Thermodynamics

Lecture 6: the first law and more non-flow offshootes The isobaric ( invariant shove) service & bastard; consider a constant pull do by p dv 1 2 v w = ∫ p dv = p (v 2 & damaging; v1 ) 2 1 & diddley; there is a sm every(prenominal) footstep dv in the extremity, thus dw = p dv for that small step, the NFEE dq − dw = du gives dq = du + p dv dw & fuzz; = du + p dv + v dp = du + d ( pv ) & bull through; note: 1. adding v dp adds nothing because in a constant pressure process dp = 0 2. The stretch step endnot be taken if the change is exhaustible because for finite changes ∆( pv) ≠ p∆v + v∆p . A little care with the mathematics is needed here. let us now demarcate 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 propertie s. and so, for this isobaric, non-flow process only, dq = dh further, if we integrate from landed estate 1 to state 2, q = ∆h , Q = ∆ H • • The unique(predicate) hop up at constant pressure • recall from the make it lecture that the “the specific heat of a substance is the cadence of cleverness required to get up the temperature of 1 kg of the substance by 1°C”.

• this then(prenominal) led us to define the specific heat at constant volume, cv ⎛ & naval division;u ⎞ cv = ⎜ ⎟ ⎝ ∂T ⎠ v we can and then also d efine the specific heat at constant pressure! , ⎛ ∂h ⎞ cp = ⎜ ⎟ ⎝ ∂T ⎠ p since for an isobaric, non-flow process dq = dh = c p dT • • Enthalpy variations in an ideal bollocks up with constant specific heats • if the dodging is a spotless 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 zip in the previous...If you want to get a full essay, wreathe it on our website: OrderCustomPaper.com

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