1.2 THE GAS LAWS 11 Substitution of the data then gives that the pressure is actually 183 atm under these conditions,so heo hthperectdter What temperature oldesut in the same sampleexerting The perfect gasequation is of the greatest importance in physical chemistry because it is used to derivea wide range of relations that are used throughout thermodynamics. However,it is also of considerable practical utility for calculating the properties of a under the condion call nce,the molar volume,V=VIn,of a per stan t temperature and p RTIP obe 24 789 d ). y calcul and pressure(STP),was 0C and I atm:at STP,the molar vol 22.414dm mol.Among otherapplications.can 18 can be used to discuss proceses in the atmosphere that give rise to the weather. The biggest sample of gas readily accessible to us is the atmosphere,a mixture of gases with the composition summarized in Table 1.3.The composition is maintained mod. callebudision and convecon (winds.partcuarly the local turbulenc and with the loca pm Table1 The composition of dry air at sea level Percentage Component By volume By mass 20.95 2314 128 Carbon dioxide.CO 003 0047 Hydrogen,H, 5.0×10- 20×10- Neon.Ne 1.8×10- 1.3×10- Helium,He 52×10 7.2×10 Methane,CH 2.0×10- 11×10 Krypton,Kr 1.1×10 3.2×10 Nitric oxide.NO 5.0x10 Xenon,Xe 87×101
1.2 THE GAS LAWS 11 Table 1.3 The composition of dry air at sea level Percentage Component By volume By mass Nitrogen, N2 78.08 75.53 Oxygen, O2 20.95 23.14 Argon, Ar 0.93 1.28 Carbon dioxide, CO2 0.031 0.047 Hydrogen, H2 5.0 × 10 −3 2.0 × 10 −4 Neon, Ne 1.8 × 10 −3 1.3 × 10 −3 Helium, He 5.2 × 10 −4 7.2 × 10 −5 Methane, CH4 2.0 × 10 −4 1.1 × 10 −4 Krypton, Kr 1.1 × 10 −4 3.2 × 10 −4 Nitric oxide, NO 5.0 × 10 −5 1.7 × 10 −6 Xenon, Xe 8.7 × 10 −6 1.2 × 10 −5 Ozone, O3: summer 7.0 × 10 −6 1.2 × 10 −5 winter 2.0 × 10 −6 3.3 × 10 −6 Substitution of the data then gives Experiment shows that the pressure is actually 183 atm under these conditions, so the assumption that the gas is perfect leads to a 10 per cent error. Self-test 1.3 What temperature would result in the same sample exerting a pressure of 300 atm? [900 K] The perfect gas equation is of the greatest importance in physical chemistry because it is used to derive a wide range of relations that are used throughout thermodynamics. However, it is also of considerable practical utility for calculating the properties of a gas under a variety of conditions. For instance, the molar volume, Vm = V/n, of a perfect gas under the conditions called standard ambient temperature and pressure (SATP), which means 298.15 K and 1 bar (that is, exactly 105 Pa), is easily calculated from Vm = RT/p to be 24.789 dm3 mol−1 . An earlier definition, standard temperature and pressure (STP), was 0°C and 1 atm; at STP, the molar volume of a perfect gas is 22.414 dm3 mol−1 . Among other applications, eqn 1.8 can be used to discuss processes in the atmosphere that give rise to the weather. IMPACT ON ENVIRONMENTAL SCIENCE I1.1 The gas laws and the weather The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases with the composition summarized in Table 1.3. The composition is maintained moderately constant by diffusion and convection (winds, particularly the local turbulence called eddies) but the pressure and temperature vary with altitude and with the local conditions, particularly in the troposphere (the ‘sphere of change’), the layer extending up to about 11 km. p2 500 300 =× = ( ) 100 167 K K atm atm
12 1 THE PROPERTIES OF GASES 30 In the troposphere the average temperature is 15C at sea level,falling to-57C at the bottom ofthe tropopause at 11 km.This variation is much less pronounced when 20 expressed on the Kelvin scalc,ranging,an average of 268 K.If we the ropopause. 15 P=poe-h 10 where p is the pressure at sea level and his a constant approximately equal to 8 km 6/ More specifically,H=RT/Mg,where Mis the average molar mass of air and Tis the temperature.The barometric formula fits the observed pressure distribution quite well even for regions well above the troposphere (see Fig.1.10).It implies that the Pressure,p air is less dense than the s l of c lair.As cools.Cool air can absorb lower concentrations of water vapour than warm air,so the variation of temperature moisture forms clouds.Cloudy skies can therefore be associated with rising air and with altitude. clear skies are often associated with descending air. Exploraion How would the graph The motion or air th upper a titudes may I ofm the shown in the s and a om othe reg ons.Th regions C Thes)ar ne the accompanving wea er map (Fi LI).The lines of constan sure differin by 4 mbar (400 Pa.about 3 Torr)- marked on it are called isobars The elongated regions of high and low pressure are known,respectively,as ridges and troughs comection.Horizontal pre ure t in the flo that we call wind（(see Fig-1.12 Wind north in the r and from th e the Earth ng slo wly(a the Winds tr rallel to the isoba ng most ra left in the Northern hemisphere and to the right in the Southern hemisphere.At the surface,where wind speeds are lower,the winds tend to travel perpendicular to the isobars from high to low pressure.This differential motion results in a spiral outward case flow of air clockwise in the Northern hemisphere around a high and an inward counter gh pressure is 。