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Putting Gases to Work Cracking under pressure.

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Presentation on theme: "Putting Gases to Work Cracking under pressure."— Presentation transcript:


2 Putting Gases to Work Cracking under pressure


















20 Pumps 1 ½ water stream 260 feet.



23 Gas : Alteration of Latin chaos space, chaos Date: 1779

24 The gas engine is one of the wonders of the 19th century. Now, within three years of the 20th century, it is a novel machine, eagerly sought by many people. It is thought by persons who have not studied its principles that it is a steam-engine, using gas or gasoline as fuel for the purpose of making steam. This is erroneous. Gas and gasoline in specific proportion with air are explosive material.

25 The expansive force derived from explosion of these materials in the cylinder is the force that is substituted for the expansive force of steam. Hence, owing to the economy of this method as a means of deriving power, the steam engine and boiler are fast disappearing, and the Gas Engine is taking their place for small power.

26 After careful and intelligent tests by experts, it is generally admitted that the perfect steam engine does not convert more than 10 per cent of the heat efficiency into indicated work, and that ordinary engines and boilers do not realize over 4 per cent. Compare this with the effective energy produced by … the combustion of gas.... We have immediately gained from 10 to 20 per cent of effectual energy from the heat units produced.

27 It is only a matter of time when the prejudice that, as usual, exists against any innovation, the ungrounded fear of explosions and other difficulties, will be overcome and the superiority of the gas engine over the wasteful steam engine and boiler will be established.

28 The unsightly smoke stacks, belching forth smoke and soot, will be relegated to the scrap heap. The atmosphere of our manufacturing cities will be as clear as that of the country.


30 In the next decade the steam engine will occupy the same relative position to the gas engine that the flint and steel now do to the lucifer match.

31 the tallow dip to the electric light...

32 ...the stage coach to the modern electric street cars, and civilization will record another grand stride toward the millennium.




36 9 0 18 9 hits 9 sec 1 hit sec = 18 hits 9 sec 2 hits sec = 00 01 02 03 04 05 06 07 08 09 Seconds


38 V=0.5, P=2 V=0.1, P=10 V=10, P=0.1 V=6, P=5 V=3, P=10 V=30, P=2

39 9 0 18 9 hits 9 sec 1 hit sec = 18 hits 9 sec 2 hits sec = 00 01 02 03 04 05 06 07 08 09 Seconds 27 ºC = 300 K 0 K

40 9 0 18 9 hits 4.5 sec 2 hits sec = 00 01 02 03 04 05 06 07 08 09 Seconds 27 ºC = 300 K 327 ºC = 600 K 0 K

41 15 psi, 300 K 30 psi 3 psi 1 psi 600 K 60 K 20 K

42 15 psi, 2 L, 300 K 30 psi 6 psi 3 psi 2 psi 600 K 60 K 20 K 2 L 1 L 2 L 1 L





47 1.4 x 1.4 = 2 doubles 2 x 2 = 2 1.5 x 1.33 = 2

48 P is pressure measured in atmospheres. V is volume measured in Liters n is moles of gas present. R is a constant that converts the units. It's value is 0.0821 atmL/molK T is temperature measured in Kelvin. Simple algebra can be used to solve for any of these values. P = nRT V = nRT n = PV T = PV R = nT V P RT nR PV


50 Pressure=1 atmosphere Volume=1 Liter n = 1 mole R=0.0821 What is the temperature?



53 Facts: 2 Liter bottle ¼ lb = 454 g ÷ 4 = 114g PV = nRT or P = nRT What pressure could be reached when ¼ lb of dry ice is placed in this 2 liter bottle? Temperature that night was 86 °F (30 °C) V CO 2 = 12g/mol + 2*16g/mol = 44 g/mol 114 g = mol 44 g 2.61 mol

54 2.6 mol x 0.0821 atm*L x 303 K mol*K P = 2.0 L n R T V P = 32.3 atmospheres 32.3 atm = psi 1 atm 475 14.7 psi

55 760 mm of Hg 760 torr 29.9 in. of Hg 1 Atmosphere 14.7 lbs. per sq. in. Temperature conversions: Kelvin = Celsius + 273 O C = ( O F -32) x 5/9 O F= O C x 9/5 + 32 CONVERSIONS All Equal Evangelista Torricelli


57 Single-condition Problems: One set of conditions is given How many moles of oxygen gas is present if the cylinder holds 25 Liters, has a pressure of 1.2 atmospheres, and a temperature of 25 O C? PV = nRT is solved for n n = PV RT Also change 25 O C to Kelvin K=25+273 K=298 Plug in the amounts: n = 1.2 atm x 25 L n = 1.23 moles O 2 0.0821atmL/molK x 298 K


59 When a pressure cooker is used, what When a pressure cooker is used, what causes the increased pressure? PV=nRT PV=nRT P=nRT V Temperature goes from 25 o C to 100 o C Turn to Kelvin by adding 273 to Celsius 297K to 373K 75K/297K=25% increase in pressure


