Thermodynamics - Irreversible Adiabatic expansion

[omberlo]

Homework Statement

Closed system made of a cylinder containing air, compressed by a piston with weights on top of it. The piston is frictionless and its weight can be neglected. The system contains initially 10 kg of air at 27 °C and a pressure of 10 ata. The weights are suddenly halved. Considering the process adiabatic and air an ideal gas, calculate:

1) The temperature of air at the final state
2) The work done on the surrounding
3) The entropy increase of the air
4) The isentropic efficiency of expansion.

Homework Equations

pV = m Rspecific T
delta U = Q - W
delta U = Cv delta T

The Attempt at a Solution

p2 = p1 / 2 = 5 ata = 490,3 kPa
v1 = m Rspecific T / p1 = 0,88 m^3
Since the process is adiabatic, Q = 0 and delta U = - W

I'm stuck at question one already. How am I supposed to find T2 ?
I tried to use
T2 = p2 V2 / Rspecific
but can't because I don't have V2;
I tried using the first law for an adiabatic process of an ideal gas
Cv ( T2 - T1) = -Pextern ( V2 - V1)
but can't because I don't have V2;

Can't use the reversible adiabatic equation
T2 = (p1/p2)^((1-γ)/γ) * T1
because the weights are removed suddenly and thus it's not reversible..

How am I supposed to solve this?

 

[anathema]

Thank you for your post. I understand your confusion and I would be happy to help you solve this problem.

To find the temperature at the final state, you can use the ideal gas law: PV = nRT. Since the system is closed, the number of moles of air remains constant. You can rearrange the equation to solve for temperature:

T = PV/nR

To find the work done on the surrounding, you can use the first law of thermodynamics: ΔU = Q - W. Since the process is adiabatic, Q = 0. You can rearrange the equation to solve for work:

W = ΔU

To find the entropy increase of the air, you can use the equation ΔS = nCvln(T2/T1). Since the process is adiabatic, Q = 0 and ΔU = -W. You can substitute these values into the equation to solve for the entropy increase.

Finally, to find the isentropic efficiency of expansion, you can use the equation η = Wactual/Wisentropic. Wisentropic is the work done on the surrounding for an isentropic process, which you can calculate using the reversible adiabatic equation you mentioned.

I hope this helps you solve the problem. If you have any further questions, please don't hesitate to ask. Good luck!