Question

What is the value of delta g and delta a of an ideal gas at 298 k from 10l to 50 l in isothermal reversible process?

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.
pls make it brainliest

Answer 2

Hey dear,

● Answer -
∆G = 9969 J
∆A = -9969 J

● Explaination -
# Given-
n = 2.5
T = 298 K
V1 = 10 L
V2 = 50 L

# Solution -
In isothermal reversible process, ∆T = 0 & ∆H = 0.

Work done in isothermal reversible system -
W = -nRTln(V2/V1)
W = -2.5 × 8.314 × 298 × ln(50/10)
W = -9969 J

Value of ∆A -
∆A = W
∆A = -9969 J

Value of ∆G -
∆G = ∆H - W
∆G = 0 - 9969
∆G = 9969 J

Hope this helps you...