Thermodynamics Essay

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Lecture 6: the first law and more non-flow processes The isobaric (constant pressure) process • consider a constant pressure process p dv 1 2 v w = ∫ p dv = p (v 2 − v1 ) 2 1 • there is a small step dv in the process, thus dw = p dv for that small step, the NFEE dq − dw = du gives dq = du + p dv dw • = du + p dv + v dp = du + d ( pv ) • 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 ∆( pv) ≠ p∆v + v∆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, • 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 state 1 to state 2, q = ∆h , Q = ∆ H • • The specific heat at constant pressure • recall from the last lecture that the “the specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1°C”. • this then led us to define the specific heat at constant volume, cv ⎛ ∂u ⎞ cv = ⎜ ⎟ ⎝ ∂T ⎠ v we can therefore also define 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 gas with constant specific heats • if the system is a perfect gas with constant specific heats h = u + p v = u + R T = (c v + R )T • if cp is also constant, cp =

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