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Dec 9, 2020 at 12:45 comment added John Darby I do not follow your edit. Regarding p and V, it can be calculated using the external pressure, or for a reversible process or a slow process using the internal pressure. See physics.stackexchange.com/questions/243614/… on this exchange.
Dec 9, 2020 at 6:28 comment added nice life to exist-questioner Any mistake in my edit?
Dec 9, 2020 at 5:07 comment added nice life to exist-questioner Just one thing here d(PV) is being pressure external did?
Dec 8, 2020 at 16:56 history edited John Darby CC BY-SA 4.0
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Dec 8, 2020 at 16:44 comment added John Darby Internal energy is always the energy of the mass within an open or closed system, for the open system enthalpy in /out affects the internal energy. I will update my earlier answer to clarify this for a closed system.
Dec 8, 2020 at 16:38 comment added John Darby For a reversible process the work of expansion is $\int_{}^{}p \enspace dV$,.See other questions on this exchange regarding reversible/irrerversible work, e.g., physics.stackexchange.com/questions/451176/…
Dec 8, 2020 at 16:36 comment added Brian Could you explain why we use enthalpy rather than internal energy in open systems? Both are measuring energy total only , right? Or does internal energy not actually measure the 'full energy' in an open system, as in you gotta add PV to get the full picture?
Dec 8, 2020 at 16:35 comment added nice life to exist-questioner Ok So d(PV) is being mass exchange work and PdV forms work due to heat supplied!
Dec 8, 2020 at 16:29 history edited John Darby CC BY-SA 4.0
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Dec 8, 2020 at 16:22 history answered John Darby CC BY-SA 4.0