Hi everyone,

Lets say we are looking at an Irreversible isothermal process:

Quote, "An isothermal process is a thermodynamic process in which the temperature of the system stays constant; ?T = 0. This typically occurs when a system is in contact with an outside thermal reservoir, and the system changes slowly enough to allow it to adjust to the temperature of the reservoir. The opposite extreme in which a system exchanges no heat with its surroundings is known as an adiabatic process." end quote.

en.wikipedia.org/wiki/Isothermal

so, if an isothermal process means that Delta T =0, does that mean delta P=0 also, and if so how can the volume change.

hear is the question im working on and why im so confused by this.

A sample of 1.0 mol He (assume ideal) with a Cv = 3/2 R, initially at 298K and 10L is expanded, with the surroundings maintained at 298K, to a final volume of 20L, in two ways:

n=1mol, Cv= 12.471 JK, T_initial = 298K, V_initial = 10L, V_final = 20L

T_surr = constant = 298K  (therefore the T_system should be constant yes )

(a.) Isothermally and reversibly

(b.) Isothermally against a constant external pressure of 0.50atm

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

ok, so i found it fairly easy to answer part a, and am confident with my answer:

(a.)isothermally and reversibly

w_rev=-nRT ln(V_final/V_initial) = -1717.322 J

* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:

q = -w = 1717.332 J

Delta S = qrev / T = 5.763 J/K

If: PiVi = PfVf = nRT

then: Delta(pV) = 0

Delta H = Delta U + Delta(pV) = 0 + 0 = 0

Delta T = 0 (isothermal)

Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q )

Delta A = Wrev = -1717.322 J

my working for  (b.) is as follows, I get stuck at Delta T.

(b.)Isothermally against Pext = 0.50atm
     (Isothermally and Irreversibly then isnt it )

w_irrev = -Pext . Delta V = -506.6 J

Delta U = q + w = 0,  (q = -w)

Delta H = Delta U + Delta (pV) = 0 + 0 = 0

Delta S = nR ln (V_final / V_initial) = 5.763 J/K

now for Delta T, it should equal to zero being an isothermal process shouldnt it?

Or is it:

wirrev = -Pext . Delta V = Cv . Delta T

so: Delta T = w_irrev / Cv = -40.623K

and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?

If anyone can check my work for part (a.) and the work attempted for part (b.) it would be very helpfull.

cheers,

madscientist :albert:

(a.) Isothermally and reversibly

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

w_rev=-nRT ln(V_final/V_initial) = -1717.322 J

* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:

q = -w = 1717.332 J

Delta S = qrev / T = 5.763 J/K

If: PiVi = PfVf = nRT

then: Delta(pV) = 0

Delta H = Delta U + Delta(pV) = 0 + 0 = 0

Delta T = 0 (isothermal)

Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q )

Delta A = Wrev = -1717.322 J

Yes, you are correct for part (a)

state functions are function whose values are path-independent.

the notation for capital is not for highlighting state functions. small capital serves to highlight the value of the state function per unit mass

eg. we use h or H to represent enthalpy. however, the unit for h is kJ/kg whereas the unit for H is kJ.

so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.

Or is it:

wirrev = -Pext . Delta V = Cv . Delta T

so: Delta T = w_irrev / Cv = -40.623K

and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?

We are considering an isothermal process, so this approach is wrong.

(b.) Isothermally against a constant external pressure of 0.50atm

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

(b.)Isothermally against Pext = 0.50atm
     (Isothermally and Irreversibly then isnt it )

w_irrev = -Pext . Delta V = -506.6 J

Delta U = q + w = 0,  (q = -w)

Delta H = Delta U + Delta (pV) = 0 + 0 = 0

Delta S = nR ln (V_final / V_initial) = 5.763 J/K

now for Delta T, it should equal to zero being an isothermal process shouldnt it?

We are considering the case for an isothermal & irreversible process.
dU = Q + W
since dU = n.Cv.dT where dT = 0, then dU = 0 => Q = -W
W = -p_ext.dV = -50650 J
Q = -W = 50650J
deltaS = n.R.ln(V_final/V_ini) = (1.0)(8.314)ln(2) = 5.763J/mol
A = U - TS => dA = dU -TdS => deltaA = 0 - T.deltaS = -298*5.763 = 1717J
G = H - TS => dG = dH - T.dS
dH = n.Cp.dT => dH = 0 for isothermal process
dG = -T.dS => deltaG = -T.deltaS = deltaA = 1717J

so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.

The small capital denotes the value of the state function per unit mass. mass can be expressed in either in kg or moles. w can refer to amount of work done by the system per mole of substance or per kg of substance in the system.

With regards to your question on isothermal process, we will be working on a molar basis and not kg because you are provided the molar quantity (n=1.0mol) of the gas in interest.

It actually matters if the state function is express in terms of kg or moles, because the resultant equation will not be exactly the same.

let's consider molar basis for enthalpy of a perfect gas
h = u + pv = u + RT
unit of h = J/mol
unit of u = J/mol
unit of R = J/mol.K
unit of T = K

let's consider mass basis for enthalpy of a perfect gas
h = u + pv = u + RT/M where M is the molar mass of the gas
unit of h = J/kg
unit of u = J/kg
unit of R = J/mol.K
unit of T = K
unit of M = kg/mol
unit of RT/M = J/kg

from the above example, we can see that the equations for mass and molar basis are not almost entirely the same.

Hi everyone,
Would like to seek your help to clarify some doubts with regards to the questions posted by madscientist.

(a) isothermal, reversible expansion of an ideal gas
Question 1: For a reversible process, shouldn't delta G = 0?

(b) isothermal expansion of an ideal gas against a constant external pressure
Question 2: In this case, delta H = 0 but q is not equal to 0.
Since the external pressure is kept constant, shouldn't delta H = q_p = q?

Question 3: G is a state function. Hence the value of delta G is path-independent. For (a) and (b), are the delta G values calculated the same?

Thank you.

Regards,
eutectic6002

After some though, I have a better answer for (b):

You have to make a distinction between P_ext and P_int, that is the pressure of the surroundings and the pressure of the system.  delta H = q_p only when the pressure of the system (P_int) stays constant.  For a reversible process, P_ext = P_int, so you need not make the distinction here.  But, for irreversible proceses the two are not necessarily the same.  In the irreversible, isothermal expansion at constant external pressure, P_ext stays constant, but P_int decreases throughout the expansion, until P_ext = P_int.

Reversible processes are also called equilibrium processes. The idea is that in a reversible process, such as a gas expansion, the system is always in balance (the external and internal pressures are only infinitesimally different) and so essentially in equilibrium. [Source: Page 21 of Energetics and Equilibria, Lecture notes from University of Cambridge]

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.

http://en.wikipedia.org/wiki/Quasistatic_equilibrium

Reversible processes are not necessary at equilibrium, but equilibrium is a good approximation to describe reversible processes. If these processes are really at equilibrium, then dG will actually be zero. We apply the same equations on reversible and equilibrium processes, but dG isn't always zero. Does this not indicate that reversible processes are not necessary at equilibrium?