NO. 3 One mole of an ideal gas with Cv=1.5R undergoes an irreversible adiabatic

Question

NO. 3

One mole of an ideal gas with Cv=1.5R undergoes an irreversible adiabatic expansion from an initial state described by T=300K and P=1.00Bar against a constant external pressure ofr 0.500bar until the final temperature of 240K (Surroundings at 300K, 0.500bar). Calculate with explanation of that indicated properties for this change of state.

a) Find Q=?

b)Find W=?

c) Find Delta H=?

d) FInd Delta S=?

e)Find Delta S surroundings

f) Can you calculate dG for this process? Why? And if the process spontaneous?

Explanation / Answer

For an adiabatic process,

a) Q = 0

b) w = dU = nCv(T2 - T1)

                 = 1 x 1.5 x 8.314(240 - 300)

                 = -748.26 J

c) dH = nCpdT

Cp = Cv + R

      = 1.5R + R

       = 2.5R

dH = 1 x 2.5 x 8.314(240 - 300)

      = -1247.1 J

d) dS = dH/T = -1247.1/-60 = 20.785 J/K

e) dS surrounding = 0

f) dG = dH - TdS

         = -1.2471 - (-60 x 0.0200785)

         = 0.424 kJ

the process is non-spontaneous as sign of dG is +ve.

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