1 The properties of gases The perfect gas

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1 1 The properties of gases 1A The perfect gas Answers to discussion questions 1A. The partial pressure of a gas in a ixture of gases is the pressure the gas would exert if it occupied alone the sae container as the ixture at the sae teperature. Dalton s law is a liiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the liit of zero pressure where the olecules of the gas are very far apart. Hence, Dalton s law holds exactly only for a ixture of perfect gases; for real gases, the law is only an approxiation. Solutions to exercises 1A.1(b) 1A.(b) 1A.3(b) The perfect gas law [1A.5] is pv n, iplying that the pressure would be p n V All quantities on the right are given to us except n, which can be coputed fro the given ass of Ar. 5 g n.66 ol g ol so p (.66 ol) ( d 3 bar K 1 ol 1 ) (3 + 73) K 1.5bar 1.5 d 3 So no, the saple would not exert a pressure of. bar. Boyle s law [1A.4a] applies. pv constant so p f V f p i V i Solve for the initial pressure: (i) p i p V f f (1.97 bar) (.14d3 ) 1.7 bar V i ( )d 3 (ii) The original pressure in Torr is 1 at p i (1.7 bar) 1.13 bar 76 Torr 1 at 83 Torr The relation between pressure and teperature at constant volue can be derived fro the perfect gas law, pv n [1A.5] so p T and p i T i p f T f The final pressure, then, ought to be p f p T i f (15 kpa) (11+ 73)K 1 kpa T i (3+ 73)K 1A.4(b) 1A.5(b) According to the perfect gas law [1.8], one can copute the aount of gas fro pressure, teperature, and volue. pv n so n pv (1. at) ( Pa at 1 ) ( ) ol ( J K 1 ol 1 ) ( + 73)K Once this is done, the ass of the gas can be coputed fro the aount and the olar ass: ( ol) ( g ol ) g kg The total pressure is the external pressure plus the hydrostatic pressure [1A.1], aking the total pressure 1

2 p p ex + ρgh. Let p ex be the pressure at the top of the straw and p the pressure on the surface of the liquid (atospheric pressure). Thus the pressure difference is 3 p p ex ρgh (1. g c ) 1 kg 1 3 g 1 c Pa at 3 (9.81 s ) (.15) 1A.6(b) The pressure in the apparatus is given by p p ex + ρgh [1A.1] where p ex 76 Torr 1 at Pa, and ρgh g c 3 1 kg 1 3 g 1 c s Pa p Pa Pa Pa 115 kpa 1A.7(b) Rearrange the perfect gas equation [1A.5] to give R pv nt p T All gases are perfect in the liit of zero pressure. Therefore the value of p /T extrapolated to zero pressure will give the best value of R. The olar ass can be introduced through pv n M which upon rearrangeent gives M V p ρ p The best value of M is obtained fro an extrapolation of ρ/p versus p to zero pressure; the intercept is M/. Draw up the following table: p/at (p /T)/(d 3 at K 1 ol 1 ) (ρ/p)/(g d 3 at 1 ) Fro Figure 1A.1(a), R li p Figure 1A.1 (a) p T.8 6 d3 at K 1 ol 1

3 (b) ρ Fro Figure 1A.1(b), li p p g d-3 at 1 M li p ρ p (.86 d3 at K 1 ol 1 ) (73.15 K) ( g d -3 at 1 ) 1A.8(b) g ol 1 The value obtained for R deviates fro the accepted value by.5 per cent, better than can be expected fro a linear extrapolation fro three data points. The ass density ρ is related to the olar volue by V n n M ρ where M is the olar ass. Putting this relation into the perfect gas law [1A.5] yields pm p so ρ Rearranging this result gives an expression for M; once we know the olar ass, we can divide by the olar ass of phosphorus atos to deterine the nuber of atos per gas olecule. M ρ p ( Pa 1 3 ol ) [(1 + 73) K] (.6388kg 3 ) Pa.14 kg ol 1 14 g ol 1 1A.9(b) The nuber of atos per olecule is 1 14 g ol g ol suggesting a forula of P 4. Use the perfect gas equation [1A.5] to copute the aount; then convert to ass. pv n so n pv We need the partial pressure of water, which is 53 per cent of the equilibriu vapour pressure at the given teperature and standard pressure. (We ust look it up in a handbook like the CRC or other resource such as the NIST Cheistry WebBook.) p (.53) ( Pa) Pa 3

