Question

1) Compute the following values (in Joules) of w,q, AE, AH, AS, AA, and AG for 2.00 mole of an ideal gas that undergoes a reversible isothermal (300K) expansion from P = 5.00 atm to P = 0.500 atm.

Answer 1

The formulae needed are:

(P = pressure, R = Gas constant, T = temperature, q = heat change, w = work done, E = internal energy, H = enthalpy, A = helmholtz free energy, G = gibbs free energy, S = entropy, n = number of moles:

of ideal gas enou Yeve 1) The formulae needed are : - For reversible iso thermal expansion :) W = - 2.303 nRT logo 2. 3o 3 n Dros ü) DE = 0 for isothermal procesa -> 9=-W. ü) = q +w iv) AHE DE + ACPV) =) SHE ACPV). For ideal gae, PV = nRT so, DH = SCART) . But, T = constant for iso thermal. So, DH=0 om r) As = q / vi) DA = AE - A(+5) =) AA = - TAS vii) AG = AE - D(ts) televy FDG = -tas (So, AA = AG)

(in equ III, q and w represent heat taken up and work done on the system)

So, putting the values in each of the equations:

Pi = 5.00 atm, Pf = 0.500 atm, T = 300 K, R = 8.314 J K-1 mol-1 , n = 2.00 moles

i) w = -2.303 (2)(8.314 J K-1 mol-1 )(300 K)log(10) = - 11,488.3 J

ii) delE = 0

iii) q = -(- 11,488.3 J) = 11,488.3 J

iv) del H = 0

v) delS = 11,488.3 J / 300K = 38.3 J K-1

vi)del A = -(300 K)(38.3 J K-1) = - 11,488.3 J

vii) del G = - 11,488.3 J

You'll find that q, delA , delG and also w(if you consider magnitude) have same values.

These are because:

AG - DE - A(15) + ACPr) oo, DG = AA + ACPV) For ideal gas, under isothermal conditione, ACPU)=d. So, PAG - DA] -TAS DA = Also, ar, SA - - or, AA = -9 And, since AE=0 (q=-w]