A by compresion as it descends so regions of high pres Wind sure are associated with high surface temperatures.In winter,the cold surface air may prevent the complete fall of air,and result in a temperature inversion,with a layer of warm air over a layer of cold air.Geographical conditions may also trap cool air,as otation in Los Angeles,and the photochemical pollutants we know as smog may be trapped under the warm layer. (b)Mixtures of gases When dealing with gaseous mixtures,we often need to know the contribution that each component makes to the total pressure ofthe sample.The partialpressure.Pof in the a gas J ina mixture(any gas,not just a perfect gas),is defined as PI=xP [1.13
12 1 THE PROPERTIES OF GASES In the troposphere the average temperature is 15°C at sea level, falling to –57°C at the bottom of the tropopause at 11 km. This variation is much less pronounced when expressed on the Kelvin scale, ranging from 288 K to 216 K, an average of 268 K. If we suppose that the temperature has its average value all the way up to the tropopause, then the pressure varies with altitude, h, according to the barometric formula: p = p0e−h/H where p0 is the pressure at sea level and H is a constant approximately equal to 8 km. More specifically, H = RT/Mg, where M is the average molar mass of air and T is the temperature. The barometric formula fits the observed pressure distribution quite well even for regions well above the troposphere (see Fig. 1.10). It implies that the pressure of the air and its density fall to half their sea-level value at h = H ln 2, or 6 km. Local variations of pressure, temperature, and composition in the troposphere are manifest as ‘weather’. A small region of air is termed a parcel. First, we note that a parcel of warm air is less dense than the same parcel of cool air. As a parcel rises, it expands adiabatically (that is, without transfer of heat from its surroundings), so it cools. Cool air can absorb lower concentrations of water vapour than warm air, so the moisture forms clouds. Cloudy skies can therefore be associated with rising air and clear skies are often associated with descending air. The motion of air in the upper altitudes may lead to an accumulation in some regions and a loss of molecules from other regions. The former result in the formation of regions of high pressure (‘highs’ or anticyclones) and the latter result in regions of low pressure (‘lows’, depressions, or cyclones). These regions are shown as H and L on the accompanying weather map (Fig. 1.11). The lines of constant pressure—differing by 4 mbar (400 Pa, about 3 Torr)—marked on it are called isobars. The elongated regions of high and low pressure are known, respectively, as ridges and troughs. In meteorology, large-scale vertical movement is called convection. Horizontal pressure differentials result in the flow of air that we call wind (see Fig.1.12). Winds coming from the north in the Northern hemisphere and from the south in the Southern hemisphere are deflected towards the west as they migrate from a region where the Earth is rotating slowly (at the poles) to where it is rotating most rapidly (at the equator). Winds travel nearly parallel to the isobars, with low pressure to their left in the Northern hemisphere and to the right in the Southern hemisphere. At the surface, where wind speeds are lower, the winds tend to travel perpendicular to the isobars from high to low pressure. This differential motion results in a spiral outward flow of air clockwise in the Northern hemisphere around a high and an inward counterclockwise flow around a low. The air lost from regions of high pressure is restored as an influx of air converges into the region and descends. As we have seen, descending air is associated with clear skies. It also becomes warmer by compression as it descends, so regions of high pressure are associated with high surface temperatures. In winter, the cold surface air may prevent the complete fall of air, and result in a