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Physical Chemistry 7th ed [Solutions Manual]

# Physical Chemistry

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Part 1: Equilibrium

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1 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. E1.2(b) 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 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.) E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic 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 the number 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 T c . 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. Numerical exercises E1.4(b) Boyle’s law applies. pV = constant so p f V f = p i V i p f = p i V i V f = ( 104 kPa ) × ( 2000 cm 3 ) ( 250 cm 3 ) = 832 kPa E1.5(b) (a) The perfect gas law 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 . 62 6 mol so p = ( 0 . 62 6 mol ) × ( 8 . 31 × 10 2 L bar K 1 mol 1 ) × ( 30 + 273 K ) 1 . 5 L = 10 . 5 bar not 2 . 0 bar.

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4 INSTRUCTOR S MANUAL (b) The van der Waals equation is p = RT V m b a V 2 m so p = ( 8 . 31 × 10 2 L bar K 1 mol 1 ) × ( 30 + 273 ) K ( 1 . 5 L / 0 . 62 6 mol ) 3 . 20 × 10 2 L mol 1 ( 1 . 337 L 2 atm mol 2 ) × ( 1 . 013 bar atm 1 ) ( 1 . 5 L / 0 . 62 ¯ 6 mol ) 2 = 10 . 4 bar E1.6(b) (a) Boyle’s law applies. pV = constant so p f V f = p i V i and p i = p f V f V i = ( 1 . 48 × 10 3 Torr ) × ( 2 . 14 dm 3 ) ( 2 . 14 + 1 . 80 ) dm 3 = 8 . 04 × 10 2 Torr (b) The original pressure in bar is p i = ( 8 . 04 × 10 2 Torr ) × 1 atm 760 Torr × 1 . 013 bar 1 atm = 1.07 bar E1.7(b) Charles’s law applies.
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