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- Question : Q A.1 - The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. Dalton
- Question : E A.1 - The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be p=nRT V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. n= 25 g 39.95 g mol?1 = 0.626 mol so p = (0.626 mol)
- Question : E A.2 - Boyle
- Question : E A.3 - The relation between pressure and temperature at constant volume can be derived from the perfect gas law, pV = nRT [1A.5] so p ?T and pi Ti = pf Tf The final pressure, then, ought to be pf = piTf Ti = (125 kPa)
- Question : E A.4 - According to the perfect gas law [1.8], one can compute the amount of gas from pressure, temperature, and volume. pV = nRT so n= pV RT = (1.00 atm)
- Question : E A.5 - The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the total pressure p = pex + ?gh . Let pex be the pressure at the top of the straw and p the pressure on the surface of the liquid (atmospheric pressure). Thus the pressure difference is p ? pex =?gh = (1.0gcm?3)
- Question : E A.6 - The pressure in the apparatus is given by p = pex + ?gh [1A.1] where pex = 760 Torr = 1 atm = 1.013
- Question : E A.7 - Rearrange the perfect gas equation [1A.5] to give R= pV nT = pVm T All gases are perfect in the limit of zero pressure. Therefore the value of pVm/T extrapolated to zero pressure will give the best value of R. The molar mass can be introduced through pV = nRT = m M RT which upon rearrangement gives M=m V RT p =?RT p The best value of M is obtained from an extrapolation of ?/p versus p to zero pressure; the intercept is M/RT. Draw up the following table: From Figure 1A.1(a), R = lim p?0 pVm T ? ?? ? ?? = 0.082 062 dm3 atm K?1 mol?1 Figure 1A.1 (a) From Figure 1A.1(b), lim p?0 ? p ? ?? ? ?? =1.427 55 g dm-3 atm?1 M = lim p?0 RT ? p ? ?? ? ?? = (0.082062 dm3 atm K?1 mol?1)
- Question : E A.8 - The mass density ? is related to the molar volume Vm by Vm =V n =V m
- Question : E A.9 - Use the perfect gas equation [1A.5] to compute the amount; then convert to mass. pV = nRT so n= pV RT We need the partial pressure of water, which is 53 per cent of the equilibrium vapour pressure at the given temperature and standard pressure. (We must look it up in a handbook like the CRC or other resource such as the NIST Chemistry WebBook.) p = (0.53)
- Question : E A.10 - (i) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V we have (assuming a perfect gas, eqn. 1A.5) V= nJ RT p J We have the pressure of neon, so we focus on it nNe = 0.225 g 20.18 g mol?1 = 1.115
- Question : E A.11 - This exercise uses the formula, M = ?RT p , which was developed and used in Exercise 1A.8(b). First the density must first be calculated. ?= 33.5
- Question : E A.12 - This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature: V = V0 + ?? where V0 = 20.00 dm3 and ? = 0.0741 dm3
- Question : E A.13 - (i) Mole fractions are xN = nN ntotal [1A.9]= 2.5 mol (2.5 + 1.5) mol = 0.63 Similarly, xH = 0.37 According to the perfect gas law ptotV = ntotRT so ptot = ntotRT V = (4.0 mol)
- Question : P A.2 - Solving for n from the perfect gas equation [1A.5] yields pV n RT = . From the definition of molar mass n m M = , hence ?=m V =Mp RT . Rearrangement yields the desired relation, namely p RT M = ? . Therefore, for ideal gases p ? =RT M and M= RT p /?. For real gases, find the zero-pressure limit of p ?by plotting it against p. Draw up the following table. p/(kPa) 12.223 25.20 36.97 60.37 85.23 101.3 ?/(kg m
- Question : P A.4 - The mass of displaced gas is ?V, where V is the volume of the bulb and ? is the density of the displaced gas. The balance condition for the two gases is m(bulb) = ?V(bulb) and m(bulb) = ??V(bulb) which implies that ? = ??. Because [Problem 1.2] ?=pM RT the balance condition is pM = p?M? , which implies that M?= p p?
