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Posted: Wed Mar 13, 2019 9:24 pm
Why is isothermal deltaU = 0? A formula explanation would be nice if there is one!
Posted: Wed Mar 13, 2019 9:32 pm
I believe it is just a law. there can be no change in U if temperature does not change.
Posted: Wed Mar 13, 2019 10:01 pm
But if work is done, can't delta U change? Because delta U = q + w. And isothermal means there is no heat exchanged. Therefore q is 0 and delta U = w. If work is any value non zero, delta U will be too, so even if temperature is constant U will change with work.
*Edit: Please ignore what I've said above as it is wrong. What is said below is correct: q is only 0 if the system is adiabatic. Therefore if it's isothermal q=-w.
Posted: Wed Mar 13, 2019 10:19 pm
In isothermal dU=0, and we know that work is done so U=w, so what this means is that 0=q+w, q=-w/w=-q, while there's no temp change heat is released but counteracted by work. heat being measured in joules. If there was no heat across the system then it would be adiabatic system not isothermal.
Posted: Thu Mar 14, 2019 1:10 am
deltaU = 0 doesn't mean theres no heat just that theres no difference in the amount of heat in the beginning compared to the end (no change). Just like said above its just a way to tell you to use q = -w since q+w = 0
Posted: Thu Mar 14, 2019 1:25 am
Delta U for an isothermal reversible expansion is zero. These types of reactions usually take place in a heat reservoir. The energy that is lost when the system does work is replaced by heat energy from the surroundings. Therefore q and w are equal in magnitude and opposite in sign; q is positive since heat is coming into the system and w is negative since work is being done by the system. So when you add q and w to get delta U, it is zero.