Entropy increase in adiabatic irreversible compression

1. Feb 8, 2015

jorgdv

Hi there,

I was wondering if you could help me, I think I may have some concepts wrong or incomplete.

1. The problem statement, all variables and given/known data

We have an adiabatic cylinder of volume $V_1$ filled with a gas of pressure $p_1$ and temperature $T_1$ in thermal equilibrium, closed with a piston. All of a suden we rise the pressure on the outside to $p_2>p_1$, so the cylinder compress in a non reversible way. What is the change in entropy and temperature of the system?

2. Relevant equations

3. The attempt at a solution

All right, first of all, we are told that the cylinder is adiabatic, so $\Delta Q=0$ right? But change of entropy is defined as $\Delta S=T \Delta Q$, so according to that definition it should also be 0. But that sounds strange, since it's an irreversible process. I guess that definition of entropy does not apply to irreversible process, which one would apply and why?

Assuming that $\Delta Q=0$ we have

$\Delta U=c_v (T_2-T_1)=-p_2 (V_2-V_1)=-p_2 nR(\frac{T_2}{p_2}-\frac{T_1}{p_1})$

And from that equation we could get $T_2$ and thus $\Delta U$. Is this right?

Thank you very much

2. Feb 8, 2015

Staff: Mentor

That is not the equation for the entropy change. The equation for the entropy change involves integrating dS=dQ/T over a reversible path between the initial and final equilibrium states. So, to get the entropy change, you need to first solve for the temperature and volume in the final equilibrium state, and then dream up a reversible path between these same initial and final equilibrium states. For the reversible path you dream up, dQ won't be zero, and T won't be constant over the entire path.
Yes. Very nicely done, except that there should be an n in front of the Cv, assuming by Cv you mean the molar heat capacity.

After solving for T2, start thinking of how you would get from the initial equilibrium state to the final equilibrium state using a reversible path. That's what you need to do to find the entropy change.

Chet

3. Feb 9, 2015

jorgdv

But which could be the reversible path? Clearly it's not isothermal since the final temperature changes. It is neither isobaric and it is not adiabatic since then the entropy change would then be 0 through all the path. I know that a reversible path is one in which the equilibrium is conserved in all instants, so that means that in an ideal gas $PV=nRT$ should hold in all the path. Does that mean that I can choose an arbitrary path for example in pressure, as long as that equation holds all the time? Do I have to parametrize it?

4. Feb 9, 2015

Staff: Mentor

Yes. You could do what you said and integrate dU=TdS-PdV (which, after you divide by T gives you an exact differential for dS), or you could break down the path into, say, two segments, one at constant volume in which you increase the temperature and then one at constant temperature in which you decrease the volume. Either way, you get the same answer.

Chet

5. Feb 9, 2015

jorgdv

Okay, I understand now. Thank you!