- Question : P A.6 - We assume that no H2 remains after the reaction has gone to completion. The balanced equation is N2 + 3 H2 ? 2 NH3 . We can draw up the following table N2 H2 NH3 Total Initial amount n n? 0 n + n? Final amount n?1 3 n? 0 2 3 n? n+1 3 n? Specifically 0.33 mol 0 1.33 mol 1.66 mol Mole fraction 0.20 0 0.80 1.00 p=nRT V = (1.66mol)
- Question : P A.8 - The perfect gas law is pV = nRT so n= pV RT At mid-latitudes n = (1.00atm)
- Question : P A.10 - The perfect gas law [1A.5] can be rearranged to n= pV RT The volume of the balloon is V=4? 3 r3 = 4? 3
- Question : P A.12 - Avogadro
- Question : Q B.1 - The formula for the mean free path [eqn 1B.13] is ?=kT ?p In a container of constant volume, the mean free path is directly proportional to temperature and inversely proportional to pressure. The former dependence can be rationalized by noting that the faster the molecules travel, the farther on average they go between collisions. The latter also makes sense in that the lower the pressure, the less frequent are collisions, and therefore the further the average distance between collisions. Perhaps more fundamental than either of these considerations are dependences on size. As pointed out in the text, the ratio T/p is directly proportional to volume for a perfect gas, so the average distance between collisions is directly proportional to the size of the container holding a set number of gas molecules. Finally, the mean free path is inversely proportional to the size of the molecules as given by the collision cross section (and therefore inversely proportional to the square of the molecules
- Question : E B.1 - The mean speed is [1B.8] vmean = 8RT ?M ? ?? ? ?? 1/2 The mean translational kinetic energy is Ek = 12 mv2 = 12 m v2 = 12 mvr2ms= m 2 3RT M ? ?? ? ??[1B.3]= 32kT The ratios of species 1 to species 2 at the same temperature are vmean,1 vmean,2 = M2 M1 ? ?? ? ?? 1/2 and Ek 1 Ek 2 = 1 (i) vmean,H2 vmean,Hg = 200.6 4.003 ? ?? ? ?? 1/2 = 7.079 (ii) The mean translation kinetic energy is independent of molecular mass and depends upon temperature alone! Consequently, because the mean translational kinetic energy for a gas is proportional to T, the ratio of mean translational kinetic energies for gases at the same temperature always equals 1.
- Question : E B.2 - The root mean square speed [1B.3] is vrms = 3RT M ? ?? ? ?? 1/2 For CO2 the molar mass is M = (12.011 + 2
- Question : E B.3 - The Maxwell-Boltzmann distribution of speeds [1B.4] is f(v)=4? M 2?RT ? ?? ? ?? 3/2 v2e?Mv2/2RT and the fraction of molecules that have a speed between v and v+dv is f(v)dv. The fraction of molecules to have a speed in the range between v1 and v2 is, therefore, f (v) dv v1 v2 ? . If the range is relatively small, however, such that f(v) is nearly constant over that range, the integral may be approximated by f(v)?v, where f(v) is evaluated anywhere within the range and ?v = v2
- Question : E B.4 - The most probable, mean, and mean relative speeds are, respectively vmp = 2RT M ? ?? ? ?? 1/2 [1B.9] vmean = 8RT ?M ? ?? ? ?? 1/2 [1B.8] vrel = 8RT ?