temperature inversion, with a layer of warm air over a layer of cold air. Geographical conditions may also trap cool air, as in Los Angeles, and the photochemical pollutants we know as smog may be trapped under the warm layer. (b) Mixtures of gases When dealing with gaseous mixtures, we often need to know the contribution that each component makes to the total pressure of the sample. The partial pressure, pJ , of a gas J in a mixture (any gas, not just a perfect gas), is defined as pJ = xJp [1.13] p0 Pressure, p 0 Altitude, /km h 6 10 15 20 30 0 H L H H H Fig. 1.10 The variation of atmospheric pressure with altitude, as predicted by the barometric formula and as suggested by the ‘US Standard Atmosphere’, which takes into account the variation of temperature with altitude. Exploration How would the graph shown in the illustration change if the temperature variation with altitude were taken into account? Construct a graph allowing for a linear decrease in temperature with altitude. Fig. 1.11 A typical weather map; in this case, for the United States on 1 January 2000. L L Wind Rotation N S Fig. 1.12 The flow of air (‘wind’) around regions of high and low pressure in the Northern and Southern hemispheres
1.2 THE GAS LAWS 13 wherex is the mole fraction of the component J,the amount of expressed as a frac- tion of the total amount of molecules,n,in the sample: n=n+na+.· [1.14 When no J molecules are present,=0;when only I molecules are present,=1. It follows from the definition ofx that,whatever the composition of the mixture .=I and therefore that the sum of the partial pressures is equal to the tota p+p+.=(++.p=p (1.15) artial pr ressure as defined in can 1.13 is also the pressure that each gas would occupy if it occupied the same container alone at the same temperature.The latter is the original meaning of 'partial pressure'.That identification was the basis of the original formulation of Dalton's law: The pressure exerted by a mixture of gases is the sum ofthe pressures that each one would exist if it occupied the container alone. ssure (as definec in e qn1.13)and tota gases and the id that the gas would itsown is alid only for a per Example 1.3 Calculating partial pressures 0x:23. Method weexpect species with a high mole fraction to hay ortionally high partial pressure.Partial pressures are defined by ean 1.13.To use the equation.we need the mole fractions of the components.To calculate mole fractions,which are defined by eqn 1.14,we use the fact that the amount of molecules J of molar mass M in a sample of mass m is n=m/M The mole fractions are independent of the of the sample,so we can choose the latter to be 100 g(v hich make pem00g山7网5可Thus,thea N2 present is Answer The amounts ofeach tvpe of molecule present in 100 g of air,in which the masses of N,O,and Ar are 75.5g.23.2g.and 1.3 g.respectively,are 755g 755 n(N2) 2802gmo-2802mal 23.2g 23.2 n(Ar) 1.3g 13 39.95gmo39.95mol These three amounts work out as2.69 mol,0.725 mol,and 0.033 mol,respectively, for a total of 3.45 mol.The mole fractions are obtained by dividing each of the
1.2 THE GAS LAWS 13 where xJ is the mole fraction of the component J, the amount of J expressed as a fraction of the total amount of molecules, n, in the sample: n = nA + nB + · · · [1.14] When no J molecules are present, xJ = 0; when only J molecules are present, xJ = 1. It follows from the definition of xJ that, whatever the composition of the mixture, xA + xB + · · · = 1 and therefore that the sum of the partial pressures is equal to the total pressure: pA + pB + · · · = (xA + xB + · · · )p = p (1.15) This relation is true for both real and perfect gases. When all the gases are perfect, the partial pressure as defined in eqn 1.13 is also the pressure that each gas would occupy if it occupied the same container alone at the same temperature. The latter is the original meaning of ‘partial pressure’. That identification was the basis of the original formulation of Dalton’s law: The pressure exerted by a mixture of gases is the sum of the pressures that each one would exist if it occupied the container alone. Now, however, the relation between partial pressure (as defined in eqn 1.13) and total pressure (as given by eqn 1.15) is true for all gases and the identification of partial pressure with the pressure that the gas would exert on its own is valid only for a perfect gas. Example 1.3 Calculating partial pressures The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2; Ar: 1.3. What is the partial pressure of each component when the total pressure is 1.00 atm? Method We expect species with a high mole fraction to have a proportionally high