Question

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.

Answer 1

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, $\Delta U=q+w$

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

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

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

$\Delta U=(\Delta U)_{step2}-(\Delta U)_{step1}=-56J+w-37J=0$

or, w = 93 J