61 75K / 297K= 25% 297K 373K


63 You are on a camping trip and one tire has a slow leak. Finally it goes flat and you dont have a spare tire. You suggest crushing some of the dry ice you had brought along and funneling it into the tire through the tire valve. How many grams of dry ice would you need to blow up a tire with a volume of 80 liters and pressure of 32 psi? Current temp is 25 O C. Change 32 psi to atm and 25 O C to Kelvin 32 psi x 1 atm = 2.177 atm 14.7 psi 25 O C= 273+25=298 K Solve PV=nRT for n (moles) n = PV RT n = 2.177 atm x 80 Liters 0.0821 atmL/molK x 298K n = 7.118 moles > 7.118 mol x 44.01 g/mol = 313.3 g or 310 g 310 g x 1 lbs per 453 grams = 0.68 lbs.

64 P 1 V 1 =n 1 RT 1 P 2 V 2 =n 2 RT 2 n 1 T 1 n 1 T 1 n 2 T 2 n 2 T 2 P 1 V 1 = R P 2 V 2 = R n 1 T 1 n 2 T 2 P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

65 Single condition problem 3 pounds (1,362 g) of dry ice (frozen CO 2 ) is packed in a 1 gallon (3.785 L) glass jar. What will be the pressure in the jar after the dry ice turns to gas and warms to 20 O C? (report pressure in psi and assume the jar doesn't explode) Change 1,362g to moles 1,362g x 1 mole = 30.95 moles 44.01 g Change 20 O C to 293 K solve PV=nRT for P P = nRT P = 30.95 moles x 0.0821 atmL/molK x 293K V 3.785 Liters P = 196.7 atm. Change to psi 196.7 atm x 14.7 psi = 2,891 psi 1 atm P = 2,891 psi

66 Molar mass = gRT PV 12. In 1984 in a village in the African nation of Cameroon, 1,700 people died from carbon dioxide poisoning when the nearby lake released huge amounts of CO 2. As a early warning device, a pump could be designed to pump samples of air into a 4 liter container until the mass of the container grew by 22 grams. At that point a pressure reading could be taken as well as the temperature. Lets say the pressure is 3.1 atmospheres and the temperature is 30 O C (86 O F). What molar mass would the device calculate? Plug in the values into the equation Molar mass = 22 g x 0.0821 atmL/molK x (273+30) K 3.1 atm x 4 L Molar mass = 44.13 g/mole

