10. Gases. P= g h Pressure. Pressure is defined as the force across a unit area. Force N

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1 0. Gases 0. ressure ressure is defined as the force across a unit area. Force N ascal, a Area m In chemistry, the SI unit for pressure, the ascal (a), is typically too small to be of practical use. Typically the pressure in represented in units of kiloascals, ka. There are several devices used to measure pressure. One such device, the barometer, was invented in 643 by Evangelista Torrecelli. A schematic diagram of the barometer is presented below. acuum h g h External ressure Hg The height of the mercury in the column is proportional to the pressure exerted on the bowl of mercury. The ospheric pressure varies according to climatic conditions and also to altitude (height above the earth s surface). At sea level, on the average, the height of a column of mercury will be. While this is not an appropriate pressure unit, it is proportional to pressure according to the formula: gh density of the liquid (i.e., Hg), kgm 3 (at 0 C) g acceleration due to gravity, ms h height of the column of liquid, m Since the density of mercury and gravity are constants under normal conditions, the height is linearly proportional to pressure through known constants, we can (improperly) refer to pressure by the height of the column of mercury. Thus, page 87

2 ( kgm 3 ) ( ms ) ( m) kgm s a 0.3 ka Therefore a column of mercury high is equivalent to a pressure of 0.3 ka. In the British System of units, pressure is measured in pounds per square inch, psi. The standard osphere then is: Standard osphere: ka 4.70 psi torr. As an exercise, show that in the conversion of SI to British units, 0.3 ka 4.70 psi. The unit is also referred to as a torr in honour of Torricelli. The various expressions for the standard osphere can be used as conversion factors. Example: If the ospheric pressure in the lab is found to be 753.3, what is the pressure in units of and in units of ka? 0. The Gas Laws ka ka The systematic study of gases began in around the mid-seventeenth century and continues today. This is in part because so many reactions studied in chemistry occur in the gas phase and because much of our understandings of fields such as thermodynamics (the study of energy transformations) have their origins in gas theory. Finally gas theory was instrumental in establishing the existence of atoms. 0.. BOYLE S LAW In 66, the British chemist, Robert Boyle, published a manuscript describing his studies on the relationship between the pressure and the volume of a gas. In summary, Boyle found that the volume of a sample of gas varied inversely with an externally applied pressure as long as the temperature is held constant. That is: which means that the product of pressure and volume is invariant as long as the temperature and number of es are held constant: This formula means if we know the pressure and volume of a sample of gas at a given temperature, then if the pressure is changed (to ) we can predict the volume ( ) that the gas would occupy at this new pressure. 0.. CHARLES S LAW In 787, Jacques Charles studied the effect of temperature on the volume of a gas. Fifteen years later, in 80, Joseph Gay-Lussac, carried out careful similar studies and published his work. What these scientists found was that volume increases in proportion to temperature: T Similarly to Boyles Law, given the relation between the volume and temperature of a sample of gas at constant pressure, if the temperature of the gas is changed (to T ), we can predict the volume the gas ( ) that the gas would occupy at this new temperature: page 88

3 T Note that the absolute Kelvin temperature is always used in gas law calculations because in a plot of Celsius temperature gives the following formula: a T C + b It turns out that when experimental data for gas volume is plotted as a function of Celsius temperature, regardless of the gas, the temperature intercept is always 73.5 C. T Gas olume 73.5 C 0 C 0..3 AOGADRO S LAW In 8, the Italian scientist, Amadeo Avogadro proposed that for a gas at constant pressure and temperature, the volume of gas is directly proportional to the number of es of gas present. Thus if the number of es of gas is changed from n to n, the volume will correspondingly be changed from a volume to, 0.3 The Ideal Gas Law n n n We have seen that over a period of 50 years various relationship between the volume of a gas and variables such pressure, temperature and the number of es have been developed: Boyle Law: Charles and Gay-Lussac: T Avogadro s Law: n Combining these laws yields the Combined Gas Law: The proportionality can be removed as usual: Temperature ( C) T and n are constant and n are constant and T are constant nt page 89

