Chap05 soln

Physical Chemistry

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern