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Part 1: Equilibrium

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The properties of gases

Solutions to exercises
Discussion questions
E1.1(b) 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. It is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other.This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation. 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 thatabove 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 fluids in the so-called supercritical region have both liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical state.) The van der Waals equation is a cubic equation in the volume, V . Anycubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 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 thenumber of real roots of V passes from n(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc . 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.

E1.2(b)

E1.3(b)Numerical exercises
E1.4(b) Boyle’s law applies. pV = constant pf = E1.5(b) so pf Vf = pi Vi pi Vi (104 kPa) × (2000 cm3 ) = 832 kPa = Vf (250 cm3 )

(a) The perfect gas law is pV = nRT implying that the pressure would be nRT V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. 25 g = 0.626 mol n= 39.95 g mol−1 p= (0.626 mol) × (8.31 ×10−2 L bar K−1 mol−1 ) × (30 + 273 K) = 10.5 bar 1.5 L not 2.0 bar. so p =

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INSTRUCTOR’S MANUAL

(b) The van der Waals equation is p= so p = RT a − 2 V m − b Vm (8.31 × 10−2 L bar K−1 mol−1 ) × (30 + 273) K (1.5 L/0.626 mol) − 3.20 × 10−2 L mol−1 − E1.6(b) (1.337 L2 atm mol−2 ) × (1.013 bar atm−1 ) = 10.4 bar ¯ (1.5 L/0.626 mol)2 so pf Vf = pi Vi

(a) Boyle’s law applies. pV = constantand pi =

pf Vf (1.48 × 103 Torr) × (2.14 dm3 ) = = 8.04 × 102 Torr Vi (2.14 + 1.80) dm3 (b) The original pressure in bar is pi = (8.04 × 102 Torr) × E1.7(b) Charles’s law applies. V ∝T and Tf = so Vi Vf = Ti Tf 1 atm 760 Torr × 1.013 bar 1 atm = 1.07 bar

E1.8(b)

Vf Ti (150 cm3 ) × (35 + 273) K = = 92.4 K Vi 500 cm3 The relation between pressure and temperature at constant volume can bederived from the perfect gas law pV = nRT so p∝T and pi pf = Ti Tf

The final pressure, then, ought to be pf = E1.9(b) pi Tf (125 kPa) × (11 + 273) K = 120 kPa = Ti (23 + 273) K

According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT so n =pV (1.00 atm) × (1.013 × 105 Pa atm−1 ) × (4.00 × 103 m3 ) = 1.66 × 105 mol = RT (8.3145 J K−1 mol−1 ) × (20 + 273) K

and m = (1.66 × 105 mol) × (16.04 g mol−1 ) = 2.67 × 106 g = 2.67 × 103 kg E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm /T will give the best value of R.

THE PROPERTIES OF GASES

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m RT M m RT RT which upon...
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