4 ( Pa) (5 3 ) so n 151 ol ( J K 1 ol 1 ) (3+ 73) K and (151 ol) (18. 1 g ol ) g.7 kg 1A.1(b) (i) The volue occupied by each gas is the sae, since each copletely fills the container. Thus solving for V we have (assuing a perfect gas, eqn. 1A.5) V n J p J We have the pressure of neon, so we focus on it.5 g n Ne.18 g ol ol Thus V ol Pa 3 K 1 ol 1 3 K d Pa (ii) The total pressure is deterined fro the total aount of gas, n n CH4 + n Ar + n Ne. and.3 g n CH g ol g 1 1 ol n Ar g ol ol n ( ) 1 ol ol p n V ol Pa 3 K 1 ol 1 3 K Pa 8. kpa 1A.11(b) This exercise uses the forula, M ρ, which was developed and used in Exercise p 1A.8(b). First the density ust first be calculated. ρ g 13 c 3 5c 3 d g d 3 M (.134 g d 3 ) (6.36d 3 torr K 15 torr 1 1 ol ) (98 K) 16.4 g ol 1 1A.1(b) This exercise is siilar to Exercise 1.1(a) in that it uses the definition of absolute zero as that teperature at which the volue of a saple of gas would becoe zero if the substance reained a gas at low teperatures. The solution uses the experiental fact that the volue is a linear function of the Celsius teperature: V V + αθ where V. d 3 and α.741 d 3 C 1. At absolute zero, V V + αθ so θ(abs.zero) V α. d 3 Ğ7 C.741 d 3 1 C which is close to the accepted value of 73C. 1A.13(b) (i) Mole fractions are x N n N.5 ol [1A.9] n total ( ) ol.63 Siilarly, x H.37 According to the perfect gas law p tot V n tot 4

5 so p tot n tot V (4. ol) (.86 d3 at ol 1 K 1 ) (73.15 K).4 d 3 4. at (ii) The partial pressures are p N x N p tot (.63) (4. at).5 at and p H (.37) (4. at) 1.5 at (iii) p p H + p N [1A.1] ( ) at 4. at Solutions to probles pv 1A. Solving for n fro the perfect gas equation [1A.5] yields n. Fro the definition of olar ass n, hence ρ M V Mp Rearrangeent yields the desired relation, naely. p ρ. M Therefore, for ideal gases p ρ and M. For real gases, find the zero-pressure M p / ρ liit of p by plotting it against p. Draw up the following table. ρ p/(kpa) ρ/(kg 3 ) p / ρ 1 3 s Bear in ind that 1 kpa 1 3 kg 1 s. p is plotted in Figure 1A.. A straight line fits the data rather well. The extrapolation to p ρ yields an intercept of s. Then M s ( J K 1 ol 1 ) (98.15K) s.459 kg ol g ol Figure 1A. 1 5

6 Coent. This ethod of the deterination of the olar asses of gaseous copounds is due to Cannizarro who presented it at the Karlsruhe Congress of 186. That conference had been called to resolve the proble of the deterination of the olar asses of atos and olecules and the olecular forulas of copounds. 1A.4 The ass of displaced gas is ρv, where V is the volue of the bulb and ρ is the density of the displaced gas. The balance condition for the two gases is (bulb) ρv(bulb) and (bulb) ρ V(bulb) which iplies that ρ ρ. Because [Proble 1.] ρ pm the balance condition is pm p M, which iplies that M p p M This relation is valid in the liit of zero pressure (for a gas behaving perfectly). In experient 1, p 43. Torr, p 37.1 Torr; 43. Torr hence M 37.1 Torr 7.14 g ol g ol 1 In experient, p 47. Torr, p 93. Torr; 47. Torr hence M 93. Torr 7.14 g ol 1 1. g ol 1 In a proper series of experients one should reduce the pressure (e.g. by adjusting the balanced weight). Experient is closer to zero pressure than experient 1, so it is ore likely to be close to the true value: M 1 g ol 1 The olecules CH FCF 3 and CHF CHF have olar ass of 1 g ol 1. Coent. The substantial difference in olar ass between the two experients ought to ake us wary of confidently accepting the result of Experient, even if it is the ore likely estiate. 1A.6 We assue that no H reains after the reaction has gone to copletion. The balanced equation is N + 3 H NH 3. We can draw up the following table N H NH 3 Total Initial aount n n n + n Final aount n 1 n 3 n 3 n + 1 n 3 Specifically.33 ol 1.33 ol 1.66 ol Mole fraction p n V (1.66 ol) (.86 d3 at ol 1 K 1 ) (73.15K).4d at p(h ) x(h )p p(n ) x(n )p at.33 at p(nh 3 ) x(nh 3 )p at 1.33 at 1A.8 The perfect gas law is pv n so n pv At id-latitudes n (1.at) {(1.d ) (5 1 3 c) / 1 c d 1 } ol (.86d 3 at K 1 ol 1 ) (73K) In the ozone hole n (1.at) {(1.d ) (1 1 3 c) / 1c d 1 } ol (.86d 3 at K 1 ol 1 ) (73 K) The corresponding concentrations are 6