- Question : E B.5 - (i) vmean = 8RT ?M ? ?? ? ?? 1/2 [1B.8] = 8(8.3145 J K?1 mol?1)(298 K) ?(2
- Question : E B.6 - The collision diameter is related to the collision cross section by ? = ?d2 so d = (?/?)1/2 = (0.36 nm2/?)1/2 = 0.34 nm . The mean free path [1B.13] is ?=kT ?p Solve this expression for the pressure and set ? equal to 10d: p=kT ?? = (1.381
- Question : E B.7 - The mean free path [1B.13] is ?=kT ?p = (1.381
- Question : P B.2 - The number of molecules that escape in unit time is the number per unit time that would have collided with a wall section of area A equal to the area of the small hole. This quantity is readily expressed in terms of ZW, the collision flux (collisions per unit time with a unit area), given in eqn 19A.6. That is, dN dt = ?ZWA = ?Ap (2?mkT)1/2 where p is the (constant) vapour pressure of the solid. The change in the number of molecules inside the cell in an interval ?t is therefore ?N = ?ZW A?t , and so the mass loss is ?w= m?N = ?Ap m 2?kT ? ?? ? ?? 1/2 ?t=?Ap M 2?RT ? ?? ? ?? 1/2 ?t Therefore, the vapour pressure of the substance in the cell is 1/ 2 w 2 RT p A t M = ?? ?? ??
- Question : P B.4 - We proceed as in Justification 1B.2 except that, instead of taking a product of three one- dimensional distributions in order to get the three-dimensional distribution, we make a product of two one-dimensional distributions. f (vx,vy )dvxdvy = f (vx2 ) f (vy2 )dvxdvy = m 2?kT ? ?? ? ?? e?mv2/2kTdvxdvy where v2 = vx2 + v2y . The probability f(v)dv that the molecules have a two-dimensional speed, v, in the range v to v + dv is the sum of the probabilities that it is in any of the area elements dvxdvy in the circular shell of radius v. The sum of the area elements is the area of the circular shell of radius v and thickness dv which is ?(?+d?)2
- Question : P B.6 - The distribution [1B.4] is f (v) = 4? 3/2 M 2?RT ? ?? ? ?? v2e?Mv2/2RT . The proportion of molecules with speeds less than vrms is P = f (v)dv 0 vrms ? = 4? 3/2 M 2?RT ? ?? ? ?? v2e? Mv2/2RT dv 0 vrms ? Defining a ? R / 2RT , P = 4? 3/2 a ? ? ?? ? ?? v2e?av2 dv 0 vrms ? = ?4? 3/2 a ? ? ?? ? ?? d da e?av2 dv 0 vrms ? Defining ?2 ? av2. Then, dv = a?1/2d? and P = ?4? 3/2 a ? ? ?? ? ?? dda a11/2 e??2 d? 0 vrmsa1/2 { ? } = ?4? 3/2 a ? ? ?? ? ?? ?21 3/2 1 a () e??2d? 0 vrmsa1/2 ? + 1/2 1 a ()d da e??2 d? 0 vrmsa1/2 ? ? ? ? ?? ? ? ? ?? Then we use the error function [Integral G.6]: e??2 d? 0 vrmsa1/2 ? = (?1/2 / 2)erf (vrmsa1/2 ) . d da e??2 d? 0 vrmsa1/2 ? = dvrmsa1/2 da ? ? ? ? ? ?
- Question : P B.8 - The average is obtained by substituting the distribution (eqn 1B.4) into eqn 1B.7: vn = vn f (v)dv 0 ? ? =4? M 2?RT ? ?? ? ?? 3/2 vn+2e? Mv2/2RT dv 0 ? ? For even values of n, use Integral G.8: vn =4? M 2?RT ? ?? ? ?? 3/2 (n + 1)!! 2 n+4 2 ? ?? ? ?? 2RT M ? ?? ? ?? n+2 2 ? ?? ? ?? 2?RT M ? ?? ? ?? 1/2 =(n+1)!! RT M ? ?? ? ?? n 2 ? ?? ? ?? where (n+1)!! = 1
- Question : P B.10 - Dry atmospheric air is 78.08% N2, 20.95% O2, 0.93% Ar, 0.04% CO2, plus traces of other gases. Nitrogen, oxygen, and carbon dioxide contribute 99.06% of the molecules in a volume with each molecule contributing an average rotational energy equal to kT. (Linear molecules can rotate in two dimensions, contributing two
- Question : P B.12 - The fraction of molecules (call it F) between speeds a and b is given by F(a,b) = f (v)dv a b ? where f(v) is given by eqn 1B.4. This integral can be approximated by a sum over a discrete set of velocity values. For convenience, let the velocities vi be evenly spaced within the interval such that vi+1 = vi + ?v: F(a,b) ? ? f (vi )?v On a spreadsheet or other mathematical software, make a column of velocity values and then a column for f(v) [1B.4] at 300 K and at 1000 K. Figure 1B.1 shows f(v) plotted against v for these two temperatures. Each curve is labeled with the numerical value of T/K, and each is shaded under the curve between the speeds of 100 and 200 m s
- Question : Q C.2 - The critical constants represent the state of a system at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature 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.