partial pressure. Partial pressures are defined by eqn 1.13. To use the equation, we need the mole fractions of the components. To calculate mole fractions, which are defined by eqn 1.14, we use the fact that the amount of molecules J of molar mass MJ in a sample of mass mJ is nJ = mJ /MJ . The mole fractions are independent of the total mass of the sample, so we can choose the latter to be 100 g (which makes the conversion from mass percentages very easy). Thus, the mass of N2 present is 75.5 per cent of 100 g, which is 75.5 g. Answer The amounts of each type of molecule present in 100 g of air, in which the masses of N2, O2, and Ar are 75.5 g, 23.2 g, and 1.3 g, respectively, are These three amounts work out as 2.69 mol, 0.725 mol, and 0.033 mol, respectively, for a total of 3.45 mol. The mole fractions are obtained by dividing each of the n( ) . . . . Ar g g mol = = mol − 1 3 39 95 1 3 39 95 1 n( ) . . . . O g g mol mol 2 1 23 2 32 00 23 2 32 00 = = − n( ) . . . . N g g mol mol 2 1 75 5 28 02 75 5 28 02 = = − x n n J J =
1 THE PROPERTIES OE GASES above amounts by 3.45 mol and the partial pressures are then obtained by multi- plying the mole fraction by the total pressure(1.00 atm): N, 02 Ar Mole fraction: 0.210 Partial pressure/atm: 021 Se-test1.4 When carbon dioxide is taken into account,the mass percentages are .15(02 Real gases Real gases do not obey the perfect gas law exactly.Deviations from the law are particu larly importanres and low temperatures,especially when a gas is on the point of con 13 Molecular interactions molecules in ct with assist compression. Repulsive forces are significant only when molecules are almost in contact:they are short-range interactions,even on a scale measured in molecular diameters(Fig 1.13) theyar short-range interacti ns can be dobe importan e average se at ng d forces havea relatively lon h and are effective ny molecu lar diameters they are important when the molecules are fairly close together but not 0 necessarily touching (at the intermediate separations in Fig.1.13).Attractive forces Separation are ineffective when the molecules are far apart (well to the right in Fig.1.13) Interm are also important when the temperatures so low that the mo low mean speed t they can be captured Iby one anot arge volun the gas beha A derate forces pwhe tion of the molecules is onl a few molccular diameters.theattractive forces dominate the repulsive forces.In this case,the gas can be expected to be more compressible than a perfect gas because the forces help to draw the molecules together.At high pressures. hen the average separation of the molecules is small,the repulsive forces dominate repulsive at these distances.At the gas can e expected to be less compressible because now the forces help to active nteractions dominate.At large separations (a)The compression factor ergy is The compression factor,Z,of a gas is the ratio of its measured molar volume,V VIn,to the molar volume of a perfect gas,V at the same pressure and temperature:
14 1 THE PROPERTIES OF GASES Separation Potential energy 0 Contact Attractions dominant Repulsions dominant Fig. 1.13 The variation of the potential energy of two molecules on their separation. High positive potential energy (at very small separations) indicates that the interactions between them are strongly repulsive at these distances. At intermediate separations, where the potential energy is negative, the attractive interactions dominate. At large separations (on the right) the potential energy is zero and there is no interaction between the molecules. above amounts by 3.45 mol and the partial pressures are then obtained by multiplying the mole fraction by the total pressure (1.00 atm): N2 O2 Ar Mole fraction: 0.780 0.210 0.0096 Partial pressure/atm: 0.780 0.210 0.0096 We have not had to assume that the gases are perfect: partial pressures are defined as pJ = xJp for any kind of gas. Self-test 1.4 When carbon dioxide is taken into account, the mass percentages are 75.52 (N2), 23.15 (O2), 1.28 (Ar), and 0.046 (CO2). What are the partial pressures when the total pressure is 0.900 atm? [0.703, 0.189, 0.0084, 0.00027 atm] Real gases Real gases do not obey the perfect gas law exactly. Deviations from the law are particularly important at high pressures and low temperatures, especially when a gas is on the point of condensing to liquid. 