7 n V ol (1.d ) (4 1 3 ) (1d 1 ) old 3 n and V ol (1.d ) (4 1 3 ) (1d 1 ) old 3 respectively. 1A.1 The perfect gas law [1A.5] can be rearranged to n pv The volue of the balloon is V 4π 3 r 3 4π 3 (3. ) (1.at) ( d 3 ) (a) n (.86 d 3 at ol 1 K 1 ) (98 K) ol (b) The ass that the balloon can lift is the difference between the ass of displaced air and the ass of the balloon. We assue that the ass of the balloon is essentially that of the gas it encloses: (H ) nm(h ) ( ol) (. g ol ) g Mass of displaced air (113 3 ) (1. kg ) kg Therefore, the ass of the axiu payload is 138 kg 9.33 kg kg (c) For heliu, nm(he) ( ol) (4. g ol 1 ) 18kg The axiu payload is now 138 kg 18kg 1. 1 kg 1A.1 Avogadro s principle states that equal volues of gases contain equal aounts (oles) of the gases, so the volue ixing ratio is equal to the ole fraction. The definition of partial pressures is p J x J p. The perfect gas law is pv n so n J V p J x p J (a) n(ccl 3 F) ( ) (1.at) V (.86d 3 at K 1 ol 1 ) (1 + 73) K old -3 and n(ccl F ) ( ) (1.at) V (.86d 3 at K 1 ol 1 ) (1 + 73) K old -3 (b) n(ccl 3 F) ( ) (.5at) V (.86d 3 at K 1 ol 1 ) ( K) old -3 and n(ccl F ) ( ) (.5at) V (.86d 3 at K 1 ol 1 ) ( K) old B The kinetic odel Answers to discussion questions 1B. The forula for the ean free path [eqn 1B.13] is λ kt σ p In a container of constant volue, the ean free path is directly proportional to teperature and inversely proportional to pressure. The forer dependence can be rationalized by noting that the faster the olecules travel, the farther on average they go between collisions. The latter also akes sense in that the lower the pressure, the less frequent are collisions, 7

8 1B.1(b) 1B.(b) 1B.3(b) and therefore the further the average distance between collisions. Perhaps ore fundaental than either of these considerations are dependences on size. As pointed out in the text, the ratio T/p is directly proportional to volue for a perfect gas, so the average distance between collisions is directly proportional to the size of the container holding a set nuber of gas olecules. Finally, the ean free path is inversely proportional to the size of the olecules as given by the collision cross section (and therefore inversely proportional to the square of the olecules radius). Solutions to exercises The ean speed is [1B.8] v ean 8 1/ π M The ean translational kinetic energy is E k 1 v 1 v 1 v rs 3 3kT M [1B.3] The ratios of species 1 to species at the sae teperature are v ean,1 M 1/ E k v and 1 1 ean, E k M 1 v ean,h 1/ (i).6 v ean,hg (ii) The ean translation kinetic energy is independent of olecular ass and depends upon teperature alone! Consequently, because the ean translational kinetic energy for a gas is proportional to T, the ratio of ean translational kinetic energies for gases at the sae teperature always equals 1. The root ean square speed [1B.3] is v rs 3 1/ M For CO the olar ass is M ( ) 1 3 kg ol kg ol 1 so v rs 3( J K 1 ol 1 1/ )( + 73) K kg ol 1 48 s 1 For He v rs 3( J K 1 ol 1 1/ )( + 73) K kg ol s k s 1 The Maxwell-Boltzann distribution of speeds [1B.4] is 3/ M f (v) 4π π v e Mv / and the fraction of olecules that have a speed between v and v+dv is f(v)dv. The fraction of olecules to have a speed in the range between v 1 and v is, therefore, v v 1 f (v)dv. If the range is relatively sall, however, such that f(v) is nearly constant over that range, the integral ay be approxiated by f(v) v, where f(v) is evaluated anywhere within the range and v v v 1. Thus, we have, with M kg ol 1 [Exericse 1B.(b)], v v kg ol f( v)d v f( v) v 4 π (4.5 s ) 1 1 π ( J K ol )(4 K) 3/ 1 8

