Gases PDF

Chapter 2 Gases  Elemental Gases Elements that exist as gases at 25°C and 1 atmosphere  Common Gases Table 5.1 Some Substances Found as Gases at 1 atm and 25°C Elements H2 (molecular hydrogen)...

Chapter 2 Gases  Elemental Gases Elements that exist as gases at 25°C and 1 atmosphere  Common Gases Table 5.1 Some Substances Found as Gases at 1 atm and 25°C Elements H2 (molecular hydrogen) N2 (molecular nitrogen) O2 (molecular oxygen) O3 (ozone) F2 (molecular fluorine) Cl2 (molecular chlorine) He (helium) Ne (neon) Ar (argon) Kr (krypton) Xe (xenon) Rn (radon) Compounds HF (hydrogen fluoride) HCL (hydrogen chloride) HBr (hydrogen bromide) HI (hydrogen iodide) CO (carbon monoxide) CO2 (carbon dioxide) CH4 (methane) C2H2 (acetylene) NH3 (ammonia) NO (nitric acid) NO2 (nitrogen dioxide) N2O (nitrous oxide) SO2 (sulfur dioxide) SF6 (sulfur hexafluoride) H2S (hydrogen sulfide) HCN (hydrogen cyanide) *The boiling point of HCN is 26° C, but it is close enough to qualify as a gas at ordinary atmospheric conditions. What is the difference between Hydrogen chloride and Hydrochloric acid?  Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. g/L g/mL  Factors affect gases Pressure Volume (atm) (L) Temperature n (°K) (Mole)  Pressure (1 of 2) Force Pressure = Area (force = mass × acceleration) Units of Pressure 1 pascal ( pa ) = 1N m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa  Example 5.1 (1 of 2) The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized to protect the passengers. What is the pressure in atmospheres in the cabin if the barometer reading is 688 mmHg?  Example 5.1 (2 of 2) Strategy Because 1 atm = 760 mmHg, the following conversion factor is needed to obtain the pressure in atmospheres: 1 atm 760 mmHg Solution The pressure in the cabin is given by 1 atm → 760 mmHg X → 688 1 atm pressure = 688 mmHg  760 mmHg = 0.905 atm  Example 5.2 (1 of 3) The atmospheric pressure in San Francisco on a certain day was 732 mmHg. What was the pressure in kPa?  Example 5.2 (2 of 3) Strategy Here we are asked to convert mmHg to kPa? Because 1 atm = 1.01325  105 Pa = 760 mmHg the conversion factor we needs is 1.01325  105 Pa 760 mmHg  Example 5.2 (3 of 3) Solution The pressure in kPa is 1.01325 x 105 Pa → 760 mmHg X → 732 1.01325 × 105 Pa pressure = 732 mӍ mӍ HӍ g Ӎ × 760 mӍ mӍ HӍ g Ӎ = 9.76 104 Pa = 97.6 kPa  Manometers Used to Measure Gas Pressures a) Closed tube b) Open tube Apparatus for Studying the Relationship Between Pressure and Volume of a Gas  Boyle’s Law P 1V 𝟏 𝑷 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 × 𝑽 Constant temperature P  V = constant Constant amount of gas P1  V1 = P2  V2 Variation in Gas Volume with Temperature at Constant Pressure (1 of 2) Variation of Gas Volume with Temperature at Constant Pressure (2 of 2) Charles’ & Gay-Lussac’s Law VT V = constant  T Temperature must be in Kelvin 𝑽 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 T (K) = t (°C) + 2.73.15 𝑻 V1 T1 = V2 T2  Avogadro’s Law (1 of 2) V  number of moles ( n ) Constant temperature V = constant  n Constant pressure V1 n1 = V2 n2  Summary of Gas Laws Boyle’s Law Charles’s Law Avogadro’s Law (2 of 2)  Ideal Gas Equation 1 Boyle’s law : P  (at constant n and T ) v Charles’s law :V ∝ T (at constant n and P ) Avogadro’s law : V ∝ n (at constant P and T) nT V P nT nT V = constant  =R R is the gas constant P P PV = nRT  Standard Temperature and Pressure The conditions 0 °C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT PV (1 atm ) ( 22.414 L ) R= = nT (1 mol ) ( 273.15 K ) R = 0.08205 L  atm ( mol  K )  Example 5.3 (1 of 3) Sulfur hexafluoride (SF6) is a colorless and odorless gas. Due to its lack of chemical reactivity, it is used as an insulator in electronic equipment. Calculate the pressure (in atm) exerted SF6 by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5 °C  Example 5.3 (2 of 3) Strategy The problem gives the amount of the gas and its volume and temperature. Is the gas undergoing a change in any of its properties? What equation should we use to solve for the pressure? What temperature unit should we use?  