HW Key _12 s problems part a

HW Key _12 s problems part a - S45. One mole of an ideal...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
S45. One mole of an ideal gas is allowed to expand reversibly and isothermally (25 °C) at a pressure of 1 atm to a pressure of 0.1 atm. a. What is the change in free energy Starting with the Gibbs equation for dG dG = SdT + VdP For constant temperature we obtain dG = VdP Using the ideal gas law and integrating from P 1 to P 2 we obtain dG = nRT dP P Δ G = nRT ln P 2 P 1 This is the expression for finding the change in Gibbs Free Energy for an isothermal reversible expansion of an ideal gas. Δ G = (1 mole)(8.31451 J/mol-K)(298 K)ln 0.1 atm 1.0 atm = 5.7 kJ b. What would be the change in free energy if the process occurred irreversible? Δ G = 5.7 kJ It doesn’t matter if the expansion is reversible or not because G is a state function; therefore, it is independent of path. S46. Consider the spontaneous process for one mole of water going from H 2 O ( ) at 110 °C and 1.00 atm to H 2 O (g) at 110 °C and 1.00 atm. Calculate H , and S and show that S > Q /T. The enthalpy of vaporization of water at 100 °C is found on p. 746 while the heat capacities of liquid water and steam can be assumed constant at 4.18 J/g K and
Background image of page 1

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

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

Page1 / 4

HW Key _12 s problems part a - S45. One mole of an ideal...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online