- #1
madtaz
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2 litres of N2 held in a piston at a pressure of 5 atm initially held at
T=273K expands isothermally until the final pressure is 1 atm. What is
the work done on the gas in this expansion under (i) irreversible and (ii)
reversible conditions? Comment on the magnitude of your answers to
(i) and (ii). (You may assume that N2 behaves as an ideal gas.)
My attempt:
for irreversible we have : w' = p(ext) (Vf -Vi)
I've taken p(external) to be 5 atm = 5.05x10^5 Pa
V (initial) is 2 litres = 0.002 m^3
to find Vfinal i used V = P x V (initial)/ P final
to get V final = 0.01
Putting these numbers into w' = p(ext) (Vf -Vi) to get 4040 J, but for the work done on the system this is -4040j
For reversible is used w' = nRT ln (Vf / Vi) to get 1625 J, so for work done on the system = -1625 J
But the Reversible process should be higher as maximum work occurs so the value for the irreversible should be lower.
Where have I gone wrong?
T=273K expands isothermally until the final pressure is 1 atm. What is
the work done on the gas in this expansion under (i) irreversible and (ii)
reversible conditions? Comment on the magnitude of your answers to
(i) and (ii). (You may assume that N2 behaves as an ideal gas.)
My attempt:
for irreversible we have : w' = p(ext) (Vf -Vi)
I've taken p(external) to be 5 atm = 5.05x10^5 Pa
V (initial) is 2 litres = 0.002 m^3
to find Vfinal i used V = P x V (initial)/ P final
to get V final = 0.01
Putting these numbers into w' = p(ext) (Vf -Vi) to get 4040 J, but for the work done on the system this is -4040j
For reversible is used w' = nRT ln (Vf / Vi) to get 1625 J, so for work done on the system = -1625 J
But the Reversible process should be higher as maximum work occurs so the value for the irreversible should be lower.
Where have I gone wrong?
