ok so i've spent HOURS on this and am still confused, my textbook is no help either....
so here's the problem:

In the previous homework we compared isothermal one-step, irreversible work with
reversible isothermal work. We also compared a one-step isothermal process and a
one-step adiabatic process. Now we compare a one-step adiabatic, irreversible
process with a reversible, adiabatic process. Letís consider that both of these
processes start from the same P,V,T, called P1, V1, T1 , for both processes and end at the
new pressure is: P2 = (1/3) * P1, for both the reversible and irreversible adiabatic expansions.
. Consider this gas to be an ideal gas with heat capacity Cv = (3/2) *nR
The questions we consider are will the two systems end up at
the same final state? And will the processes give us different amounts of heat, work and
energy change?

Tell me if i'm correct with the following because we had to figure these out for the processes

Equation of State: PV = nRT |||PV = nRT
First Law A: deltaU = w = -nR*ln(V2/V1)|||deltaU = w = Pexternal * (V2-V1)
Work: w = -nR*ln(v2/v1) ||| w = -Pext(V2-V1)
Heat q = 0 |||| q = 0
First Law B: deltaU = Cv*ln(T2/T1) ||| deltaU = Cv*(T2-T1)
Relate T to V: Cv*ln(T2/T1) = -nR*ln(V2/V1) ||| Cv*(T2-T1) = -Pext(V2-V1)

a) Explain why the work cannot be greater than an upper bound: |w| < Cv*T1

ok so i don't understand this at all...what if T2 = 3T1 then wouldn't work have to be Cv*2*T1? isn't the upperbound of work just Cv*(T2-T1)???

cool.gif Using the fact that P2 < Pext , which means the pressure along the reversible path is always greater than the one on the irreversible path, to explain, qualitatively, why one will get more work from one case than the other.

c) Use the premise that the work obtained from the system following the reversible
path is greater than that from the irreversible path (i.e.
−w(reversible) > −w(irreversible) ) to explain qualitatively why the final temperature of the irreversible expansion must be larger than that of the reversible expansion.

d) Qualitatively explain why the volume of the irreversible expansion will be different
from that of the reversible expansion. Will the volume expansion for the reversible
process be larger than that of the irreversible process?

thank you! i really don't know where to start on this...