- Question : Q C.4 - The van der Waals equation is a cubic equation in the volume, V. Every cubic equation has some values of the coefficients for which the number of real roots passes from 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 some values of temperature and pressure for which the number of real roots of V passes from n to 1. That is, the multiple values of V converge from n to 1 as the temperature approaches the critical temperature. This mathematical result is consistent with passing from a two phase region (more than one volume 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 experimental result as the critical point is reached.
- Question : E C.1 - The van der Waals equation [1C.5a] is p= nRT V ? nb ?an2 V2 From Table 1C.3 for H2S, a = 4.484 dm6 atm mol
- Question : E C.2 - The conversions needed are as follows: 1 atm = 1.013
- Question : E C.3 - The compression factor Z is [1C.1] Z=Vm Vm
- Question : E C.4 - (i) According to the perfect gas law Vm o= RT p = (8.3145 J K?1 mol?1)
- Question : E C.5 - The molar volume is obtained by solving Z= pVm RT [1C.2], for Vm , which yields Vm =ZRT p = (0.86)
- Question : E C.6 - Equations 1C.6are solved for b and a, respectively, and yield b = Vc/3 and a = 27b2pc = 3Vc2pc . Substituting the critical constants b= 148 cm3 mol?1 3 = 49.3 cm3 mol?1 = 0.0493 dm3 mol?1 and a = 3
- Question : E C.7 - The Boyle temperature, TB, is the temperature at which the virial coefficient B = 0. In order to express TB in terms of a and b, the van der Waals equation [1C.5b] must be recast into the form of the virial equation. p= RT Vm ?b ?a Vm 2 Factoring out RT Vm yields p=RT Vm 1 1?b/Vm ? a RTVm ? ? ? ?? ? ? ? ?? So long as b/Vm < 1, the first term inside the brackets can be expanded using (1
- Question : E C.8 - States that have the same reduced pressure, temperature, and volume [1C.8] are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25
- Question : E C.9 - The van der Waals equation [1C.5b] is p= RT Vm ?b ?a Vm 2 which can be solved for b b=Vm ? RT p+ a Vm 2 = 4.00
- Question : P C.2 - From the definition of Z [1C.1] and the virial equation [1C.3b], Z may be expressed in virial form as ? Z =1+B 1 Vm ? ?? ? ??+C 2 1 Vm ? ?? ? ?? +L Since Vm = RT p (by assumption of approximate perfect gas behavior), 1 Vm = p RT ; hence upon substitution, and dropping terms beyond the second power of 1 Vm ? ?? ? ?? Z =1+B p RT ? ?? ? ?? +C RpT ? ?? ? ?? 2 =1+ (?21.7
- Question : P C.4 - Since B?(TB) = 0 at the Boyle temperature [Topic 1.3b]: B?(TB) = a + be?c/TB2 = 0 Solving for TB: ( ) 1/2 1/2 2 2 B 1 1 (1131K ) 5.0 10 K ln ( 0 1993bar ) ln (0 2002bar ) c T a b ? ? ? ? ? ? ? ? = ?? ? ?? = ?? ? ?? =
- Question : P C.6 - From Table 1C.4 Tc = 2 3 ? ?? ? ??
