One mole monoatomic ideal gas was takenthroughprocess ABCD as show in. Calculate(i)W_(AB), W_(BC), W_(CD),W_(DA)(ii)q_(AB), q_(BC), q_(CD),q_(DA) (iii)DeltaH_(AB), DeltaH_(BC), DeltaH_(CD),DeltaH_(DA) [Use :ln(3//2) = 0.40 : In (4//3) = 0. 29]

One mole monoatomic ideal gas was takenthroughprocess ABCD as show in.

1.	One mole monoatomic ideal gas was takenthroughprocess ABCD as show in. Calculate(i)W_(AB), W_(BC), W_(CD),W_(DA)(ii)q_(AB), q_(BC), q_(CD),q_(DA) (iii)DeltaH_(AB), DeltaH_(BC), DeltaH_(CD),DeltaH_(DA) [Use :ln(3//2) = 0.40 : In (4//3) = 0. 29]
Answer» Solution :(i) `W_(AB)` : Along 'AB' pressure is variable from graph :`(V-10)/(T- 300) = (15- 10)/( 600-300) rArr60 V- 600 = T - 300 rArr 60 V - 300 =T` sub value ofT in ideal gas equation `PV= R (60 V - 300)""rArr""P= 60R-(300R)/(V)` `W_(AB) = - underset(10)overset(15)int PdV = -underset(10)overset(15)int[60R-(300R)/(V)] DV= - 60R(V_(2)- V_(1)) + 300R ln((V_(2))/(V_(1)))` `= - 300R + 300 R(0.4)""("taken ln"(3//2)= 0.10 "data")` `W_(AB) = - 180 R = - 1496.52 J` `W_(BC) = - RT ln ((V_(3))/(V_(2))) = -RT(4//3) "" ("taken ln "(4//3)= 0.29"data")` `= - 8.314 xx 600 xx 0.29 = - 1446.63 "Joule"` `W_(CD)= 0` `W_(DA) = - NRT ln ((V_(1))/(V_(2))) = - 1 xx 8.314 xx 300 ln ((10)/(20)) = + 1728.84"Joule"` (ii) `q_(AB)= DeltaE_(AB)- W =(3)/(2) R xx 300 + 1496.52=450R +180R=630R=5237.82"Joule"` `q_(BC) = - W_(BC) = +1446.63"Joule"` `q_(CD)= DeltaU = nC_(V)DeltaT = 1 xx (3)/(2)R xx(300-600)= -3741.3"Joule"` `q_(DA) = - W_(DA) = 1728.84"Joule"` (iii)`DeltaH_(AB) = C_(P) (T_(2) - T_(1)) =(5)/(3) R xx 300 = 750R = 6235.5"Joule"` `DeltaH_(BC) =0 ` `DeltaH_(CD)= DeltaU_(CD) + nR DeltaT = nC_(p) DeltaT = 1 xx (5)/(2) R xx (-300)= - 6235.5"Joule"` `DeltaH_(DA)= DeltaU_(DA) + nRDelta T = 0"Joule"`