Explained: Combined 1st & 2nd Laws of Thermodynamics

In this video we will derive two forms of the combined first and second laws of thermodynamics (energy and enthalpy forms).
We will start with a statement of the first law, then move to a statement of the second law. then define the reversible heat addition and reversible work done on the system, and finally solve for the combined law (energy form). Then we will use the definition of enthalpy to derive the combined law in terms of enthalpy.

[@huffypuff0]

Very good! Thank you so much!

[@JoshTheEngineer]

No problem!

[@sohamprajapati3035]

Great explanation, Thank You

[@JoshTheEngineer]

You're welcome!

[@IKMCDANIEL]

Josh, what if your process isn't constant area - i.e. not a piston? Do you then have to had another term to the combined law: de = Pdv + VdP + TdS?

[@tiddlywinks497]

Probably a silly question, but in the enthalpy form what is the differential with respect to? Thanks!

[@nikan4now]

thanks for the great video. one question ..in the 1st law the term dq could be for an irreversible process in which case dq irrev would not equal Tds..I think it would equal Tds+dsgenerated...is this correct? or can we simply substitute dq irrev by Tds like you did? if so why? thanks a lot.

[@JoshTheEngineer]

Right, in the general first law equation, dq can be for any process, and in this case it would not equal Tds. It would be larger than the reversible dq, which is why you'll usually see it written as an inequality, where ds >= dq/T, where the equality applies to the reversible case, and the inequality applies to irreversible cases.
The reason we can substitute dq in the first law by Tds is because we have defined the process we want to look at as being reversible. Because we said the process is reversible, the equality (ds = dq/T) applies, and we can substitute that into the first law. The main goal of assuming a reversible process is because we can then express a combined first and second law in terms of only state variables that don't depend on the path. And when we have an equation in terms of only state variables, then it applies to any process.

[@hazazihashim3856]

thank you eden hazard

[@JoshTheEngineer]

I had to look up who that was. I can definitely see the resemblance.