1
$\begingroup$

Let's say we have two completely identical cylinders of some ideal gas (same $P,V,T,n$, etc...), and we just want to double each one's volume.

First Cylinder

The first cylinder undergoes a simple reversible isothermal expansion, from $V_i$ to $2V_i$ (and likewise, $P_i \to P_i/2$). The work done is just $$W = P_i V_i \ln \left(V_f/V_i\right) = P_i V_i \ln 2 = nRT_i \ln 2$$ But, since this is an isothermal process, $\Delta U = 0$, and so therefore by the first law, the environment has to supply $Q_1 = W = P_i V_i \ln 2$ to the system in order for this process to occur.

Second cylinder

The second cylinder undergoes two steps:

  1. A reversible adiabatic expansion from $V_i \to 2V_i$. So, the work done will be $$W = \frac{P_iV_i^\gamma \left(V_f^{1-\gamma} - V_i^{1-\gamma}\right)}{1-\gamma} = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$ And, since this is an adiabatic process, the environment transfers no heat/energy, and so the system loses $\Delta U = -W = -P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$ from its internal energy to perform this work.

  2. A reversible isochoric heating back up to the original initial temperature $T_i$. Since, due to the earlier step, the system just lost $$ \Delta U = -W = -P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$ then all we need to do to get the system back up to its original internal energy/temperature is have the environment supply $Q_2 = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$ back into the system.

Question

In both processes, we have some cylinder migrate from some initial state $\left(P_i,V_i\right)$ to a final state of $\left(P_i/2, 2V_i\right)$. In both processes, the starting and end points are the same, and neither system has a net increase or decrease in its internal energy. So, for both cases, the environment must supply all the energy anyway (in the form of heat) to do everything. But, Mayer's Relation states that $\gamma = C_p/C_v = \left(C_v + R\right)/C_v$, which implies $$Q_1 = P_i V_i \ln 2 \ne Q_2 = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$

Why is this so? Why does the energy supplied by the environment to expand cylinder 1 have to be different than the energy supplied to expand cylinder 2? There's no net change in either system's internal energy and the starting and end points for each system are identical! So why does the environment have to supply different amounts of energy for one versus the other? Is there a hidden "irreversibility" here that I'm missing? I think all of the processes I mentioned can be done reversibly.

Edit: in fact, in order to make $Q_1 = Q_2$, we must violate Mayer's Relation and have $$\gamma = 1 - \lg \left(1 -\frac{R\ln 2}{C_v}\right) \ne \frac{C_v + R}{C_v}$$ It seems to me that the environment "has to" supply the same amount of energy to both cylinders for the sake of energy conservation (since neither cylinder has a net change in internal energy but goes from identical starting to identical end states). Oddly enough, the entropy transfer for both cylinder 1 (isothermal) and cylinder 2 (adiabatic + isochoric) is identical however if Mayer's Relation is true.

$\endgroup$
5
  • $\begingroup$ What makes you think that putting that amount of heat back in at constant volume will get you exactly back to the original temperature and P/2? $\endgroup$ – Chet Miller Oct 30 '20 at 0:35
  • $\begingroup$ Since no work is done at constant volume, then by the first law all of the added heat just goes into increasing the internal energy of the system, by the same amount of energy as the amount of heat added. So if the system adiabatically some lost $W$ amount of internal energy before, we can add that same amount of energy in the form of heat at constant volume to get the system back to its original internal energy. $\endgroup$ – ManRow Oct 30 '20 at 0:45
  • $\begingroup$ And since the internal energy of an ideal gas only depends on its temperature, then having the same internal energy as initially means it will have the same temperature as initially too. And if it has the same final temperature and volume as cylinder 1, it must therefore have the same final pressure too. $\endgroup$ – ManRow Oct 30 '20 at 0:45
  • $\begingroup$ Thanks. That makes sense. So the net effect in the 2nd case is that you've done less work and added less heat. That doesn't seem surprising. $\endgroup$ – Chet Miller Oct 30 '20 at 0:55
  • 1
    $\begingroup$ As Wolphram jonny pointed out, I was under the false impression that work & heat are state functions independent of the path taken. This is not the case (and as you also mentioned, one could just "do less work and add less heat"). $\endgroup$ – ManRow Oct 30 '20 at 1:28
2
$\begingroup$

You need more heat in the isothermal expansion because the gas does more work, due to the fact that the pressure is larger than during the adiabatic expansion.

$\endgroup$
2
  • $\begingroup$ Ahh, so if I understand correctly, while "internal energy" is a state function, the amount of heat and work can actually vary depending on what path a system takes as long as the first law $\Delta U = Q - W$ is satisfied? For that matter, then, one cannot therefore derive quantities for heat transferred and/or work done simply just by only looking at the start & end points even if everything is reversible, is that correct? The irreversible case seems straightforward, but just to confirm that this even applies to the fully-reversible case as well! $\endgroup$ – ManRow Oct 30 '20 at 0:31
  • 1
    $\begingroup$ yes correct, the path matters, for instance, remember that work is the area below the P-V diagram $\endgroup$ – Wolphram jonny Oct 30 '20 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.