Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in tv diagram

Three moles of an ideal gas are taken through a cyclic process ABCA as shown in T V indicator diagram. The gas loses 2510 J of heat in the complete cycle. If TA = 100 K and TB = 200 K, the work done by the gas during the process BC is line AB if extended passes through origin

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Text Solution

Solution : For the process A - B, it is given that <br> PT = constant <br> Differentiating above equation partially, we have <br> PdT + TdP = 0 <br> Equation of state for two moles of a gas <br> `PV = 2 RT` or `P = (2RT)/(V)` <br> After differentiating Eq. (ii) partially, we get <br> PdV + VdP = 2R dT <br> From Eq. (ii) partially, we get <br> PdV + VdP = 2R dT <br> From Eqs. (i) and (ii), we have <br> `((2RT)/(V)) dT + T dP = 0` <br> or 2RT dT + VTdP = 0 <br> VdP = - 2 RdT <br> Now form Eqs. (iii) and (iv), we have <br> `- 2RdT + VdP = 2RdT` <br> or `PdV = 4 RdT` <br> a. The work done in the process AB <br> `W_(AB) int PdV = int_(600)^(300) 4 RdT` <br> `= 4 R |T|_(600)^(300) = 4 R (300 - 600)` <br> `= - 1200 R` <br> b. As process `B rarr C` is isobaric, so <br> `Q_(BC) = nC_(P) Delta T = 2 xx (5R)/(2) xx (600 - 300)` <br> = 1500 R <br> ii. Process `C rarr A` is isothermal, so `Delta U = 0` <br> `Q_(CA) = Delta U + W_(CA) = W-(CA)` <br> `W_(CA) = nRT 1n (P_(C ) // P_(A))` <br> `2R xx 600 1n (2P_(1) // P_(1)) = 1200 R1n 2` <br> `Q_(CA) = 1200 R 1n 2` <br> Again for process `A rarr B` <br> `Q_(AB) = Delta U + W_(AB)` <br> `= nC_(V) Delta T + W_(AB)` <br> `= 2 xx ((3R)/(2)) xx (300 - 600) - 1200 R` <br> `= - 900 R - 1200 R = - 2100 R`