: A process in which there are changes in the state of a thermodynamic system Heat Q added to the system Q>0 taken away from the system Q<0 (through conduction, convection, radiation) Work done by the system onto its surroundings W>0 done by the surrounding onto the system W<0 Energy change of the system is Q + (-W) or Q-W Gaining energy: +; Losing energy: - Liu UCD Phy9B 07 2 19-2. Work Done During Volume Changes Area: A Pressure: p Force exerted on the piston: F=pA Infinitesimal work done by system dW=Fdx=pAdx=pdV V Work done in a finite volume change W = final pdV ∫V initial Liu UCD Phy9B 07 3 Graphical View of Work Gas expands Gas compresses Constant p dV>0, W>0 dV<0, W<0 W=p(V2-V1) Liu UCD Phy9B 07 4 19-3. Paths Between Thermodynamic States Path: a series of intermediate states between initial state (p1, V1) and a final state (p2, V2) The path between two states is NOT unique. V2 W= p1(V2-V1) +0 W=0+ p2(V2-V1) W = pdV ∫V 1 Work done by the system is path-dependent. Liu UCD Phy9B 07 5 Path Dependence of Heat Transfer Isotherml: Keep temperature const. Insulation + Free expansion (uncontrolled expansion of a gas into vacuum) Heat transfer depends on the initial & final states, also on the path. -
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.