9 exp ( kg ol 1 )(4.5 s 1 ) ( J K 1 ol 1 )(4 K) (45 4) s 1.17, just over 1% 1B.4(b) The ost probable, ean, and ean relative speeds are, respectively v p 1/ M [1B.9] v ean 8 1/ π M [1B.8] v rel 8 πµ The teperature is T (+73) K 93 K. so v p ( J K 1 ol 1 1/ )(93 K) kg ol s 1 1/ [1B.1b] and v ean 8( J K 1 ol 1 )(93 K) π( kg ol 1 ) s 1 For any purposes, air can be considered as a gas with an average olar ass of 9. g ol 1. In that case, the reduced olar ass [1B.1b] is µ M A M B M A + M B (9. g ol 1 )( 1.8 g ol 1 ) ( ) g ol g ol 1 1/ and v rel 8( J K 1 ol 1 )(93 K) π( kg ol 1 ) s 1 Coent. One coputes the average olar ass of air just as one coputes the average olar ass of an isotopically ixed eleent, naely by taking an average of the species that have different asses weighted by their abundances. Coent. Note that v rel and v ean are very nearly equal. This is because the reduced ass between two very dissiilar species is nearly equal to the ass of the lighter species (in this case, H ). 1/ 1/ 1B.5(b) (i) v ean 8 π M [1B.8] 8( J K 1 ol 1 )(98 K) π( kg ol 1 ) (ii) The ean free path [1B.13] is λ kt σ p kt π d p ( J K 1 )(98 K) π( ) (1 1 9 Torr) 1 Torr Pa 1/ 475 s k The ean free path is uch larger than the diensions of the puping apparatus used to generate the very low pressure. (iii) The collision frequency is related to the ean free path and relative ean speed by [1B.1] 1B.6(b) λ v rel z so z v rel λ 1/ v ean λ z 1/ (475 s 1 ) s 1 [1B.1a] The collision diaeter is related to the collision cross section by σ πd so d (σ/π) 1/ (.36 n /π) 1/.34 n. The ean free path [1B.13] is λ kt σ p Solve this expression for the pressure and set λ equal to 1d: 9

10 1B.7(b) p kt σλ ( J K 1 )(93 K).36 (1 9 ) ( ) J MPa Coent. This pressure works out to 33 bar (about 33 at), conditions under which the assuption of perfect gas behavior and kinetic odel applicability at least begins to coe into question. The ean free path [1B.13] is λ kt σ p ( J K 1 )(17 K).43 (1 9 ) ( Pa at 1 ) Solutions to probles 1B. The nuber of olecules that escape in unit tie is the nuber per unit tie that would have collided with a wall section of area A equal to the area of the sall hole. This quantity is readily expressed in ters of Z W, the collision flux (collisions per unit tie with a unit area), given in eqn 19A.6. That is, dn Ap Z dt W A (π kt ) 1/ where p is the (constant) vapour pressure of the solid. The change in the nuber of olecules inside the cell in an interval t is therefore N Z A t, and so the ass loss is M w N Ap π kt t Ap π t Therefore, the vapour pressure of the substance in the cell is 1/ w π p A t M For the vapour pressure of geraniu 1/ kg p π( )(7 s) π( J K 1 ol 1 )(173 K) kg ol Pa 7.3 Pa 1B.4 We proceed as in Justification 1B. except that, instead of taking a product of three onediensional distributions in order to get the three-diensional distribution, we ake a product of two one-diensional distributions. f (v x,v y )dv x dv y f (v x ) f (v y )dv x dv y π kt e v /kt dv x dv y where v v x + v y. The probability f(v)dv that the olecules have a two-diensional speed, v, in the range v to v + dv is the su of the probabilities that it is in any of the area eleents dv x dv y in the circular shell of radius v. The su of the area eleents is the area of the circular shell of radius v and thickness dv which is π(ν+dν) πν πνdν. Therefore, f (v) kt ve v /kt M ve Mv / M R k The ean speed is deterined as v ean vf (v)dv kt v e v /kt dv Using integral G.3 fro the Resource Section yields v ean kt π 1/ 4 1B.6 The distribution [1B.4] is kt 3/ π kt W 1/ 1/ π M 1/ 1/ 1

