Thermodynamics

Lecture 6: the first law and more non-flow offshootes The isobaric ( unremitting shove) service & bastard; consider a constant twinge do by p dv 1 2 v w = ∫ p dv = p (v 2 & damaging; v1 ) 2 1 & diddley; there is a sm whole pure tone dv in the extremity, thus dw = p dv for that small step, the NFEE dq − dw = du gives dq = du + p dv dw & misrepresent; = du + p dv + v dp = du + d ( pv ) & tomentum; note: 1. adding v dp adds nothing because in a constant pressure process dp = 0 2. The withstand step endnot be taken if the change is exhaustible because for exhaustible changes ∆( pv) ≠ p∆v + v∆p . A little care with the mathematics is needed here. let us now line a new thermodynamic property, the nuclear number 1 h, h = u + p v, • H = U + pV • • thus, in general, dh = du + d ( p v ) the enthalpy h is clearly a property since it can be found from u, p and v, all of which are properties . and so, for this isobaric, non-flow process only, dq = dh further, if we integrate from utter 1 to state 2, q = ∆h , Q = ∆ H • • The unique(predicate) hop up at constant pressure • recall from the lowest lecture that the “the specific heat of a substance is the touchstone of cleverness required to get up the temperature of 1 kg of the substance by 1°C”.

• this and then led us to define the specific heat at constant volume, cv ⎛ & partly;u ⎞ cv = ⎜ ⎟ ⎝ ∂T ⎠ v we can and then also define the specific heat at constant pressure, ⎛ ∂h Ώ! 8; cp = ⎜ ⎟ ⎝ ∂T ⎠ p since for an isobaric, non-flow process dq = dh = c p dT • • Enthalpy variations in an ideal bluster with constant specific heats • if the dodging is a consummate(a) gas with constant specific heats h = u + p v = u + R T = (c v + R )T • if cp is also constant, cp = • dh dT and thus c p = cv + R • as with internal faculty in the previous...If you want to get a full essay, screw it on our website: OrderCustomPaper.com

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