Example 5.3 (3 of 3) Solution Because no changes in gas properties occur, we can use the ideal gas equation to calculate the pressure. Rearranging Equation (5.8), we write PV = nRT nRT P= V = (1.82 mol ) ( 0.0821 L  atm K  mol ) ( 69.5 + 273) K 5.43 L = 9.42 atm  Example 5.4 (1 of 3) Calculate the volume (in L) occupied by 7.40 g of NH 3 at STP. NH3 PV = nRT 𝒏𝑹𝑻 𝑽 = 𝑷 V=?? , R=0.082 P= 1 atm T= 0 + 273 = 273 K n= m/Mwt n= 7.4 / 17.03= 0.4345 mole V = nRT / P V= (0.4353× 0.0821× 273)/1 = 9.74 L  n= m/Mwt n= 7.4 / 17.03 = 0.4345 mole 1 mole → 22.41 L 0.4345 → X X= (0.4353× 22.41)/ 1= 9.74 L  Example 5.5 (1 of 3) An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? A scientific research helium balloon.  Example 5.5 (2 of 3) Strategy The amount of gas inside the balloon and its temperature remain constant, but both the pressure and the volume change. What gas law do you need? Solution We start with Equation (5.9) PV PV 1 1 = 2 2 n1T1 n2T2 Because n1 = n2 and T1 = T2 , 1 1 = PV PV 2 2 Which is Boyle’s law [see Equation (5.2)].  Example 5.5 (3 of 3) 1 1 = PV PV 2 2 The given information is tabulated: Intial Conditions Final Conditions P1 = 1.0 atm P2 = 0.40 atm V1 = 0.55L V2 = ? Therefore, P1 V2 = V1  P2 1.0 atm = 0.55 L  0.40 atm Check = 1.4 L When pressure applied on the balloon is reduced (at constant temperature), the helium gas expands and the balloon’s volume increases. The final volume is greater than the initial volume, so the answer is reasonable.  Example 5.6 (1 of 3) Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure (in atm). Electric lightbulbs are usually filled with argon.  Example 5.6 (2 of 3) Strategy The temperature and pressure of argon change but the amount and volume of gas remain the same. What equation would you use to solve for the final pressure? What temperature unit should you use? Solution Because n1 = n2 and V1 = V2 , Equation ( 5.9 ) becomes P1 P2 = T1 T2 Which is Charles’s law [see Equation (5.6)].  Example 5.6 (3 of 3) Next we write Intial Conditions Final Conditions P1 = 1.20 atm P2 = ? T1 = (18 + 273) K = 291 K T2 = (85 + 273)K = 353 K The final pressure is given by T1 P2 = P1  T2 358 K = 1.20 atm  291 K = 1.48 atm Check At constant volume, the pressure of a given amount of gas is directly proportional to its absolute temperature. Therefore the increase in pressure is reasonable  Example 5.7 (1 of 5) A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressure is 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.  Example 5.7 (2 of 5) Strategy In solving this kind of problem, where a lot of information is given, it is sometimes helpful to make a sketch of the situation, as shown here: What temperature unit should be used in the calculation?  Example 5.7 (3 of 5) Solution According to Equation (5.9) PV PV 1 1 = 2 2 n1T1 n2T2 We assume that the amount of air in the bubble remains constant, that is, n1 = n2 so that PV PV 1 1 = 2 2 T1 T2 which is Equation (5.10).  Example 5.7 (4 of 5) The given information is summarized: Intial Conditions Final Conditions P1 = 6.4 atm P2 = 1.0 atm V1 = 2.1 mL V2 = ? T1 = (8 + 273) K = 281 K T2 = (25 + 273) K = 298 K Rearranging Equation (5.10) gives P1 T2 V2 = V1   P2 T1 6.4 atm 298 K = 2.1 mL   1.0 atm 281 K = 14 mL  Example 5.7 (5 of 5) Check We see that the final volume involves multiplying the initial volume by a ratio of pressures ( P1 P2 ) and a ratio of temperatures ( T2 T1). Recall that volume is inversely proportional to pressure, and volume is directly proportional to temperature. Because the pressure decreases and temperature increases as the bubble rises, we expect the bubble’s volume to increase. In fact, here the change in pressure plays a greater role in the volume change.