11 M f (v) 4π π v e Mv /. The proportion of olecules with speeds less than v rs is P v rs Defining a R /, a P 4π π 3/ M f (v)dv 4π π 3/ v rs 3/ v rs a v e av dv 4π π v e Mv / dv 3/ d da 1/ Defining χ av. Then, dv a d χ and 3/ a v P 4π d 1 rs a π e χ dχ da{ a 1/ 1/ } 3/ a 4π π 1 3/ v 1 () rs a 1/ 1/ e χ dχ a + 1 () d a da Then we use the error function [Integral G.6]: v rs a 1/ e χ dχ ( π 1/ / )erf(v rs a 1/ ). d da v rs a 1/ e χ dχ dv rs a1/ da (e av rs ) 1 z v rs c e av dv v rs a 1/ a 1/ e av rs e χ dχ where we have used dz d f ( y)d y f (z) Substituting and cancelling we obtain P erf(v rs a 1/ ) ( v rs a 1/ / π 1/ )e av rs Now v rs 3 M and P erf 3 1/ so v rs a 1/ 3 M 1/ 1/ 6 π e 3/ M 1/ 3 Therefore, (a) 1 P 39% have a speed greater than the root ean square speed. (b) P 61% of the olecules have a speed less than the root ean square speed. (c) For the proportions in ters of the ean speed v ean, replace v rs by ( ) 1/ ( 8 / 3π ) 1/ v rs so v ean a 1/ /π 1/. v ean 8kT / π Then P erf(v ean a 1/ ) ( v ean a 1/ / π 1/ ) (e av ean ) erf ( / π 1/ ) ( 4 / π)e 4/π That is, 53% of the olecules have a speed less than the ean, and 47% have a speed greater than the ean. 1B.8 The average is obtained by substituting the distribution (eqn 1B.4) into eqn 1B.7: v n M v n f (v)dv 4π π For even values of n, use Integral G.8: where v n 3/ M (n + 1)!! 4π π n+4 (n+1)!! (n+1) Thus v n 1/n (n + 1)!! M 1/ 3/ M n+ v n+ e Mv / dv even n π M 1/ 1/ (n + 1)!! M n 11

12 For odd values of n, use Integral G.7: n + 1 3/ M! v n 4π π 1/n M n+3 π 1/ Thus v n 1/n n/ 1/ π 1/ M 1/n π 1/n M odd n Question. Show that these expressions reduce to v ean and v rs for n 1 and respectively. 1B.1 Dry atospheric air is 78.8% N,.95% O,.93% Ar,.4% CO, plus traces of other gases. Nitrogen, oxygen, and carbon dioxide contribute 99.6% of the olecules in a volue with each olecule contributing an average rotational energy equal to kt. (Linear olecules can rotate in two diensions, contributing two quadratic ters of rotational energy, or kt by the equipartition theore [Topic B.3(b)]. The rotational energy density is given by ρ R E R V.996N ε R.996NkT.996 p V V.996( Pa) J 3.14 J c 3 The total energy density is translational plus rotational (vibrational energy contributing negligibly): M ρ tot ρ T + ρ R.15 J c J c 3.5 J c 3 1B.1 The fraction of olecules (call it F) between speeds a and b is given by F(a,b) b a f (v)dv where f(v) is given by eqn 1B.4. This integral can be approxiated by a su over a discrete set of velocity values. For convenience, let the velocities v i be evenly spaced within the interval such that v i+1 v i + v: Fab (, ) f( vi )Δv On a spreadsheet or other atheatical software, ake a colun of velocity values and then a colun for f(v) [1B.4] at 3 K and at 1 K. Figure 1B.1 shows f(v) plotted against v for these two teperatures. Each curve is labeled with the nuerical value of T/K, and each is shaded under the curve between the speeds of 1 and s 1. F(a,b) is siply the area under the curve between v a and v b. One should take soe care to avoid double counting at the edges of the interval, that is, not including both endpoints of the interval with full weight. exaple, beginning the su with the area under the curve at those speeds. Using a spreadsheet that evaluates f(v) at 5- s 1 intervals, and including points at both 1 and s 1 with half weight, F(1 s 1, s 1 ).81 at 3 K and.66 at 1 K. Figure 1B.1 n/ 1

