Question

Four and one-half moles of an ideal diatomic gas is caused to pass through the cycle shown, where V_2 = 2.25 V_1, V_3 = 5.25V_1, p_1 = 4.50 atm and T_1 = 300 K. [gamma = 1.400] (a) Compute: p_3, p_4, and T_2, T_3 and T-4, (b) Compute: W, Q, delta U, and delta S for each process adiabatic and for the whole cycle and (c) What is the efficiency of an engine using this cycle?

Answer 1

a) P_1 V_1^r = P_4V_4^r = P_4V_3^r P_4 = P_1(V_1/squareroot 3) ^r = 4.5(1/5.25) ^1.4 = .449 tm P_@V_2 = P_3 V_3 P_3 = P_2(V_2/squareroot 3) = 4.5 times (2.25/5.25) = 1.939tm T_1 V_1^r - 1 = T_4 V_4 ^r - 1 T_4 = T_1(V_1/V_4) r - 1 = 300(1/5.25) ^.4 = 154.4K P_4/T_4 = P_3/T_3 T_3 = T_4 times P_3/P_4 = 154.5 times 1.93/.44 = 677.89K P_$ = .449tm: P_3 = 1.93 atm T_2 = T_3 = 677.89K & T_4 = 154.5K \r\n b) - > W_34 = 0 (isocharic) W_41 = nR(T_1 - T_4) /r - 1 = 4.5 times 8.314 times 145.4/.4 = 13599.63J W_12 = nc_p deltaT = 4.5 times 7/2 times 8.314 times (677.89 - 300) = 49482.995j W_23 = nRT ln(squareroot 3/squareroot 2) = 4.5 times 8.314 times 677.89 ln(5.25/2.25) = 21489.08y Q_23 = W_23 = 21489.08J Q_34 = nC_r Delta T = 4.5 times 5/2 R times(T_4 - T_3) = 4.5 times 2.5R times(154 - 677.83) = -48953.97J Delta