Hi chemkid01. Welcome to Physics Forums!chemkid01 said:One mole of an ideal monatomic gas initially at 298 k expands from 1.0 L to 10.0 L. Assume the expansion is irreversible, adiabatic, and no work is done. Calculate delta S of the gas and delta S of the surroundings.
I know that delta dS = dq/T but q = 0 in adiabatic processes right? So does dS = 0 in all adiabatic processes?
Yes. Nice analysis. That's the one I was thinking of.chemkid01 said:Ok, so for this case, since the process is adiabatic and q = 0 and no work is done, then delta U = q + w = 0. And since it's an ideal gas, U only depends on T so T is constant as well. So could my dreamed up reversible process be a reversible isothermal expansion from 1.0 L to 10.0 L at 298 k?
No, because you won't be able to hold the temperature constant over that reversible path. So, the initial and final equilibrium states won't be the same as for the original irreversible path.chemkid01 said:Ok, thanks for the help! One more thing, I'm not exactly sure what makes a process reversible/irreversible. For example, the question says that it is an irreversible, adiabatic expansion from 1.0 L to 10.0 L. Why can't my alternate reversible path simply be a reversible, adiabatic expansion from 1.0 L to 10.0 L. Is that possible?
chemkid01 said:Hmm, ok, I don't think I quite understand reversibility.
In the case you are studying, no work is being done by the expanding gas, because the gas is essentially expanding against a vacuum. To carry out the process reversibly (and isothermally), you need to provide a force against the expansion, and very gradually back off on this force. This allows the system to do work against the force. To hold the temperature constant during this reversible path, you need to add heat Q. If you try to do it adiabatically, ΔU=-W, and the temperature will drop. So the reversible path between the initial and final states (which, for your problem are at the same temperature) can't be accomplished unless you add heat.Could you explain what it is about the path being reversible that makes it so you won't be able to hold the temperature constant?
Also, on a side note, since a reversible, isothermal expansion does bring you from the same initial state to the same final state as the irreversible, adiabatic expansion, does that mean that in this case, on the p-v diagram, the isotherm and the adiabat are the same line?
Entropy change in an irreversible adiabatic process refers to the change in the amount of disorder or randomness in a system as it undergoes a non-equilibrium, adiabatic process. It is a measure of the energy dispersal and the number of microstates available to a system.
The entropy change in an irreversible adiabatic process can be calculated using the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred, and T is the temperature. In an adiabatic process, there is no heat transfer, so the entropy change is equal to zero.
The entropy change and energy transfer in an irreversible adiabatic process are inversely proportional. This means that as the energy transfer increases, the entropy change decreases and vice versa. In an adiabatic process, there is no energy transfer, so the entropy change is equal to zero.
The second law of thermodynamics states that the total entropy of an isolated system always increases over time. In an irreversible adiabatic process, the system is isolated and there is no energy transfer, so the entropy change is always equal to or greater than zero, in accordance with the second law.
No, the entropy change in an irreversible adiabatic process cannot be negative. This is because in an adiabatic process, there is no energy transfer and the system is isolated, so the entropy cannot decrease. The entropy change can only be zero or positive in an irreversible adiabatic process.