1 kg of air (ideal gas) undergoes a thermodynamic cycle in a piston-cylinder arr

Question

1 kg of air (ideal gas) undergoes a thermodynamic cycle in a piston-cylinder arrangement. Process 1-2 is an adiabatic expansion (PV^1.4 = constant) in which the volumes doubles. Process 2-3 is a constant pressure process in which the volume is reduced to its initial value. Process 3-1 is a constant volume prcess to the initial state. If P1 = 500 KPa and T1= 150 C and R= .287 KJ/Kg * K what is the work done during each process, the total work for the cycle and the change in internal energy during process 1-2

Explanation / Answer

may b hipfull.. (a) The total work done is the sum of the work done in each step. Work done on the gas is given by the integral W = - ? p dV from initial to final volume Since the the volume does not change in first step, no work is done on the gas: W1 = 0 In second step gas expands at constant pressure, so the work integral simplifies to: W2 = - p · ? dV from initial to final volume = - p · (V_final - V_initial) = - 1.4 · 101325Pa · (0.0093m³ - 0.0068m³) = -355.6Pam³ = -355.6J So the total work in the two-step process is: W = W1 + W2 = 0J + -355.6J = 355.6J (b) Assuming ideal gas behavior, internal energy depends solely on temperature (and not on pressure and volume): U = n·Cv·T So the change of internal energy is: ?U = n·Cv·?T Because we return to initial temperature after second step, the change of internal energy is zero ?U = 0 (c) Change of internal energy equals work done On the gas plus heat transferred to the gas: ?U = W + Q => Q = ?U - W = 0J - (-355.6J) = 355.6J

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