- Question : P C.8 - Substitute the van der Waals equation [1C.5b] into the definition of the compression factor [1C.2] Z= pVm RT = 1 1? b Vm ? ?? ? ?? ? a RTVm [Exercise 1C.7(a)] which upon expansion of 1 2 m m m 1 b 1 b b V V V ? ?? ? ?? = + + ?? ?? + ? ? ? ? ? yields 2 2 m m 1 1 1 a Z b b RT V V = + ?? ? ??
- Question : P C.10 - The Dieterici equation is p= RTe ?a/RTVm Vm ?b [Table 1C.4] At the critical point the derivatives of p with respect to Vm equal zero along the isotherm defined by T = Tc . This means that (?p / ?Vm )T = 0 and (?2 p / ?Vm2 )T = 0 at the critical point. T ?p ?Vm ? ?? ? ?? = p aVm ?ab?RTVm2 Vm 2 (Vm ? b)(RT) ? ? ? ?? ? ? ? ?? and ( ) { } 2 2 2 3 2 m m m m m 2 2 3 2 m m m m m m 2 4 2 T T ( )( ) ( ) ( ) p p aV ab RTV aV V ab RTV ab p V V V V b RT V V b RT ? ? ? ? ? ? ? ? ? ? ? + + ? ? ? = ? ? ? ? + ? ? ? ? ? ? ? ? ? ?? ? ?? Setting the Dieterici equation equal to the critical pressure and making the two derivatives vanish at the critical point yields three equations: and pc = RTce ?a/RTcVc Vc ?b aVc ? ab? RTcVc2 = 0 ?2aVc2 + 4Vcab + RTcVc3 ? 2ab2 = 0 Solving the middle equation for Tc, substitution of the result into the last equation, and solving for Vc yields the result Vc = 2b or b = Vc / 2 (The solution Vc = b is rejected because there is a singularity in the Dieterici equation at the point Vm = b.) Substitution of Vc = 2b into the middle equation and solving for Tc gives the result Tc = a / 4bR or a = 2RTcVc Substitution of Vc = 2b and Tc = a / 4bR into the first equation gives pc =ae?2 4b2 = 2RTce?2 Vc The equations for Vc, Tc, pc are substituted into the equation for the critical compression factor [1C.7] to give Zc = pcVc RTc = 2e?2 = 0.2707 . This is significantly lower than the critical compression factor that is predicted by the van der Waals equation: Zc(vdW) = pcVc / RTc = 3 / 8 = 0.3750 . Experimental values for Zc are summarized in Table 1C.2 where it is seen that the Dieterici equation prediction is often better
- Question : P C.12 - m 1 2 pV B p C p RT = + ? + ? +? [1C.3a] m 2 m m 1 pV B C RT V V = + + +? [1C.3b] Thus 2 2 m m Bp C p B C V V ? + ? +? = + +? Multiply through by Vm, replace pVm by RT{1+(B/Vm) + ...}, and equate coefficients of powers of 1/Vm: 2 2 m m B RT BB RT C R T B C V V ? + ? + ? +?= + +? Hence, B?RT = B, implying that B?= B RT Also BB?RT + C?R2T2 = C = B2 + C?R2T2, implying that C?=C?B2 R2T2
- Question : P C.14 - Write Vm = f(T, p); then dVm = p ?Vm ?T ? ?? ? ?? dT+ T ?Vm ?p ? ?? ? ?? dp Restricting the variations of T and p to those which leave Vm constant, that is dVm = 0, we obtain p ?Vm ?T ? ?? ? ?? =? T ?Vm ?p ? ?? ? ??
- Question : P C.16 - Z=Vm Vm o [1C.1], where Vm
- Question : P C.18 - The virial equation is m 2 m m 1 B C pV RT V V ? ? = ? + + + ? ? ? ? [1C.3b] or m 2 m m 1 pV B C RT V V = + + +? (a) If we assume that the series may be truncated after the B term, then a plot of pVm RT vs 1 Vm will have B as its slope and 1 as its y-intercept. Transforming the data gives p/MPa Vm/(dm3 mol
- Question : P C.20 - The perfect gas equation [1A.5] gives Vm =RT p = (8.3145 J K?1 mol?1)(250 K) 150

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