HW7_Answer - Chem340 Physical Chemistry for Biochemists Dr...

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1 Chem340 Physical Chemistry for Biochemists Dr. Yoshitaka Ishii Homework 7 Due Date Mar. 9, 2011 Ch 6 P6.2-6.8, P6.10, P6.13, P6.14, P6.16-6.20. P6.24, P6.28, P6.30, P6.38 P6.2) Calculate A for the isothermal compression of 2.00 mol of an ideal gas at 298 K from an initial volume of 35.0 L to a final volume of 12.0 L. Does it matter whether the path is reversible or irreversible? At constant T, we consider the reversible process. Because is a state function, any path whether reversible or irreversible, between the same initial and final states will give the same r dA SdT PdV A  11 3 esult. 12.0 L ln 2.00 mol 8.314 J mol K 298 K ln 5.30 10 J 35.0 L f i V f i V V A PdV nRT V   P6.3) Calculate G for the isothermal expansion of 2.50 mol of an ideal gas at 350 K from an initial pressure of 10.5 bar to a final pressure of 0.500 bar. At constant T, we consider the reversible process. Because G is a state function, any path between the same initial and final states will give the same result. ln 2.50 mol f i P f i P dG SdT VdP P G VdP nRT P 3 0.500 bar 8.314 J mol K 350 K ln 22.1 10 J 10.5 bar 
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2 P6.4) A sample containing 2.50 mol of an ideal gas at 298 K is expanded from an initial volume of 10.0 L to a final volume of 50.0 L. Calculate G and A for this process for (a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure of 0.750 bar. Explain why G and A do or do not differ from one another. a) for the isothermal reversible path 11 3 ln ln 10.0 L 2.50 mol 8.314 J mol K 298 K ln 9.97 10 J 50.0 L f i P f i if P P V G VdP nRT nRT PV    3 ln 50.0 L 2.50 mol 8.314 J mol K 298 K ln 9.97 10 J 10.0 L f i V f i V V A PdV nRT V  b) Because A and G are state functions, the answers are the same as to part a) because the systems go between the same initial and final states, T , V i T , V f . G A = H U = PV ) = nRT ). Therefore, G = A for an ideal gas if T is constant. P6.5) The pressure dependence of G is quite different for gases and condensed phases. Calculate G m (C, solid, graphite, 100 bar, 298.15 K) and G m (He, g, 100 bar, 298.15 K) relative to their standard state values. By what factor is the change in G m greater for He than for graphite?
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3       3 5 3 For a solid or liquid, ( , ,100 bar) ( , ,1 bar) ( , ,1 bar) 12.011 10 kg 0 99.0 10 Pa = 52.8 J 2250 kg m f i P fi P mm m f i m f i GV d P V P P M GC s s VP P GC s P P    11 3 Treating He as an ideal gas, , ,100 bar , ,1 bar 100 bar 0 ln 1 mole 8.314 J mol K 298.15 K ln =11.4 10 J 1 bar f i P P f i GH e g e g V d P P RT P  This result is a factor of 216 greater than that for graphite. P6.6) Assuming that H f is constant in the interval from 275 to 600 K, calculate G for the process (H 2 O, g, 298 K) (H 2 O, g, 525 K).
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HW7_Answer - Chem340 Physical Chemistry for Biochemists Dr...

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