InterviewSolution
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InterviewSolution
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An ideal gas is taken round a cyclci thermodynamic process ABCA as shown if Fig. If the internal energy of the gas at point A is assumed zero while at B it is 50 J. The heat absorbed by the gas in the process BC is 90 J. (a) What is the internal energy og the gas at point C ? (b) How much heat energy is absorbed by the gas in the process AB? (c ) Find the heat energy rejected or absorbed by the gas in the process CA. (d) What is the net work done by the gas in the complete cycle ABCA ? |
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Answer» Given that `U_(A) = 0, U_(B) = 50 J` and `Q_(BC) = 90 J` Also `P_(A) = P_(B) = 10 Nm^(-2), P_(C) = 300 Nm^(-2), V_(A) = 1m^(3)` and `V_(B) = V_(C ) = 3 m^(3)` a. In process BC as volume of gas remains constant, work done by gas in this process is zero, thus `W_(BC) = 0` Heat abosrbed by the gas is `Q_(BC_ = 90 J`. From the first law of thermodynamics. `(Delta U) B_(C ) = U_(C ) - U_(B) = Q_(BC) - W_(BC) = 90 J - 0 = 90 J` `U_(C ) = (Delta U)_(BC) + U_(B) = 90 J + 50 J = 140 J` b. In process AB, we have `(Delta U)_(AB) = U_(B) - U_(A)` `= 50 - 0 = 50 J` work done is given as `W_(AB)` = area under AB in P - V diagram = area of rectangle ABED `= AB xx AD = (3 m^(3) = 1 m^(3)) xx 10 Nm^(-2)` `= 20 J` Thus heat abosrbed by the system is `Q_(AB) = (Delta U)_(AB) + W_(AB) = 50 + 20 = 70 J` c. For process CA `(Delta U)_(CA) - U_(A) - U_(C ) = 0 - 140 = - 140 J` Work done is given as `W_(CA) = ` area ACED = area of triangle ACB + area of rectangle ABED `1//2 xx AB xx BC + AB xx AD` `= 1//2 xx (3 - 1) m^(3) xx (30 - 10) Nm^(-2) + 20` `= 20 + 20 = 40 J` In this process, the volume decreases, the work is done on the gas, Hence, the work done is negaitve, Thus, `W_(CA) = - 40 J` Thus heat rejected by the gas is `Q_(CA) = (Delta U)_(CA) + W_(CA) = - 140 - 40 = - 180 J` Net work done in the complete cyclic process ABCA is `W = "area of triangle" ABC = (1)/(2) xx 2 xx 20 = 20 J` As the cycle is anticlockwise, net work is done on the gas. |
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