13 1C Real gases Answers to discussion questions 1C. The critical constants represent the state of a syste at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical teperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though supercritical fluids have both liquid and vapour characteristics. 1C.4 The van der Waals equation is a cubic equation in the volue, V. Every cubic equation has soe values of the coefficients for which the nuber of real roots passes fro three to one. In fact, any equation of state of odd degree n > 1 can in principle account for critical behavior because for equations of odd degree in V there are necessarily soe values of teperature and pressure for which the nuber of real roots of V passes fro n to 1. That is, the ultiple values of V converge fro n to 1 as the teperature approaches the critical teperature. This atheatical result is consistent with passing fro a two phase region (ore than one volue for a given T and p) to a one phase region (only one V for a given T and p), and this corresponds to the observed experiental result as the critical point is reached. 1C.1(b) Solutions to exercises The van der Waals equation [1C.5a] is p n V nb an V Fro Table 1C.3 for H S, a d 6 at ol 1 and b.434 d 3 ol 1. (i) p (1. ol) (.86 d3 at ol 1 K 1 ) (73.15 K).414 d 3 (1. ol) ( d 3 ol 1 ) (4.484 d6 at ol ) (1. ol).99 at (.414 d 3 ) (ii) p (1. ol) (.86 d3 at ol 1 K 1 ) (5 K).15d 3 (1. ol) ( d 3 ol 1 ) (4.484 d3 at ol 1 ) (1. ol) (.15 d 3 ) 19 at ( sig. figures) 1C.(b) 1C.3(b) The conversions needed are as follows: 1 at Pa, 1 Pa 1 kg 1 s, 1 d 6 (1 1 ) , 1 d Therefore, a 1.3 at d 6 ol kg 1 s at d kg 5 s ol and b.46 d 3 ol ol 1 d 3 The copression factor Z is [1C.1] Z V p (i) Because V ο +.1 V ο (1.1)V ο, we have Z 1.1 (ii) The olar volue is 13

14 (1.1)V ο (1.1) p (1.1) (.86 d3 at ol 1 K 1 ) (35 K) 1at.7 d3 ol 1 Since > V o repulsive forces doinate. 1C.4(b) 1C.5(b) (i) According to the perfect gas law V o p ( J K 1 ol 1 ) (98.15K) 1 d ( bar) (1 5 1 Pa bar ) d 3 ol 1 (ii) The van der Waals equation [1C.5b] is a cubic equation in. Cubic equations can be solved analytically. However, this approach is cubersoe, so we proceed as in Exaple 1C.1. The van der Waals equation is rearranged to the cubic for V 3 b + p V + a p V ab p or x 3 b + p x + a p x ab p with x /(d 3 ol 1 ). It will be convenient to have the pressure in at: bar 1 at at 1.13 bar The coefficients in the equation are b + p ( d 3 ol 1 ) + (.86 d3 at ol 1 K 1 ) (98.15 K) at ( ) d 3 ol d 3 ol 1 a p 1.36 d6 at ol d 6 ol at ab p (1.36 d6 at ol ) ( d 3 ol 1 ) d 9 ol at Thus, the equation to be solved is x x + ( )x ( ). Calculators and coputer software for the solution of polynoials are readily available. In this case we find x.11 and.11 d 3 ol 1. The perfect-gas value is about 15 percent greater than the van der Waals result. The olar volue is obtained by solving Z p [1C.], for, which yields Z p (.86) (.86 d3 at ol 1 K 1 ) (3 K) 1.6 d 3 ol 1 at 3 (i) Then, V n ( ol) (1.6 d 3 ol 1 ) d c 3 (ii) An approxiate value of B can be obtained fro eqn 1C.3b by truncation of the series expansion after the second ter, B/, in the series. Then, pv B 1 V (Z 1) (1.6 d 3 ol 1 ) (.86 1).15 d 3 ol 1 1C.6(b) Equations 1C.6are solved for b and a, respectively, and yield b V c /3 and a 7b p c 3V c p c. Substituting the critical constants b 148 c3 ol c 3 ol d 3 ol

15 and a 3 (.148 d 3 ol 1 ) (48.at) 3.17 d 6 at ol But this proble is overdeterined. We have another piece of inforation T c 8a 7Rb If we use T c along with V c as above, we would arrive at the sae value of b along with a 7RbT c 8 9RV c T c 8 9(.86 d3 at ol 1 K 1 )(.148 d 3 ol 1 )(35.4 K) d 6 at ol Or we could use T c along with p c. In that case, we can solve the pair of equations for a and b by first setting the two expressions for a equal to each other: a 7b p c 7RbT c 8 Solving the resulting equation for b yields b c (.86 d3 at ol 1 K 1 )(35.4 K).6499 d 3 ol 1 8p c 8(48. at) and then a 7(.6499 d 3 ol 1 ) (48. at) d 6 at ol These results are suarized in the following table Using a/d 6 at ol b/d 3 ol 1 V c & p c V c & T c p c & T c One way of selecting best values for these paraeters would be to take the ean of the three deterinations, naely a 4.8 d 6 at ol and b.546 d 3 ol 1. By interpreting b as the excluded volue of a ole of spherical olecules, we can obtain an estiate of olecular size. The centres of spherical particles are excluded fro a sphere whose radius is the diaeter of those spherical particles (i.e., twice their radius); that volue ties the Avogadro constant is the olar excluded volue b 4π(r) 3 b N A 3 so r 1 1/3 3b 4π N A r 1 3(.546 d 3 ol 1 ) 4π( ol 1 ) 1/ d.139 n 1C.7(b) The Boyle teperature, T B, is the teperature at which the virial coefficient B. In order to express T B in ters of a and b, the van der Waals equation [1C.5b] ust be recast into the for of the virial equation. p b a Factoring out yields p 1 a 1 b / So long as b/ < 1, the first ter inside the brackets can be expanded using (1 x) x + x +..., which gives p 1+ b a 1 +L We can now identify the second virial coefficient as B b a 15

