Question

Im doing a question where i am trying to find the work done for an adiabatic irreversible prefaces but i don't have the final temperature or pressure.. I also don't have the force or area of the flask being used to find the second pressure. How do I find it? Given 0.35 moles of methane (Cvm = 27.38 J/Kmol) at SATP. The change is due to minimum external pressure to reach the final state.

Answer 1

For an adiabatic process

Q = 0

Work done = Change in internal energy

Cv = moles X Cvm = 0.35 X 27.38 = 9.583 J

Cp = Cv + R = 9.583 + 8.314 = 17.897

work = Cv (T2-T1)

The final temprature = T2 =[ ( Cv + R(p2/p1) ) / Cp ] T1

We may assume that as final pressure = inital pressure as hange is due to minimum external pressure to reach the final state

So T2 =[ Cv + R / Cp ] T1 = (10.583 / 17.897 ) X 298 K

T2 = 298 K

so there will not be any change in temperature hence no work done.