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As I was thinking about isobaric processes, one thing came to mind: In a reversible isobaric process, the value of Cp (molar isobaric heat capacity) is bigger than Cv (molar isochoric heat capacity) by exactly one R (ideal cas constant). For example, for argon, Cp is 2.5R while for Cv it is 1.5R.

This is due to the derivation, where under the pV integral, pV can be substituted by RT for one mole of the gas. This is only possible though, if the gas is in equilibrium with the outside pressure the whole time:

https://en.m.wikipedia.org/wiki/Isobaric_process

So my question is, would the value of the molar isobaric heat capacity change, if we applied an irreversible isobaric change?

That is, the gas is heated much more quickly that it can expand, so that the pressure in the gas is higher than the outside pressure on the piston, so that the gas and outside pressure are not in equilibrium the whole time.

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  • $\begingroup$ As you realise if equilibrium is not maintained a different value of $C$ will be measured compared to that measured while equilibrium is maintained i.e. under reversible conditions. The average heat capacity is $C_{av}\equiv q/\Delta T$ and as $q$ is different, reversible vs non-reversible conditions so is $C$. In the past rates of fast reaction were measured using this effect via the velocity of sound. $\endgroup$
    – porphyrin
    Jun 26, 2023 at 7:22
  • $\begingroup$ @porphyrin Hmm, but as $p_\text{ext}$ is constant, $W=\int{p_\text{ext}\cdot \text{d}V}= p_\text{ext}\cdot \int{\text{d}V}$ is constant for the the same $T_2$ and $T_1$ regardless of reversibility, so does $\Delta(p V)$. So does $\Delta U(\Delta T)$ and $\Delta H$ and $C_V$. Am I wrong? // UNLESS, the OP means a differential $C_V$ where heat is passed, T is locally raised but work is not being done proportionally yet. Hm, then what T would this differential $C_V$ refer to?// But in integral form from initial to final state, I think the former holds. $\endgroup$
    – Poutnik
    Jun 26, 2023 at 8:13
  • $\begingroup$ @Poutnick I didn't read the question carefully enough, was thinking of a reaction where due to rapid heating the reaction equilibrium does not have time to establish itself. Otherwise, I agree with your analysis. $\endgroup$
    – porphyrin
    Jun 26, 2023 at 14:54

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If the gas is initially in equilibrium with the surroundings and, in the final state, it is also in equilibrium with the surroundings, then $$p_{gas,init}=p_{gas, final}=P_{ext}$$then$$W=p_{ext}(V_{final}-V_{init})=p_{gas,final}V_{final}-p_{gas,init}V_{init}=\Delta (PV)_{gas}$$

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  • $\begingroup$ Okay that makes sense, now consider following case: An ideal gas is originally in equilibrium with surroundings, then the outside pressure is abruptly lowered and kept constant at that smaller value; this is commonly known as an "irreversible isobaric process" although the initial and end gas pressures are not the same; why is it called that then? $\endgroup$
    – Mäßige
    Jun 26, 2023 at 17:21
  • $\begingroup$ Well, for sure, it's a little ambiguous, but it is what people with experience normally call this. But, for sure, in that case, Q is not equal to $\Delta H=C_p\Delta T$ (even if the final state is at thermodynamic equilibrium). Run the analysis and see what you get. $\endgroup$ Jun 26, 2023 at 19:24
  • $\begingroup$ Yeah in this case, as the pressure is not constant Cp is not applicable. $\endgroup$
    – Mäßige
    Jun 26, 2023 at 20:59

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