# 090318_homework7_answer_v3 - 1 P6.2 Calculate Δ A for the...

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Unformatted text preview: 1 P6.2) Calculate Δ 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 = − − 1 1 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 − − Δ = − = − = − × × × = × ∫ P6.3) Calculate Δ 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 = − + Δ = = = ∫ 1 1 3 0.500 bar 8.314 J mol K 350 K ln 22.1 10 J 10.5 bar − − × × × = − × 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 Δ G and Δ 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 Δ G and Δ A do or do not differ from one another. a) for the isothermal reversible path 2 1 1 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 i f P P V G VdP nRT nRT P V − − Δ = = = = × × × = − × ∫ 1 1 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 − − Δ = − = − = − × × × = − × ∫ 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 . Δ G – Δ A = Δ H – Δ U = Δ （ PV ) = Δ （ nRT ). Therefore, Δ G = Δ 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? ( ) ( ) ( ) 3 5 3 For a solid or liquid, ( , ,100 bar) ( , ,1 bar) ( , ,1 bar) 12.011 10 kg 99.0 10 Pa = 52.8 J 2250 kgm f i P f i P m m m f i m f i G V d P V P P M G C s G C s V P P G C s P P ρ − − Δ = = − = + − = + − × = + × × ∫ ( ) ( ) 1 1 3 Treating He as an ideal gas, , ,100 bar , ,1 bar 100 bar ln 1 mole 8.314 J mol K 298.15 K ln =11.4 10 J 1 bar f i P m m P f i G He g G He g VdP P RT P − − = + = + = × × × × ∫ This result is a factor of 216 greater than that for graphite....
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090318_homework7_answer_v3 - 1 P6.2 Calculate Δ A for the...

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