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Let's say a cylinder has some ideal gas. We use a piston to suddenly expand the gas in isothermal condition from $1\ \mathrm{atm}$, $20\ \mathrm L$ to $0.5\ \mathrm{atm}$, $50\ \mathrm L$.

Since the gas is ideal, $pV = nRT$ implying that $nRT = 20\ \mathrm{L\ atm}$ initially. Later $pV$ changes to $25\ \mathrm{L\ atm}$. However, $nRT$ has to continue to be $20\ \mathrm{L\ atm}$ as the number of moles and the temperature remain constant.

Doesn't this violate the ideal gas equation, as $pV$ and $nRT$ are no longer equal. Does this mean that the irreversible expansion or compression of an ideal gas can not be isothermal unless the number of moles change (through some chemical reaction)?

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closed as unclear what you're asking by Mithoron, A.K., airhuff, Todd Minehardt, Tyberius Sep 10 '18 at 15:03

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    $\begingroup$ If it is isothermal, it will be 0.5atm in 40L $\endgroup$ – Raoul Kessels Sep 6 '18 at 15:28
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    $\begingroup$ You can’t control all of p, V, and T at the same time. $\endgroup$ – orthocresol Sep 6 '18 at 15:41
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    $\begingroup$ Well, there is some misunderstanding or mistake, but I'm afraid OP didn't make it clear enough to tell what it is. There may be some options: 1. misprint of volume 2. isothermal/adiabatic confusion 3. heavily non-ideal gas 4. ? $\endgroup$ – Mithoron Sep 6 '18 at 18:36
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    $\begingroup$ How would this ensure that the pressure was 0.5atm? If the piston controls the volume it doesn't also control the pressure. $\endgroup$ – matt_black Sep 6 '18 at 22:09
  • $\begingroup$ @orthocresol That's what I thought as well. I found a similar question in my chemistry textbook so I was wondering if such a process requires a chemical change (resulting in the change of $n$) or not. $\endgroup$ – Aman Vernekar Sep 7 '18 at 3:25
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The ideal gas law describes the behavior of an ideal gas exclusively for thermodynamic equilibrium states. During an irreversible expansion or compression, the gas passes through non-thermodynamic-equilibrium states, except at the very beginning and at the very end. In these irreversible non-equilibrium states, the ideal gas law does not correctly describe the gas behavior. For rapid expansions and compressions, the force that the gas exerts on the piston depends not only on the gas volume, but also on the rate of change of gas volume. So yes, in irreversible/expansion of an ideal gas, the ideal gas law is violated.

That said, if we can control the force between the gas and the piston by some external means (say by using an automatic control system in conjunction with a flush mounted pressure transducer in the inner piston face), we can calculate the amount of work that the gas does on the piston (without trying to resort to the ideal gas law) even for an irreversible expansion/compression. This then would enable us to use the first law of thermodynamics to calculate the final state of the gas (at least for an adiabatic process).

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  • $\begingroup$ Hmm, I interpreted the question to not be about the process but rather the initial/final states, but I'm probably wrong. $\endgroup$ – orthocresol Sep 6 '18 at 17:40
  • $\begingroup$ That explains what happens during the process. However, at the end of the process, $pV$ and $nRT$ are not equal either. $\endgroup$ – Aman Vernekar Sep 7 '18 at 3:21
  • $\begingroup$ If the system is again at thermodynamic equilibrium in the final state, they certainly are equal. What gave you the misconception that they wouldn't be? $\endgroup$ – Chet Miller Sep 7 '18 at 11:52
  • $\begingroup$ @ChesterMiller So does that imply that the gas is not at thermodynamic equilibrium at the final state I've mentioned (where $pV ≠ nRT$) and the process will continue for more time till it attains the equilibrium? $\endgroup$ – Aman Vernekar Sep 8 '18 at 14:02
  • $\begingroup$ The final state is thermodynamic equilibrium, and, in that state, pV=nRT $\endgroup$ – Chet Miller Sep 8 '18 at 15:03
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Doesn't this violate the ideal gas equation, as pV and nRT are no longer equal?

Yes, it does.

Does this mean that the irreversible expansion or compression of an ideal gas can not be isothermal unless the number of moles change (through some chemical reaction)?

Yes, that's right. I think it is quite misleading to speak of the process "...irreversible expansion or compression..." rather than the initial and final states, which is what you're appealing to when you use the ideal gas law.

As you've argued, regardless of process, there is no way an ideal gas with $(p, V, n, T) = (1\,\pu{atm}, 20\,\pu{L}, n_1, T_1)$ can get to $(p, V, n, T) = (0.5\,\pu{atm}, 50\,\pu{L}, n_2, T_2)$ if $n_1=n_2$ and $T_1=T_2$.

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  • $\begingroup$ Didn't you mean 50 L in the second state (in line with what the OP speculated over)? I think that with 40 L, the final state can be attained. $\endgroup$ – Chet Miller Sep 7 '18 at 0:57

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