InterviewSolution
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InterviewSolution
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A cylindercontainer of volume 4.48 liters is containing equal no .Ofmoles of a monoatomic gasin two section A and B separatedby an adiabatic frictionless piston as show in figure . The initial temperatureandpressue of gases was 273 K and 1 atm.Now gas section'A' is slowlyheated till the volumeof sectionB becomes 1//2 sqrt(2)of initinalvolume. Findtotal changein DeltaH forsectionA and B . ( in cal/mole ) "" [C_(v) "ofmonoatomic gas" =3//2 R, gamma = 5//3] [Use R = 2 cal //mol &sqrt(2)=1.4] |
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Answer» Solution :Initial volume ofsec. A = sec. B= 2.24 LITRES Find volume ofsexc.`B = [(2.24)/(2sqrt(2))]` litres Thegasin sec . 'B' compressed reversiblyand adiabatically `rArr T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) rArr T_(2) = T_(1)((V_(1))/(V_(2)))^(r-1) rArr T_(2)= T_(1) (2sqrt(2)) ^(2//3) = 2T_(1)` { Formonoatom gas`gamma = (5)/(3)`} The final pressure in sec. 'B' `P_(f)= (P_(1)V_(1))/(T_(1)),(T_(2))/(V_(2)) = P_(1) XX ((T_(2))/(T_(1)))((V_(1))/(V_(2))) P_(1) xx 2xx 2sqrt(2)"" = 4 sqrt(2) "atm` `rArr` Pressure in sec. 'A' = `4sqrt(2)"atm"``rArr` Final temperature in sec. `AT_(2) = ((P_(2))/(P_(1)))((V_(2))/(V_(1)))T_(1)` `T_(2) = ((P_(2))/(P_(1))){V_(1) + V_(1)-(V)/(2sqrt(2))} T_(1)` `4sqrt(2) (2 -(1)/(2sqrt(2)))T_(1) ""=[((4sqrt(2)-1)(4sqrt(2)))/(2sqrt2)]T_(1)` `(4 sqrt(2)-1)2 T_(1)` `[8sqrt(2)T_(1)- 2T_(1)- T_(1)]={4sqrt(2)T_(1) - 3T_(1)}` `DeltaH_(A) = (0.1) (5)/(2) R{8sqrt(2)-3} rArr DeltaH_(B) = (0.1) (5)/(2)R{273}` `""2(4sqrt(2)-1)` `DeltaH_(T) = (0.1)(5)/(2)R(273){2} (4sqrt(2)-1)""=1255.8 or 5 xx 1255.8= 6279` |
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