thank you so much!!!! So far I didn't understand why starting presure is P2, not P1 in irreversible process. your exain make me understanding. But stiill could not imagine a situation that starting presure, p1, in the piston is going down very quickly even the volume is still v1.
what a wonderfull explications !!! That all i need to understand the reversible and irreversible process that i learn at my university !!! thank you very much for your explications.
Mam one specific question Which book to study for better understanding of thermodynamics? specifically how you explain first 2 mins I don't have that type of understanding regarding Reversible and Irreversible process. Hopefully get a positive response from you mam🙏🙂
Mam, I've one doubt in irreversible expansion process i.e in the graph ( irreversible) at pressure=P1 there is a volume V1 ,then why did you consider V1 at P2.Mam please reply back
At the end of the expansion, the pressure inside the system and outside must be the same (since the piston is at equilibrium). That is, Pext (atmospheric pressure) must equal P2. For the irreversible expansion, when the system is at V1 the external pressure is just the pressure from the atmosphere (since the mass has been removed). Since Pext = P2, we plot the pressure at V1 (initial volume) as P2 (external pressure). P2 reflects the EXTERNAL pressure, not the pressure of the system itself. Does that help?
@@Chemistry123 I keep arriving at an apparent paradox - If the ideal gas equation must be satisfied at every stage of it's, then the work done by the gas can be expressed as integral (PdV) from V1 to V2 irrespective of whether my process is reversible or not (I take work done according to its normal sign in physics). Can you please correct my mistake?
5:20 why can you presume that this process happens isothermal ? Isn't it already happening isobaric and since p * V = n * R * T with constant T and constant p, V would also be constant?
The process described at 5:20 is neither isothermal nor isobaric. Calling this process isothermal was a verbal typo. I should have said "no overall change in temperature" rather than isothermal. For the irreversible process described at 5:20, the starting and ending temperature is the same, so overall the change in T is 0, but we would expect the temperature to change during the process (during the expansion the temperature would initially decrease until we give the system enough time to re-establish thermal equilibrium and rise back up to the initial temperature). For a process to be isothermal, the temperature would have to stay constant throughout the process (not just the same for the initial and final states). The irreversible process here is not isothermal but the reversible version of this state change is isothermal. I can see why it seems that the process might be isobaric (constant pressure), but this is not the case. For a process to be isobaric, the pressure of the system would have to be the same throughout the process. Here, the external pressure is constant, but the pressure of the system is changing, so the process is not isobaric.
Amazing explanation - I was so confused before but this video helped so much. Just one question: so the reversible system can do more work than the irreversible system can, but where does this difference in energy go in the irreversible case? Does it leave as heat instead of being used for work?
In both the reversible and irreversible expansions the overall change in internal energy is zero (because the temperature change is zero and we have an ideal gas with no change in the moles of gas). So, for both cases, it must be that the exact same amount of energy lost by the system as work is regained by the system as heat. In the irreversible case, the system doesn't have to do much work to expand because it lifts only against the external pressure. So, in the irreversible expansion the system does a small amount of work and the same small amount of heat flows in to keep the internal energy change as 0 overall. In the reversible expansion, the system has to do more work. Not only does the reversible system lift against the external pressure, but it also has to do work to lift the sand. So, the reversible expansion involves the system doing a relatively large amount of work, and this energy is replaced by the same relatively large amount of heat flowing in. In both cases the net change in internal energy is 0, but the energy flow in and out is smaller for the irreversible case and larger for the reversible case. Does this help?
well explained...thnx ..by the way i hav got one doubt dat why u did consider only the external pressure in case of irreversible expasion .,.. i mean like the pressure inside the system is changing ryt?? can . u plz ... elaborate on dat??
But in real life... what's the difference? I can totally understand the first example. But where do reversible expansions occur in real life? Because the work done is hugely different and I still don't understand the difference in real life applications.
Yes, that's right. Since both are involving ideal gasses with the same temperature at the beginning and the end, the change in energy is 0 and thus q = -w.
