What is the thermal efficiency of an engine that operates by taking n moles of diatomic ideal gas through the cycle shown in F? ig. P20.38??

Solution 46P Step 1: The process, 1 2 is an isochoric process. Therefore, the work done by the system will be zero. According to ideal gas equation, at point 1, P1 1nRT 1 Where, “n” is the number of moles of the gas P - Pressure at point 1 1 V 1Volume at point 1 R - Universal gas constant T 1Temperature at point 1 T = P V /nR 1 1 1 From the graph, we can see that, P = P and V = V 1 0 1 0 Therefore, T = P V1/nR 0 0 Similarly, T = 2P2 /nR 0 0 The heat change in the system, Q = nC T V Where, C - SpeVfic heat capacity at constant volume T - Change in temperature of the system In process 1 2, T = T - T = 2P V /n2- P 1 /nR = P0V0nR 0 0 0 0 For a diatomic gas, C = 5/2 R V Therefore, Q = n×(5/2)R×P V /nR = 5P V 02 0 0 0 Step 2: The process, 2 3 is an isobaric process. Therefore, the work done by the system will not be zero. Work done, W = PdV Where, dV - Change in volume dV = (V - V ) = ( 2V - V ) = V 3 2 0 0 0 Therefore, W = 2P V 0 0 Step 3: The process, 3 4 is an isochoric process. Therefore, the work done by the system will be zero. According to ideal gas equation, at point 3, P3 3nRT 3 T3 P V3nR3 From the graph, we can see that, P = 2P and V = 2V3 0 3 0 Therefore, T = 4P 3 /nR 0 0 Similarly, T = 2P V /nR 4 0 0 The heat change in the system, Q = nC T V Where, C - Specific heat capacity at constant volume V T - Change in temperature of the system In process 3 4, T = T - T = 2P V /nR - 4P V /nR = - 2P V /nR 3 4 0 0 0 0 0 0 For a diatomic gas, C = 5/2 R V Therefore, Q = n×(5/2)R× -2P V /nR = - 5P V0 0 0 0