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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.

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Answer:

Work for the gas compression is 93 J.

Step-by-step 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

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