I was seraching some good lectures for chemistry in english but everything were in hindi and i found it difficult to understand mam,But your teaching is awesome and i can easily understand it.Thank you mam🙏🙏
Mam my doubt is that for isothermal process the value of q is o and as u said that the work is wasted will the value of w would also be o and thus deltaU would be zero 🤔 please tell me mam
Free expansion is technically isothermal, but it is misleading to call it that. Isothermal usually means isothermal and reversible, if not otherwise specified. It really is called an isenthalpic process, since the internal energy (and correspondingly, enthalpy) is constant. The process is irreversible, and doesn't do work, as a reversible isothermal process would do.
Please help mam , i have a doubt In adiabatic process ,as u said in this video del q =0 but not del t But if temperature is increasing or decreasing then heat should also increase or decrease ??
This is only true when you are dealing with a condensed form of matter whose change in density is insignificant as it is heated. For gasses specifically, they are free to do work or have work done on them, as undergo any process that isn't constant volume. This means, isothermal and adiabatic are two different processes for a gas. Isothermal means constant temperature. Since temperature doesn't change, and temperature alone determines internal energy, this means all heat added to the system becomes work done by the system. (not on a closed cycle, otherwise that would violate the 2nd law of thermodynamics). For an adiabatic process, there is no heat transfer. The change in internal energy therefore is equal and opposite to the work done by the system. Adiabatic reversible processes are also called isentropic, since they happen with constant entropy of the gas.
Ma'am first of all thanks. I have a doubt ma'am. In case of rearrangement of 1st equation, how can ∆U be =0 as ∆U is always 0 in case of free expansion only and need not be zero in case of just irreversible change as you have mentioned(there is no mention that it is a case of free expansion) . Also if ∆U is 0 then q and W must also be 0 as ∆U=q+w. And if we assume that process to be free expansion then w is 0(as w=-p∆v and in free expansion p=0). How can we say q=-w? Exactly for this doubt I was searching for a video. Please help.
It isnt neccessaey that q and w should be zero, look we talk in context to isothermal reactions, this means that there is no temperature change, now q can be changed only by two factors, they are either changing temperature or by doing work. Since U = q + w nd U = 0 and temp is also constant, the only other thing contributing to q is work done(w) since the sum is 0 we get q + w = O nd hence q = -w. The reactions are only for isothermal nd not for free expansion
Ya ofcourse , in container there will be gas molecules ( pressure internal present). We use piston only to see is there is pressure or not. Hope you understand 🙂
Mam , so there should be no expansion or compression taking place during free expansion ... I am not getting the concept behind it .... Mam pls explain it when you are free
I think , in free expansion there is no external pressure so pressure inside the system causes expansion! During calculations of work done we use external pressure. So work done in free expansion is zero but system still expanding!!
