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 definition of G , which is G = U + pV T S . 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 S = 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 coefficient, φ , can be expressed in terms of an integral involving the compression factor, specifically an integral of Z 1 (see eqn 5.20). Therefore, we expect that the variation with pressure of the fugacity coefficient should be similar, in a very qualitative sense, to the variation with pressure of the compression factor itself. Comparison of figures 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 find 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 p κ T = − 1 V × ∂V ∂p T ∂S ∂p T = − ∂V ∂T p = αV E5.5(b) G = nRT ln p f p i at constant temperature, p f p i = V i V f = nRT ln V i V f = ( 2 . 5 × 10 3 mol ) × ( 8 . 314 J K 1 mol 1 ) × ( 298 K ) × ln 72 100 = − 2 . 0 35 = 2 . 0 J

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76 INSTRUCTOR S MANUAL E5.6(b) ∂G ∂T p = − S ∂G f ∂T p = − S f and ∂G i ∂T p = − S i S = S f S i = − ∂G f ∂T p + ∂G i ∂T p = − ∂(G f G i ) ∂T p = − G ∂T p = − ∂T {− 73 . 1 + 42 . 8 T / K } J = 42 . 8 J K 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 G = V p ρ = m V so V = m ρ = 25 g 0 . 791 g cm 3 = 31 . 61 cm 3 G = ( 31 . 61 cm 3 ) × 1 m 3 10 6 cm 3 × ( 99 . 9 × 10 6 Pa ) = + 3 . 2 kJ E5.8(b) (a) S = nR ln V f V i = nR ln p i p f [Boyle’s Law] Taking inverse logarithms p f = p i e S/nR = ( 150 kPa ) exp ( 15 . 0 J K 1 ) ( 3 . 00 mol ) × ( 8 . 314 J K 1 mol 1 ) = 274 kPa (b) G = nRT ln p f p i = − T S [ H = 0 , constant temperature, perfect gas] = − ( 230 K ) × ( 15 . 0 J K 1 ) = + 3450 J = 3 . 45 kJ E5.9(b) µ = µ f µ i = RT ln p f p i = ( 8 . 314 J K 1 mol 1 ) × ( 323 K ) × ln 252 . 0
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