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Posted: Sun Feb 04, 2018 2:40 pm
I have a couple of questions about situations where this equation is true.
First, why is it that the surroundings give heat to the system to make up for energy lost as work? (I understand that it keeps the temperature constant, but is this just something we observe or?)
Second, are these scenarios irreversible processes, reversible processes, or can they be both?
Posted: Sun Feb 04, 2018 2:59 pm
For the first question, we assume the system and its surrounding are at equilibrium, which means they have same temperature. When the system is doing work, it releases energy and its temperature decreases. It is a spontaneous process that the system absorbs heat from its surrounding to reach equiliruim, so the temperature remains constant.
For the second question, I think this equations is true for both reversible and irreversible process unless the process is isothermal.
Posted: Sun Feb 04, 2018 3:37 pm
If the system is at equilibrium, anything lost must be regained in another form to regain a state of equilibrium (ex: q and w).
Posted: Sun Feb 04, 2018 4:45 pm
The reason q=-w comes down to the fact that in isothermal expansions, delta U is equal to zero. (The temperature is constant, so the molecules have the same kinetic energy and same potential energy. Thus, there is no change in internal energy). We know that delta U equals q+w, so if delta U is zero, q must equal -w, or vice versa.
Also, these scenarios can be irreversible or reversible processes.
Posted: Sun Feb 04, 2018 5:44 pm
This formula is true for instances of isothermal expansions since having deltaT=0 (no change in temp), then deltaU (internal energy) would be 0 and since deltaU=q+w, then q would equal -w
Re: q=-w [ENDORSED]
Posted: Mon Feb 05, 2018 12:40 pm
Christina Bedrosian 1B wrote:This formula is true for instances of isothermal expansions since having deltaT=0 (no change in temp), then deltaU (internal energy) would be 0 and since deltaU=q+w, then q would equal -w