Question

In class we discussed a reversible, isothermal compression of an ideal gas. The initial point (1) is an ideal gas in equilibrium at pressure 1 (P1), volume 1 (V1) and temperature (T), and the final point (2) is an ideal gas in equilibrium at pressure 2 (P2), volume 2 (V2) and temperature (T), where V2

d) For each differential step, the change in entropy is given by dS=qrev/T =!!!"#! . Since T is constant , this express ion can be integrated along the path of the process to find ∆S=qrev/T. Calculate the entropy change for the process in Joules/Kelvin (J/K)

e) The state function enthalpy is H=U+PV. Calculate the enthalpy change ∆H for the process in Joules.

f) The Gibbs free energy is G = H–TS. Since this process is carried out at constant temperature, we find ∆ G=∆(H−TS)=∆H−T∆S Calculate ∆G for the process in Joules.

g) This process is carried out reversibly, hence at equilibrium throughout. Under certain constraints on the system , we found that ∆G= 0 for equilibrium processes , so you might have (incorrectly) expected ∆G= 0 for the above process In fact,∆G≠0 , as you should have found in part f . What aspect of this process violates one or more constraints that were used in deriving ∆G= 0 for equilibrium processes?

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