4 nt nt What this means is that for any gas the ratio of to nt is invariant (this means that the ratio is a constant). The constant is conventionally given the symbol R. or more conventionally R nt nrt Depending on the choice of units for pressure, the universal gas constant: R kalk LK Note that these at the same number expressed in different systems of units. In calculations, if the pressure is represented in units of ka, the appropriate form of R to use is kalk ; if the pressure is represented in units of, the appropriate form of R to use is LK. Example: A certain sample of gas occupies 0.0 L at.50 at a given temperature. What volume would the same sample of gas occupy if the pressure is increased to.0 when the temperature is held constant? Initial State (,, n, T ) Final State (,, n, T ) n R T n R T Since the sample of gas is the same n n. Since the temperature is held constant, T T. Thus: L 7.50 L.0 Example: Calculate the ecular mass of a gas if a 4.75 g sample of the gas occupies 4.84 L at torr and 5.0 C. nrt 0.3 ka torr 00.4 ka T 5.0 C K (00.4 ka)(4.84 L) n 0.96 RT (8.34 ka L K )(98. K) 4.75 g MM 7.8 g 0.96 page 90

5 Example: The density of a gas was found to be.95 gl at ST (Standard Temperature and ressure). Calculate its ecular mass. ST.000 and 0 C. m Recall that nrt m RT MM RT m MM m RT RT MM MM MM (.95 )( )(73.5 ) RT g ml L K K 43.7 g Gas Stoichiometry Example: In a reaction used for making chlorine gas, 5.4 ml of 3.4 M hydrochloric acid are added to.50 g MnO. Chlorine gas is produced according to the balanced chemical equation: MnO (s) + 4 HCl (aq) MnCl (aq) + Cl (g) + H O (g) Determine the volume of chlorine gas formed at 54 and 36.0 C. Atomic Mass: Mn O 6.00 H.008 Cl Step : Write the balanced chemical equation The balanced equation is given above. Step : Calculate the number of es of given compounds i) Molecular masses: MnO (54.94) + (6.00) HCl Step 3: Determine the Limiting Reactant. Assume MnO is the Limiting Reactant nmno.50 g g n (3.4 L )(0.054 L) HCl 4 HCl nhcl MnO 0.5 HCl are required MnO Since the actual es of HCl < 0.5 HCl required our assumption that MnO is the limiting reactant is not correct. Thus HCl is the limiting reactant. Step 4: Calculate the number of es chlorine in the chemical equation produced based on the Limiting Reactant, HCl. page 9

6 Cl n Cl 0.07 HCl 4 HCl Step 5: Calculate the volume of chlorine gas formed at 54 and 36.0 C.. nrt T K (0.07 )( L K )(309. K) L 0.73 Example: If a 3.5 g sample of acetylene, C H, is burned in 7.0 g oxygen, what volume of gas will be present at 76.7 and.5 C? What volume will the gas occupy at ST? Atomic Mass: C.0 H.008 O 6.00 C H (g) + 5 O (g) 4 CO (g) + H O (l) Step : Calculate the number of es glucose ii) Molecular masses: C H (.0) + (.008) O (6.00) 3.00 iii) Moles: n 3.5 g 0.5 CH g n 7.0 g O 3.00 g Step 3: Determine the Limiting Reactant. Assume C H is the Limiting Reactant 5 O n 0.5 C H 0.3 O O are required C H Since the Actual es of O > 0.3 O required, our assumption that acetylene is the limiting reactant is correct. Step 4: Calculate the number of es of the other compounds in the chemical equation at the end of the reaction based on the Limiting Reactant. At the end of the reaction: n 0.00 CH n O 4 CO n 0.5 C H 0.50 CO C H Since at page 9

7 and.5 C, water is essentially a liquid. Thus the total es of gas will be due to the contributions of CO and O. n n + n tot O CO Step 5: Calculate the volume of gas at the end of the reaction at 76.7 and.5 C. nrt (0.78 )( )( ) nrt L K K 8.9 L.00 The volume of gas at ST is: (0.78 )( )(73.5 ) nrt L K K 7.5 L.000 page 93