[Donaldson Tan]

Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion

Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct?

What do you mean by non-expansion work? According to the Carnot Principle, the maximum work extracted from a process corresponds to the work done if the process is reversible. This maximum work is called the availability (D) of the system. The availability of system corresponds to the change of the Gibbs' free energy of the system. Since the availlability of the system is a state function, your statement is correct as long as the 3 paths share the same initial and final conditions of the gas.

[Donaldson Tan]

2.  delta H = q_p
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = q_p, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q  is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case,  q is not q_p?

Consider the case for isothermal expansion of 1 mole of a monoatomic ideal gas:

H = U + PV
PV = RT & U = (3/2)RT
H = (5/2)RT
dH = (5/2)R.dT

Does not the above expression show that enthalpy of an ideal gas is dependent on temperature, but independent of pressure?

Since dT = 0 for isothermal processes, then dH = 0

dH = Q + W = 0
Q = -W = P.dV

This above expression means that the heat input is converted to work done corresponding to the gas expansion.

I have checked many textbooks and websites and it seems that the constant pressure associated with the expression

How can a fixed mass of gas expand if its pressure and temperature are kept constant? (V = nRT/P)

You must keep in mind that P_external can be used as a substitute for P_gas if the pressure difference is negligible. It does not actually mean the gas is at constant pressure, but the value you get from using P_externaldV is a good approximation to the actual work done.

There are alot of approximations and simplification techniques (or tricks or assumptions) used in thermodynamics.

There is no Q_p or Q.

[Yggdrasil]

Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.

If you really think about it, it would be impossible for the gas to expand if it were always in equilibrium with its surroundings (since equilibrium implies that the state of the gas remains constant).  In order to accomplish transformation reversibly, you infintesimally distrurb the equilibrium, let the system move back to equilibrium, distrub infintesimally again, etc.  it is in these infintesimally pertubations of the system where you find your changes in free energy.

Question 2
http://www.chem1.com/acad/webtext/thermeq/TE4.html
Problem example 2 in the above website deals with the reversible isothermal expansion of an ideal gas which is relevant to our discussion. In the given answers, delta G is not zero but delta S(universe) is zero. If delta S(universe) is zero, shouldn’t delta G = zero since delta G = –T delta S(universe)?

delta G_sys = delta H_sys - T[sys] delta S[sys]

Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion

Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct? 

2.  delta H = q_p
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = q_p, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q  is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case,  q is not q_p?

Again, when delta H is defined as q_p, it is assumed that the process is a reversible process, and that P_ext = P_int.

[eutectic6002]

Thanks for sharing, geodome and Yggdrasil.
Still trying hard to grasp the basics of Chemical Thermodynamics.

Cheers,
eutectic6002  

[eutectic6002]

Hi everyone,

Back again for clarification with regards to the answer given by Yggdrasil for Question 2 in the earlier posting.

Again please consider the reversible isothermal expansion of an ideal gas.
Good to refer to Problem example 2 in the website listed below: http://www.chem1.com/acad/webtext/thermeq/TE4.html

In this case,
  delta U = 0
  delta H = 0
  delta G_sys < 0 and delta S_sys > 0
  delta S_universe = 0

Both the following equations are commonly quoted in textbooks and used for calculations:
Equation (1): delta G_sys = delta H_sys - T delta S_sys
Equation (2): delta G_sys = - T delta S_universe

Equations (1) and (2) do not give the same delta G value for the above expansion process.
Shouldn't both equations give the same delta G value?

Can it be that the isothermal expansion of the ideal gas is a spontaneous process (though carried out reversibly) and hence delta S_universe is not zero?

Another question: In this case, is pressure P considered constant?
Equation (1) is only applicable at constant P and constant T right?

Thank you once again ...

Cheers,
eutectic6002

[Donaldson Tan]

Here's the problem & solution from http://www.chem1.com/acad/webtext/thermeq/TE4.html for easy reference:

Problem Example 2

One mole of an ideal gas at 300 K is allowed to expand slowly and isothermally to twice its initial volume. Calculate q, w, ?S°sys, ?S°surr, and ?G° for this process.

Solution: Assume that the process occurs slowly enough that the it can be considered to take place reversibly.

    a) The work done in a reversible expansion is

    so w = (1 mol)(8.314 J mol–1 K–1)(300K)(ln 2) = 1730 J.

    b) In order to maintain a constant temperature, an equivalent quantity of heat must be absorbed by the system: q = 1730 J.

    c) The entropy change of the system is given by Eq. 10 (Part 2):

    ?S°sys = R ln (V2/V1) = (8.314 J mol–1 K–1)(ln 2) = 5.76 J mol–1 K–1

    d) As was explained in Part 2, the entropy change of the surroundings, formally defined as qrev/T, is
    ?S°surr = q/T = (–1730 J) / (300 K) = – 5.76 J mol–1 K–1.

    e) The free energy change is
    ?G° = ?H° – T ?S°sys = 0 – (300 K)(5.76 J K–1) = –1730 J

Comment: Recall that for the expansion of an ideal gas, ?H° = 0. Note also that because the expansion is assumed to occur reversibly, the entropy changes in (c) and (d) cancel out, so that ?S°world = 0.

[Donaldson Tan]

Equation (2): delta G_sys = - T delta S_universe

Shouldn't it be dH_sys = - T.dS_universe?