67 Kinetic-Molecular Theory Explains the Gas Laws: Velocities of Molecules Since the number of molecules of gas present anywhere, N, is given by nNA, it is possible to rewrite the overall kinetic energy expression as Ek = nNAmv2/2 = nMv2/2 In this expression m is the mass of a single molecule, so the molar mass M is the product mNA. Since the kinetic energy is also 3nRT/2, the square root of the square of the mean velocity, known as the root-mean-square velocity v(rms), of the molecules of the gas is proportional to the square root of its molar mass. The root-mean-square velocity, like the actual distribution of velocities embodied in the Maxwell law, is a function only of the absolute temperature. v(rms) = (the square root of)3RT/M Example. Let us calculate the root-mean-square velocity of oxygen molecules at room temperature, 25oC. Using v(rms) = (the square root of)3RT/M, the molar mass of molecular oxygen is 31.9998 g/mol; the molar gas constant has the value 8.3143 J/mol K, and the temperature is 298.15 K. Since the joule is the kg-m2/s2, the molar mass must be expressed as 0.0319998 kg/mol. The root-mean-square velocity is then given by: v(rms) = (the square root of)3(8.3143)(298.15)/(0.0319998) = 482.1 m/s A speed of 482.1 m/s is 1726 km/h, much faster than a jetliner can fly and faster than most rifle bullets. The very high speed of gas molecules under normal room conditions would indicate that a gas molecule would travel across a room almost instantly. In fact, gas molecules do not do so. If a small sample of the very odorous (and poisonous!) gas hydrogen sulfide is released in one corner of a room, our noses will not detect it in another corner of the room for several minutes unless the air is vigorously stirred by a mechanical fan. The slow diffusion of gas molecules which are moving very quickly occurs because the gas molecules travel only short distances in straight lines before they are deflected in a new direction by collision with other gas molecules. The distance any single molecule travels between collisions will vary from very short to very long distances, but the average distance that a molecule travels between collisions in a gas can be calculated. This distance is called the mean free path l of the gas molecules. If the root-mean-square velocity is divided by the mean free path of the gas molecules, the result will be the number of collisions one molecule undergoes per second. This number is called the collision frequency Z1 of the gas molecules. The postulates of the kinetic-molecular theory of gases permit the calculation of the mean free path of gas molecules. The gas molecules are visualized as small hard spheres. A sphere of diameter d sweeps through a cylinder of cross-sectional area (pi)d2 and length v(rms) each second, colliding with all molecules in the cylinder, as shown in the Figure below. Figure is not available. The radius of the end of the cylinder is d because two molecules will collide if their diameters overlap at all. This description of collisions with stationary gas molecules is not quite accurate, however, because the gas molecules are all moving relative to each other. Those relative velocities range between zero for two molecules moving in the same direction and 2v(rms) for a head-on collision. The average relative velocity is that of a collision at right angles, which is v(rms) times the square root of 2. The total number of collisions per second per unit volume, Z1, is Z1 = (pi)d2v(rms)(the square root of)2 This total number of collisions must now be divided by the number of molecules which are present per unit volume. The number of gas molecules present per unit volume is found by rearrangement of the ideal gas law to n/V = p/RT and use of Avogadro's number, n = N/NA; thus N/V = pNA/RT. This gives the mean free path of the gas molecules, l, as (v(rms)/Z1)/(N/V) = l = RT/(pi)d2pNA(the square root of)2 According to this expression, the mean free path of the molecules should get longer as the temperature increases; as the pressure decreases; and as the size of the molecules decreases. Example. Let us calculate the length of the mean free path of oxygen molecules at room temperature, 25oC, taking the molecular diameter of an oxygen molecule as 370 pm. Using the formula for mean free path given above and the value of the root-mean-square velocity v(rms) calculated in the previous example, l = (8.3143 kg m2/s2K mol)(298.15 K)/3.14159(370 x 10-12 m)2(101325 kg/m s2) (6.0225 x 10+23 mol-1)((the square root of)2), so l = 6.7 x 10-8 m = 67 nm. The utility of SI units and of the quantity calculus in this example should be obvious. The apparently slow diffusion of gas molecules takes place because the molecules travel only a very short distance before colliding. At room temperature and atmospheric pressure, oxygen molecules travel only (6.7 x 10-8 m)/(370 x 10-12 m) = 180 molecular diameters between collisions. The same thing can be pointed out using the collision frequency for a single molecule Z1, which is the root-mean-square velocity divided by the mean free path: Z1 = (pi)d2pNA(the square root of)2/RT = v(rms)/l For oxygen at room temperature, each gas molecule collides with another every 0.13 nanoseconds (one nanosecond is 1.0 x 10-9 s), since the collision frequency is 7.2 x 10+9 collisions per second per molecule. For an ideal gas, the number of molecules per unit volume is given using pV = nRT and n = N/NA as N/V = NAp/RT which for oxygen at 25oC would be (6.0225 x 10+23 mol-1)(101325 kg/m s2)/(8.3143 kg m2/s2 K mol)(298.15 K) or 2.46 x 10+25 molecules/m3. The number of collisions between two molecules in a volume, Z11, would then be the product of the number of collisions each molecule makes times the number of molecules there are, Z1N/V, except that this would count each collision twice (since two molecules are involved in each one collision). The correct equation must be Z11 = (pi)d2p2NA2v(rms) (the square root of)2/2R2T2 If the molecules present in the gas had different masses they would also have different speeds, so an average value of v(rms) would be using a weighted average of the molar masses; the partial pressures of the different gases in the mixture would also be required. Although such calculations involve no new principles, they are beyond our scope. However, the number of collisions which occur per second in gases and in liquids are extremely important in chemical kinetics, so we shall return to this topic in other sections. Graham's Law of Effusion and Diffusion Root-mean-square velocities of gas molecules are sometimes directly useful, but the comparison of velocities explains the results of, and is useful in, studies of effusion of molecules through a small hole in a container or diffusion of molecules through porous barriers. The comparison between two gases is most conveniently expressed as: v(rms)1/v(rms)2 = (the square root of)(M2/M1) = (the square root of)(d2/d1) This equation gives the velocity ratio in terms of either the molar mass ratio or the ratio of densities d. The ratio of root-mean-square velocities is also the ratio of the rates of effusion, the process by which gases escape from containers through small holes, and the ratio of the rates of diffusion of gases. This equation is called Graham's law of diffusion and effusion because it was observed by Thomas Graham (1805-1869) well before the kinetic-molecular theory of gases was developed. As an empirical law, it simply stated that the rates of diffusion and of effusion of gases varied as the square root of the densities of the gases. Graham's law is the basis of many separations of gases. The most significant is the separation of the isotopes of uranium as the gases 238UF6 and 235UF6. Fluorine has only one isotope, so the separation on the basis of molar mass is really a separation on the basis of isotopic mass. Example. The ratio of root-mean-square velocities of 238UF6 and 235UF6 can be calculated as follows. The molar mass of 238UF6 is 348.0343 and the molar mass of 238UF6 is 352.0412. The mass ratio is 1.011513 and the ratio of root-mean-square velocities is 1.00574. Although the difference is small, many kilograms of 235U have been separated using this difference in the gas-diffusion separation plant at Oak Ridge, Tennessee, U. S. A. This plant prepared the uranium for the Manhattan Project of the Second World War and produced the uranum used in the uranium atomic bomb dropped on Japan in 1945. [NEXT: Equations of State for Gases] [PREVIOUS: Kinetic-Molecular Theory Explains the Gas Laws: Partial Pressure] Copyright 1995 James A. Plambeck ( Updated September 26, 1995.

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