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How to derive that $\delta w = - PdV$? - Physics Stack Exchange I am not understanding how to derive this particular expression, which relates the inexact differential of work to the exact differential of volume, $$\\delta w = -PdV $$ My attempt: Reversible work
dp$ work and when do I use it? - Physics Stack Exchange Thanks a lot, this clears my confusion in all the PV state diagrams people describe work done as the area under the curve projected to X-axis and not the Y-axis, Hence dw=Pdv always
$TdS = dQ$ for irreversible process? - Physics Stack Exchange Your first equation (dU=Td$-PdV) describes the changes in the parameters between two closely neighboring thermodynamic equilibrium states For an irreversible process, the intermediate states are not at thermodynamic equilibrium, and you cannot use this equation
thermodynamics - Are you supposed to use the internal or external . . . Is it valid do use W=∫p_ext⋅dV for a free expansion? Also in the case where the piston is pushed down by a constant external pressure (it is compressed until the internal and external pressure are the same) which formula is correct to use, W=∫p_ext⋅dV or ∫p⋅dV? Thank you for all of your help!
Solving an inexact differential - Physics Stack Exchange In thermodynamics, we use $-\hspace {-1ex}d W=PdV$ to evaluate the work done by a process described by a path on the PV-diagram (i e a set of equilibrium states) The result is a line-integral, which path-dependent since $-\hspace {-1ex}d W$ is inexact (One does not get something like an anti-derivative to be evaluated at the endpoints ) Note that, for a given path on the PV-diagram, the