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Consider the following cyclic process carried out in two steps on a gas. Step 1: 50. J of heat is added to the gas, and 13 J of expansion work is performed. Step 2: 56 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.

Respuesta :

Answer:

Work for the gas compression is 93 J.

Explanation:

The change in internal internal energy during the whole cycle will be 0 as internal energy is a state function.

According to first law of thermodynamics, [tex]\Delta U=q+w[/tex]

where, [tex]\Delta U[/tex] represents change in internal energy, q is the heat exchanged and w is work done.

Here, [tex](\Delta U)_{step1}=50J-13J=37J[/tex]  ( q is positive for addition of heat and w is negative for work don by system)

[tex](\Delta U)_{step2}=-56J+w[/tex]   (q is negative for removal of heat)

[tex]\Delta U=(\Delta U)_{step2}-(\Delta U)_{step1}=-56J+w-37J=0[/tex]

or, w = 93 J