Question

2.50 mole of ideal gas at 298k epands from 10l to 50l what will be deltag and deltaa

Answer 1

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, Tinitial = 298K, Vinitial = 10L, Vfinal = 20L

Tsurr = constant = 298K  (therefore the Tsystem 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

wrev=-nRT ln(Vfinal/Vinitial) = -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 )

wirrev = -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 (Vfinal / Vinitial) = 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 = wirrev / 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.
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