A heat engine takes 0.350 of a diatomic ideal gas around the cycle shown in the
ID: 2202732 • Letter: A
Question
A heat engine takes 0.350 of a diatomic ideal gas around the cycle shown in the pV-diagram of the figure (Figure 1) . Process is at constant volume, process is adiabatic, and process is at a constant pressure of 1.00 . The value of for this gas is 1.40. a. Find the pressure points at 1,2,3. b. Find the volume at points 1,2,3 c. Calculate, Q, W, delta U for the 1 to 2 process d. Calculate, Q, W, delta U for the 2 to 3 process e. Calculate, Q, W, delta U for the 3 to 1 processExplanation / Answer
No.of moles n = 0.4 mol Temperature at point 1 is T1= 300 K pressure at point 1 is P1= 1 atm = 101.3 * 10 ^ 3 Pa from ideal gas equation P1V1 = nRT1 Volume at point 1 is V1 = nRT1/P1 where R = gas constant = 8.314 J / mol K plug the values weget V1 = 9.848*10^-3 m ^ 3 At point 2 : ----------- Volume V2 = V1 = 9.848 * 10 ^ -3 m ^ 3 Temperature T2 = 600 K In Process 1--> 2 volume is constant So, P2/P1=T2/T1 from this Pressure at point 2 is P2 = P1T2 / T1 = 1 atm * ( 600 K / 300 K ) = 2 atm = 2 * 101.3 * 10 ^ 3 Pa = 202.6 * 10 ^ 3 Pa at point 3 : ----------- temperature T3 = 492 K Pressure P3 = P1 = 1 atm = 101.3 * 10 ^ 3 Pa In process 3 -->1 ,pressure is constant So, V3/V1 = T3 / T1 from this volume at point 3 isV 3 = T3 V1 / T1 = 16.15 * 10 ^ -3 m ^ 3 (a). Q for the process 1--> 2 is Q = n*Cv * dT = 0.4 mol * ( 5/2) R * ( T2-T1 ) = 0.4 * 2.5 * 8.314 * ( 600K - 300K ) = 2494.2 J (b). Q for the process 2--> 3 is Q = 0 Since it is an adiabatic process (c). Q for the process 3 --> 1 is q = n * Cp* ( T 1-T3 ) = 0.4* ( 7/2) R * ( 300 K - 492K ) = 0.4* 3.5 * 8.314* ( - 192 ) = -2234.8032 J