1.3 Molecular interactions Real gases show deviations from the perfect gas law because molecules interact with one another. Repulsive forces between molecules assist expansion and attractive forces assist compression. Repulsive forces are significant only when molecules are almost in contact: they are short-range interactions, even on a scale measured in molecular diameters (Fig. 1.13). Because they are short-range interactions, repulsions can be expected to be important only when the average separation of the molecules is small. This is the case at high pressure, when many molecules occupy a small volume. On the other hand, attractive intermolecular forces have a relatively long range and are effective over several molecular diameters. They are important when the molecules are fairly close together but not necessarily touching (at the intermediate separations in Fig. 1.13). Attractive forces are ineffective when the molecules are far apart (well to the right in Fig. 1.13). Intermolecular forces are also important when the temperature is so low that the molecules travel with such low mean speeds that they can be captured by one another. At low pressures, when the sample occupies a large volume, the molecules are so far apart for most of the time that the intermolecular forces play no significant role, and the gas behaves virtually perfectly. At moderate pressures, when the average separation of the molecules is only a few molecular diameters, the attractive forces dominate the repulsive forces. In this case, the gas can be expected to be more compressible than a perfect gas because the forces help to draw the molecules together. At high pressures, when the average separation of the molecules is small, the repulsive forces dominate and the gas can be expected to be less compressible because now the forces help to drive the molecules apart. (a) The compression factor The compression factor, Z, of a gas is the ratio of its measured molar volume, Vm = V/n, to the molar volume of a perfect gas, Vo m, at the same pressure and temperature:
1.3 MOLECULAR INTERACTIONS 15 z [1.16 Because the molar volume of a perfect gas is equal to RT/p,an equivalent expression isZ=RTIpV which we can write as pVm=RTZ (1.17) Because for a perfect gas Z=l under all conditions.deviation of Z from l is a measure of departure from perfect behaviour. Some experimental values of Z are plotted in Fig.1.14.At very low pressures,all the gases shown haveZI and behave nearly perfectly.At high pressures,all the now nant.At in forces are reducing the molar ve (b)Virial coefficients fect-gas isoth 、The small dif that the perec as law is in fac the first term in an expression of the form pVm=RTl+Bp+C'p2+.) (1.18) This e a go 140 120 50 100 H 40C 31.04t(T月 60 E DC CH p/atm 10 0.98 H 0℃ 20 NH. 0.96 0 200 600600 02 0.4 0.6 V/dm'mol-) Fle1.14 The variation of the comp 10 1.1 Experimental isotherms of carbon 0 al gases a Notice that. ugh the cu 1asp0,they do so with different slopes
1.3 MOLECULAR INTERACTIONS 15 [1.16] Because the molar volume of a perfect gas is equal to RT/p, an equivalent expression is Z = RT/pV o m, which we can write as pVm = RTZ (1.17) Because for a perfect gas Z = 1 under all conditions, deviation of Z from 1 is a measure of departure from perfect behaviour. Some experimental values of Z are plotted in Fig. 1.14. At very low pressures, all the gases shown have Z ≈ 1 and behave nearly perfectly. At high pressures, all the gases have Z > 1, signifying that they have a larger molar volume than a perfect gas. Repulsive forces are now dominant. At intermediate pressures, most gases have Z < 1, indicating that the attractive forces are reducing the molar volume relative to that of a perfect gas. (b) Virial coefficients Figure 1.15 shows the experimental isotherms for carbon dioxide. At large molar volumes and high temperatures the real-gas isotherms do not differ greatly from perfect-gas isotherms. The small differences suggest that the perfect gas law is in fact the first term in an expression of the form pVm = RT(1 + B′p + C′p2 + · · · ) (1.18) This expression is an example of a common procedure in physical chemistry, in which a simple law that is known to be a good first approximation (in this case pV = nRT) is Z V V = m m o H2 CH4 C H2 4 NH3 p/atm 2 1 0 200 400 600 800 Perfect 0Compression factor, Z 1.00 0.98 0.96 10 H2 CH4 C H2 4 NH3 p/atm 140 120 100 80 60 40 20 0 0 0.2 0.4 0.6 50°C 40°C 31.04°C ( ) Tc 0°C p/atm Vm /(dm mol ) 3 1 EDC B A F * 20°C - Fig. 1.14 The variation of the compression factor, Z, with pressure for several gases at 0°C. A perfect gas has Z = 1 at all pressures. Notice that, although the curves approach 1 as p → 0, they do so with different slopes. Fig. 1.15 Experimental isotherms of carbon dioxide at several temperatures. The ‘critical isotherm’, the isotherm at the critical temperature, is at 31.04°C. The critical point is marked with a star