16 At the Boyle teperature B b a so T B a B br 7T c 8 (i) Fro Table 1C.3, a d 6 at ol and b.434 d 3 ol 1. Therefore, (4.484 d 6 at ol ) T B (.86 L at ol 1 K 1 ) (.434 d 3 ol 1 ) 159 K (ii) As in Exercise 1C.6(b), 4π(r) 3 b N A 3 so r 1 3b 4π N A r 1 3(.434 d 3 ol 1 ) 4π( ol 1 ) 1/3 1/ d n 1C.8(b) 1C.9(b) States that have the sae reduced pressure, teperature, and volue [1C.8] are said to correspond. The reduced pressure and teperature for N at 1. at and 5 C are [Table 1C.] p r p 1.at p c 33.54at.3 and T T (5 + 73) K r T c 16.3K.36 The corresponding states are (i) For H S (critical constants obtained fro NIST Cheistry WebBook) T.36(373.3 K) 881 K p.3(89.7 at).67 at (ii) For CO T.36(34. K) 718 K p.3(7.9 at). at (iii) For Ar T.36(15.7 K) 356 K p.3(48. at) 1.4 at The van der Waals equation [1C.5b] is p b a which can be solved for b ( J K 1 ol 1 ) (88 K) b p + a ol Pa ol Pa + ( ol 1 ) ol 1 The copression factor is Z p [1C.] (4. 16 Pa) ( ol 1 ).67 ( J K 1 ol 1 ) (88 K) Solutions to probles 1C. Fro the definition of Z [1C.1] and the virial equation [1C.3b], Z ay be expressed in virial for as Z 1+ B 1 V + C 1 V +L 16

17 Since p (by assuption of approxiate perfect gas behavior), 1 p ; hence upon substitution, and dropping ters beyond the second power of 1 Z 1+ B p + C p 1 at 1+ ( d 3 ol 1 ) (.86 d 3 at ol 1 K 1 ) (73K) 1 at +( d 6 ol ) (.86 d 3 at ol 1 K 1 ) (73K) (.97) (.97) p (.86 d 3 at ol 1 K 1 )(73 K) 1 at.8 d3 Question. What is the value of Z obtained fro the next approxiation using the value of just calculated? Which value of Z is likely to be ore accurate? 1C.4 Since B (T B ) at the Boyle teperature [Topic 1.3b]: B (T B ) a + be c/t B Solving for T B : T 1/ c (1131K ) ln a ( 1993bar ) ( b ). ln 1 (. bar ) B 1 1/ 5. 1 K 1C.6 Fro Table 1C.4 T c 3 a 3bR a ( ) 1 / 3bR 1/, p c 1 1 ar 3b 3 ay be solved for fro the expression for p c and yields Thus T c 3 1 p b c R 8 3 p c V c R 1/ 1bp c R. 8 3 (4 at) ( d 3 ol 1 ).86 d 3 at ol 1 K 1 1K By interpreting b as the excluded volue of a ole of spherical olecules, we can obtain an estiate of olecular size. The centres of spherical particles are excluded fro a sphere whose radius is the diaeter of those spherical particles (i.e., twice their radius); that volue ties the Avogadro constant is the olar excluded volue b 4π(r) 3 b N A 3 so r 1 1/3 3b 4πN [Exercise 1C.6(b)] 1 1/3 V c A 4πN A r 1 16 c 3 ol 1 4π( ol 1 ) 1/ c.138 n 1C.8 Substitute the van der Waals equation [1C.5b] into the definition of the copression factor [1C.] 17

