# Thermodynamics Essay

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Lecture 6: the first law and more non-flow processes The isobaric (constant pressure) process &bull; consider a constant pressure process p dv 1 2 v w = &int; p dv = p (v 2 &minus; v1 ) 2 1 &bull; there is a small step dv in the process, thus dw = p dv for that small step, the NFEE dq &minus; dw = du gives dq = du + p dv dw &bull; = du + p dv + v dp = du + d ( pv ) &bull; note: 1. adding v dp adds nothing because in a constant pressure process dp = 0 2. The last step cannot be taken if the change is finite because for finite changes &#8710;( pv) &ne; p&#8710;v + v&#8710;p . A little care with the mathematics is needed here. let us now define a new thermodynamic property, the enthalpy h, h = u + p v, &bull; H = U + pV &bull; &bull; 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 state 1 to state 2, q = &#8710;h , Q = &#8710; H &bull; &bull; The specific heat at constant pressure &bull; recall from the last lecture that the &ldquo;the specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1&deg;C&rdquo;. &bull; this then led us to define the specific heat at constant volume, cv &#9115; &part;u &#9118; cv = &#9116; &#9119; &#9117; &part;T &#9120; v we can therefore also define the specific heat at constant pressure, &#9115; &part;h &#9118; cp = &#9116; &#9119; &#9117; &part;T &#9120; p since for an isobaric, non-flow process dq = dh = c p dT &bull; &bull; Enthalpy variations in an ideal gas with constant specific heats &bull; if the system is a perfect gas with constant specific heats h = u + p v = u + R T = (c v + R )T &bull; if cp is also constant, cp =