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Onemole ofan idealmonoatomic gas (C_(V.M)= 1.5 R)is subjected to the followingsequence of steps : (a)The gasis heated reversiblyat constantpressure of 1 atm from 298 K to 373 K. (b) Next, thegas isheatedreversiblyand isothermallyto doubleits volume. (c)Finally , thegasis cooled reversiblyandadiabatically to 308 K . Calcuated q, w, DeltaH for the overall porcess. |
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Answer» `= - nRDeltaT` `=- 1xx 8.14 XX (373 - 298)` ` = -623.55 J//mol` `DeltaH = q = nC_(P)DeltaT` `= - 2.5 xx 8.314 xx 75"" (C_(P)=2.5 R)` `= 1558.8 J//mol` `DeltaU = 1.5 xx 8.314 xx 75""(C_(V) = 1.5 R)` ` =935.5 J//mol` (b)` W=-2.303RT log ((V_(1))/(V_(2)))` ` = - 2.303 xx 8.314 xx 373 log 2` ` = - 2149. 7 J//mol` `DeltaU = 0 ,DeltaH = 0 ""q =2149.7 J//mol^(-1)` (C)`W = Deltau = nC_(V)DeltaT""("For adiabaticprocess "q = 0, DeltaU ne W)` `= 1 xx 1.5 xx 8.314 (308-373)` `= - 810.62 J//mol` `q= 0` `DeltaH = nC_(P)DeltaT=- 1 xx 2.5 xx 8.314 (308-373)` `= -1351 .03 J//mol` For overall process `q_(net) = 3708.59 J//mol` `W_("net") = - 3583. 88 J//mol""DeltaU_("net")= 124.71 J//mol` `DeltaH_("net")= 207.85 J//mol` |
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