Chap05 soln

Physical Chemistry

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5 The Second Law: the machinery Solutions to exercises Discussion questions E5.1(b) See the solution to Exercise 3.14(a) and Example 5.1, where it is demonstrated that π T = a/V 2 m for a van der Waals gas. Therefore, there is no dependence on b for a van der Waals gas. The internal pressure results from attractive interactions alone. For van der Waals gases and liquids with strong attractive forces (large a ) at small volumes, the internal pressure can be very large. E5.2(b) The relation (∂G/∂T ) p =− S shows that the Gibbs function of a system decreases with T at constant p in proportion to the magnitude of its entropy. This makes good sense when one considers the de±nition of G , which is G = U + pV TS . Hence, G is expected to decrease with T in proportion to S when p is constant. Furthermore, an increase in temperature causes entropy to increase according to 1S = Z f i d q rev /T The corresponding increase in molecular disorder causes a decline in th Gibbs energy. (Entropy is always positive.) E5.3(b) The fugacity coef±cient, φ , can be expressed in terms of an integral involving the compression factor, speci±cally an integral of Z 1 (see eqn 5.20). Therefore, we expect that the variation with pressure of the fugacity coef±cient should be similar, in a very qualitative sense, to the variation with pressure of the compression factor itself. Comparison of ±gures 1.27 and 5.8 of the text shows this to be roughly the case, though the detailed shapes of the curves are necessarily different because φ is an integral function of Z 1 over a range of pressures. So we expect no simple proportionality between φ and Z . But we ±nd φ< 1 in pressure regions where attractive forces are expected to predominate and φ> 1 when repulsive forces predominate, which in behavior is similar to that of Z . See Section 5.5(b) for a more complete discussion. Numerical exercises E5.4(b) α = 1 V × ± ∂V ∂T ² p κ T ± 1 V ² × ± ∂p ² T ± ∂S ∂p ² T ± ² p = αV E5.5(b) 1G = nRT ln ± p f p i ² at constant temperature, p f p i = V i V f = nRT ln ± V i V f ² = ( 2 . 5 × 10 3 mol ) × ( 8 . 314 J K 1 mol 1 ) × ( 298 K ) × ln ± 72 100 ² 2 . 0 35 = 2 . 0J
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76 INSTRUCTOR S MANUAL E5.6(b) ± ∂G ∂T ² p =− S ± f ² p S f and ± i ² p S i 1S = S f S i ± f ² p + ± i ² p ³ ∂(G f G i ) ´ p ± ∂1G ² p {− 73 . 1 + 42 . 8 T/ K } J = 42 . 8JK 1 E5.7(b) See the solution to Exercise 5.7(a). Without knowledge of the compressibility of methanol we can only assume that V = V 1 ( 1 κ T p) V 1 . Then 1G = V1p ρ = m V so V = m ρ = 25 g 0 . 791gcm 3 = 31 . 61 cm 3 1G = ( 31 . 61 cm 3 ) × 1m 3 10 6 cm 3 ! × ( 99 . 9 × 10 6 Pa ) = + 3 . 2kJ E5.8(b) (a) 1S = nR ln ± V f V i ² = nR ln ± p i p f ² [Boyle’s Law] Taking inverse logarithms p f = p i e 1S/nR = ( 150 kPa ) exp ( 15 . 0JK 1 ) ( 3 . 00 mol ) × ( 8 . 314 J K 1 mol 1 ) !
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Chap05 soln - 5 The Second Law: the machinery Solutions to...

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