InterviewSolution
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InterviewSolution
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Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB , pressure and temperature of the gas very such that `PT=Constant`. It `T_1=300K`, calculate (a) the work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R. |
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Answer» For the process A - B, it is given that PT = constant Differentiating above equation partially, we have PdT + TdP = 0 Equation of state for two moles of a gas `PV = 2 RT` or `P = (2RT)/(V)` After differentiating Eq. (ii) partially, we get PdV + VdP = 2R dT From Eq. (ii) partially, we get PdV + VdP = 2R dT From Eqs. (i) and (ii), we have `((2RT)/(V)) dT + T dP = 0` or 2RT dT + VTdP = 0 VdP = - 2 RdT Now form Eqs. (iii) and (iv), we have `- 2RdT + VdP = 2RdT` or `PdV = 4 RdT` a. The work done in the process AB `W_(AB) int PdV = int_(600)^(300) 4 RdT` `= 4 R |T|_(600)^(300) = 4 R (300 - 600)` `= - 1200 R` b. As process `B rarr C` is isobaric, so `Q_(BC) = nC_(P) Delta T = 2 xx (5R)/(2) xx (600 - 300)` = 1500 R ii. Process `C rarr A` is isothermal, so `Delta U = 0` `Q_(CA) = Delta U + W_(CA) = W-(CA)` `W_(CA) = nRT 1n (P_(C ) // P_(A))` `2R xx 600 1n (2P_(1) // P_(1)) = 1200 R1n 2` `Q_(CA) = 1200 R 1n 2` Again for process `A rarr B` `Q_(AB) = Delta U + W_(AB)` `= nC_(V) Delta T + W_(AB)` `= 2 xx ((3R)/(2)) xx (300 - 600) - 1200 R` `= - 900 R - 1200 R = - 2100 R` |
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