S31. Consider the following cyclic process carried out in two steps on a gas: Step 1: 45 J of heat is added to the gas, and 10. J of expansion work is performed. Step 2: 60 J of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in step 2. Because the process is cyclic, the change in energy for the two steps combined must be equal to zero. This is because energy is a state function and since the final and initial state (the states at the beginning and end of the cyclic process) are the same the energy must be unchanged after both steps are completed. Thus Δ E = W 1 + Q 1 + W 2 + Q 2 =0 W 2 = − ( W 1 + Q 1 + Q 2 ) The only thing left is to decide what the signs are for the given values. Because heat is added to the system in step 1, Q 1 = + 45 J , whereas because the work is expansive, the internal energy of the system decreases with the work performed and W 1 = − 10 J . For step 2, heat is removed from the system decreasing the internal energy of the system
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