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I have this problem. I have an ideal gas that goes through an irreversible adiabatic decompression. I have the initial state (P,T,V), and the final pressure, and I have to calculate the entropy difference of the proccess. So, what I know is that I can make up any reversible proces bewteen the initial and final state because entropy is a state function, and integrate the heat over that process, but I can't solve it. I'm aplying $P_1V_1^\lambda=P_2V_2^\lambda$ to get the final state, but as I am creating the final point of the adiabatic, I get entropy difference of $0$ for an invented reversible path. That way of getting the final state is not correct, right? I have $PV=nRT$ but I only have the final $P$ so I need another equation.

EDIT: I have tried also to use $\Delta U=W\Rightarrow C_v\Delta T=P_{ext}\Delta V$, get $T_{final}$, and go on, but I don't get the correct result either. I thought this was general: $dU=C_vdT$, when can I apply that equation?

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You are in an adiabatic process there is no "heat" –  Sankaran Jan 31 '13 at 23:41

3 Answers 3

up vote 4 down vote accepted

I'm going to assume that the setup of the problem is such that the expansion of the gas is a free expansion, namely it expands into a larger container without doing work on anything because otherwise, I'm skeptical that there is enough information to determine the final state. In this case, using the fact that the work done by the gas is zero ($W=0$), the first law tells us that $$ \Delta U = Q-W = Q = 0 $$ where the last equality comes from the fact that the process is adiabatic ($Q=0$). Since the change in internal energy is zero, this means that $\Delta T = 0$ because the internal energy of an ideal gas only depends on temperature and the number of particles in the sample. Now that you know the temperature remains constant and that the pressure is given, you can use the ideal gas law to determine the final volume of the gas, and you therefore know the entire final state.

If we weren't given enough information to determine the work done by the gas (like how I assumed a free expansion so that $W=0$), then I don't immediately see how to proceed or if it would be possible to proceed.

I'll let you attempt to do the rest; let me know if you'd like more detail and/or guidance!

Cheers!

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I'm sorry for the lack of information, the expansion is not free, it's a fast expansion to an external pressure of 1 bar, being the initial internal pressure 4 bars. Anyway, you helped and following "more or less" your same path I got the correct result. Thanks. –  MyUserIsThis Feb 1 '13 at 9:51
    
Cool, well glad to be of help all the same! –  joshphysics Feb 1 '13 at 9:59

For systems with fixed amount of gas (closed systems) you always need two variables to define any state: thermal and mechanical variables. Remember a thermodynamic state is usually defined by $n+2$ variables: a thermal variable, mechanical variable, and $n$ composition state variables.

You final state pressure is known. You know the system is adiabatic. Lets write the energy and entropy balance for this system on a differential basis (always a great way to analyze systems) \begin{align*} dU=-\delta W-\delta Q = -\delta W \quad (\text{adiabatic})\\ dS= -\frac{\delta Q}{T}+\delta S_{gen}= \delta S_{gen}\quad (\text{irreversible but adiabatic}) \end{align*} SO this clearly tells you that you need the amount of work to estimate change in internal energy. That is the only way to proceed! If you are not given any work, then the only possibility one can calculate is free-expansion ($W=0$). The answer for that case is given by Joshphysics.

You are right in the way you were headed when you tried to use $C_vdT$. \begin{align*} \int_{initial}^{final} dU= \int_{initial}^{final} C_v(T)dT=\int_{initial}^{final} \delta W =W \end{align*} Now ideal gas does not mean $C_v$ is constant. That is true only for perfect gases. For an ideal gas $C_v$ is a function of temperature. So either of the two things happened. If it is a perfect gas, i.e., $C_v$ is constant then you have made an error in calculation or you have integrate for an ideal, non-perfect gas by considering temperature dependence of $C_v$ into consideration.

Additional comment: You cannot equate $C_v\Delta T$ to $P\Delta V$. Remember the pressure is changing while the gas is expanding so $P$ is a function of $V$ and not a constant. In other words: \begin{align*} \int C_vdT=\int PdV \neq P\Delta V \end{align*} unless $P$ is a constant.

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The way to solve this is to first determine the initial state and the final state. Then construct a reversible path between the two states. Last, compute the entropy change along the reversible path.

You know the initial state. All you know about the final state is the pressure. This is not enough information to solve the problem unless (a) there is something special about the final pressure or (b) one can work out one more state variable from the given process. I do not see a way to go further without more information.

You cannot use $p_1 V_1^\gamma = p_2 V_2^\gamma$ since this equation applies ONLY to reversible paths for ideal gas and your path is not reversible.

You also know that the gas is idea. The internal energy $U$ for an ideal gas depends only on the temperature. But we have no idea of the temperature.

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