18 Z p 1 a 1 b [Exercise 1C.7(a)] 1 which upon expansion of 1 b 1 b b V + + V + V yields a 1 1 Z 1 b b V V We note that all ters beyond the second are necessarily positive, so only if a b b > + + V V V can Z be less than one. If we ignore ters beyond b, the conditions are siply stated as Z < 1 when a > b Z > 1 when a < b Thus Z < 1 when attractive forces predoinate and when there is insufficient theral energy to disrupt those forces, and Z > 1 when size effects (short-range repulsions) predoinate. 1C.1 The Dieterici equation is p e a/ [Table 1C.4] b At the critical point the derivatives of p with respect to equal zero along the isother defined by T T c. This eans that ( p / ) T and ( p / ) T at the critical point. and p p av ab V V T ( b)( ) 3 ( av + 4ab + V ab ) { } p p av ab V p + 3 V V V( V b)( ) T T V ( V b)( ) Setting the Dieterici equation equal to the critical pressure and aking the two derivatives vanish at the critical point yields three equations: p c c e a/ cv c V c b av c ab c V c and av c + 4V c ab + c V 3 c ab Solving the iddle equation for T c, substitution of the result into the last equation, and solving for V c yields the result V c b or b V c / (The solution V c b is rejected because there is a singularity in the Dieterici equation at the point b.) Substitution of V c b into the iddle equation and solving for T c gives the result T c a / 4bR or a c V c Substitution of V c b and T c a / 4bR into the first equation gives p c ae e c 4b V c The equations for V c, T c, p c are substituted into the equation for the critical copression factor [1C.7] to give Z c p c V c c e.77. This is significantly lower than the critical copression factor that is predicted by the van der Waals equation: Z c (vdw) p c V c / c 3 / Experiental values for Z c are 18

19 suarized in Table 1C. where it is seen that the Dieterici equation prediction is often better. 1C.1 Thus p 1 Bp Cp pv 1+ B + C + [1C.3b] V V Bp [1C.3a] + Cp V V B C Multiply through by, replace p by {1+(B/ ) +...}, and equate coefficients of powers of 1/ : B + BB + C R T + B + C + V V Hence, B B, iplying that B B Also BB + C R T C B + C R T, iplying that C C B R T 1C.14 Write f(t, p); then d T dt + p p dp Restricting the variations of T and p to those which leave constant, that is d, we obtain T p p Fro the equation of state and Substituting p p T T T P p T T 1 p p V T (a + bt ) V + (a + bt ) 3 3 R + b V RV + b 3 V R + b + (a + bt ) Fro the equation of state, a + bt p Then T P R + b + p R + b p T T RV + b + (a + bt ) 1C.16 Z o [1C.1], where the olar volue of a perfect gas Fro the given equation of state b + p b + V o For 1b, we have 1b b +, so 9b. Then Z 1b 9b C.18 The virial equation is pv B C [1C.3b] V V 19

20 or pv B C V V (a) If we assue that the series ay be truncated after the B ter, then a plot of p vs 1 will have B as its slope and 1 as its y-intercept. Transforing the data gives p/mpa /(d 3 ol 1 ) (1/ )/(ol d 3 ) p / Figure 1C.1(a) The data are plotted in Figure 1C.1(a). The data fit a straight line reasonably well, and the y- intercept is very close to 1. The regression yields B d 3 ol 1. (b) A quadratic function fits the data soewhat better (Figure 1C.1(b)) with a slightly better correlation coefficient and a y-intercept closer to 1. This fit iplies that truncation of the virial series after the ter with C is ore accurate than after just the B ter. The regression then yields

21 Figure 1C.1(b) B d 3 ol 1 and C d 6 ol 1C. The perfect gas equation [1A.5] gives p ( J K 1 ol 1 )(5 K) d Pa The van der Waals equation [1C.5b] is a cubic equation in. Cubic equations can be solved analytically. However, this approach is cubersoe, so we proceed as in Exaple 1C.1. The van der Waals equation is rearranged to the cubic for V 3 b + p V + a p V ab p or x 3 b + p x + a p x ab p with x /(d 3 ol 1 ). It will be convenient to have the pressure in at: 1 at 15 kpa at 11.3 kpa The coefficients in the equation are b + p (5.4 1 d 3 ol 1 ) + (.86 d3 at ol 1 K 1 ) (5 K) at ( ) d 3 ol d 3 ol 1 a p 6.6 d6 at ol 4.3 d 6 ol at ab p (6.6 d6 at ol ) (5.4 1 d 3 ol 1 ).91 1 d 9 ol at Thus, the equation to be solved is x x + 4.3x (.91 1 ). Calculators and coputer software for the solution of polynoials are readily available. In this case we find x 13.6 and 13.6 d 3 ol 1. Taking the van der Waals result to be ore accurate, the error in the perfect-gas value is % %