Question

2.00 mole of an ideal gas at 330k, 3.50 atm and 3.0L are isothermically compressed causing a change of entropy in the system equal to -25.0 J*K^-1. For such situation, you are asked to determine

(a) The final volume of the system

(b) The final pressure of the system

(c) The change in internal energy

(d) The change in enthalpy

(e) The change in Helmholtz energy

(f) The change in Gibbs energy

Answer 1

a)

It is isothermal; therefore:

dS = n R Ln(Vf/Vi)

==> Ln(Vf/Vi) = dS/(n R)

==> Vf/Vi = e^(dS/(n R))

==> Vf = Vi e^(dS/(n R))

==> Vf = 3 * e^(-25/(2*8.314))

==> Vf = 0.66706 L

==> Vf = 0.667 L = 0.667 x 10^-3 m3

b)

P V = n R T

==> Pf = n R T/Vf = 2*8.314*330/(0.66706*1e-3) = 8.226e6 = 8.23 x 10^6 Pa

c)

The change in internal energyis zero, because it is isothermal.

dE = 0

d)

dH = dE + d(PV)

==> dH = (Ef - Ei) + (Pf Vf - Pi Vi)

==> dH = 0 + ((8.226e6)*(0.66706e-3) - (3.05*1.01e5)*(3e-3))

==> dH = 4563 J

==> dH = 4.56 x 10^3 J

e)

dA = dE - d(TS)

T is constant:

==> dA = dE - T dS

==> dA = 0 - 330*(-25)

==> dA = 8250 J

f)

dG = dH - d(TS)

T is constant:

==> dG = dH - T dS

==> dG = 4563 - 330 * (-25)

==> dG = 12813

==> dG = 1.28 x 10^4 J