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The given figure shows a chnge of state A to state C by rtwo paths ABC and AC for an ideal gas. Calculate. a. The path along which work done is least. b. The internal enegry at C if the internal enegry of gas at A is 10J and amount of heat supplied to change its state to C through the path AC is 200J. c. The amount of heat supplied to the gas to go from A to B, if inetrnal enegry of gas at state B is 10J. |
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Answer» Solution :We know, Path `CA:` Isothermal compression Path `AB`: Isobaric expansion Path `BC`: Isochroic change Let `V_(i)` and `V_(f)` be initial volume and final volume at repective points. For temperature `T_(1) ("for" C): PV - nRT_(1)` `2 xx 10 = 1 xx 0.0821 xx T_(1)` `T_(1) = 243.60K` For temperature `T_(2)` (For `C` and `B): (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` `(2xx10)/(T_(1)) = (20 xx 10)/(T_(2))` `(T_(2))/(T_(1)) = 10` `T_(2) = 243.60 xx 10 = 2436.0K` Path `CA: w = + 2.303 nRT_(1) "log"(V_(i))/(V_(f))` `= 2.303 xx 1xx2xx 243.6"log" (10)/(1)` `=+1122.02 CAL` `DeltaE = -0` for isotermal compression, also `Q = W` Path `AB, w =-P (V_(f) - V_(1))` `=- 20 xx (10 - 1) =- 180 L-atm` `= (-180 xx 2)/(0.0821) =- 4384.9 cal` Pth `BC: w =- P(V_(f) - V_(1)) = 0` (`:' V_(f) - V_(1) = 0)` since volume is constant For monoatomic gas, heat chnage at constant volume `=qv = DeltaU` Thus, for path `BC, q_(v) = C_(v) xx n xx DeltaT = DeltaU` `:.q_(v) = (3)/(2)R xx1 (2436 - 243.6)` `= (3)/(2) xx2xx1xx (2192.4) = 65772.2 cal` Since process involves cooling `:.q_(v) = DeltaU =- 6577.2 cal` Also in path `AB`, the initial enegry in state `A` and state `C` is same. Thus,, during path `AB`, an INCREASE in internal enegry equivalent of change in internal enegry during path. `BC` should take place. Thus, `DeltaU` for path `AB = + 6577.2 cal` Now `q` for path `AB = DeltaU - w_(AB) = 6577.2 + 6577.2 cal` `= 10962.1 cal` Cycle: `DeltaU = 0`, `q =- w =- [w_(path CA) +w_(pathAB) +w_(pathBC)]` `=- [+1122.02 +(-4384.9+0]` `:. q =- w = + 3262.88 cal` |
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