Question

onsider the following cyclic process carried out in two steps on a gas: Step 1: 42 J of heat is added to the gas, and 14 J of expansion work is performed. Step 2: 53 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.

Answer 1

W = -25 J

Step-by-step explanation:

For a cyclic process, the variation of internal energy (ΔU) must be 0. It means that ΔU from step 1 must be equal to ΔU from step 2. For the first law of the thermodynamics, the energy must be conserved, so:

ΔU = Q - W, from each step. Q is the heat and W the work

When the heat is gained by the system, the process is endothermic, so Q >0, when the system loses heat, the process is exothermic, and Q<0. When the system is expanding W>0, and when it is compressing, W<0.

From step 1:

ΔU = 42 - 14

ΔU = 28 J

From step 2:

28 = -53 - W

W = -53 + 28

W = -25 J