plik


’ž18. (a) In an adiabatic process Q = 0. This can be done by placing the gas in a thermally insulated container whose volume can be adjusted (say, by means of a movable piston). If the volume is slowly increased from Vi to Vx, then the process is reversible. To realize the reversible, constant- volume process from x to f, we would place the gas in a rigid container, which has a fixed volume Vf and is in thermal contact with a heat reservoir. If we gradually increase the temperature of the reservoir from Tx to Tf , the gas will undergo the desired reversible process from x to f. ³ (b) For the two states i and x we have piVi/Ti = pxVx/Tx and piVi³ = pxVx . We eliminate pi and px from these equations to obtain ³-1 Tx Vi = . Ti Vx For monatomic ideal gases ³ = 5/3 (see §20-8 and §20-11), so ³ - 1 = 2/3. Also Vx = Vf . Substituting these into the equation above, we obtain Tx = Ti(Vi/Vf )2/3. (c) For an ideal gas undergoing an isothermal process, Eq. 20-45 implies "Eint = 0. And Eq. 20-14 gives W = nRT ln(Vf /Vi ) for such a process. Therefore, the first law of thermodynamics leads to Vf Qpath I ="Eint I + WI = nRTi ln . Vi And for path II, we have Qpath II = Qadiabat + Qconst vol =0 +nCV "T . 3 But CV = R (see Eq. 20-43), so we obtain 2 3 Qpath II = nR (Tf - Tx) . 2 We see that Qpath I = Qpath II . (d) Since the first part of path II is reversibly adiabatic, then the entropy changes only during the second, constant-volume, part of the path: Tf nCV dT Tf 3 Tf "S = = nCV ln = nR ln . T Tx 2 Tx Tx Entropy is a function of  where you are on the pV diagram, not  how you got there. Since the beginning and ending point of path I are the same as those of path II, then "S is the same for both. 1 (e) Using the result in part (b) with Vi/Vf = and Ti = 500 K, we find 2 2/3 1 Tx = (500 K) = 315 K . 2 For path I, Eq. 21-2 gives QI =("S) T where T = Ti = Tf and "S is the expression calculated in the part (d). Thus, 3 Tf QI = nRTf ln 2 Tx which can be alternatively derived from Eq. 20-14 and the first law of thermodynamics. With n =1 mol, Tf = Ti = 500 K, we find 3 500 QI = (1)(8.31)(500) ln = 2880 J . 2 315 For path II, Q = Qconstant volume = nCV "T and we obtain 3 QII =(1) (8.31) (500 - 315) = 2306 J . 2

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