Question

An ideal gas is confined to a cylinder by a piston. The piston is slowly pushed in so that the gas temperature remains at 20 degrees Celsius. During the compression, 730 J of work is done on the gas. Part A) Calculate the entropy change of the gas. Part B) Describe clearly why isn’t the result a violation of the entropy statement of the second law, ΔS  0 ?

Answer 1

Answer:

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Explanation:

(a) It is given that the gas is ideal. Formula for change in entropy of the gas according to the first law of thermodynamics is as follows.

       $\Delta U = dQ - dW$

For isothermal process, $\Delta U = 0$ at constant temperature.

So,   $\Delta U = dQ - dW$

      $0 = dQ - dW$

or,       dQ = dW = 730 J

Now, according to the second law of thermodynamics the entropy change is as follows.

           $\Delta S = \frac{dQ}{dT}$

                        = $\frac{730 J}{293.15 K}$

                        = 2.490 J/K

Therefore, the entropy change of the gas is 2.490 J/K.

(b)  In the given process, at constant temperature the gas will be compressed slowly because then kinetic energy of the gas molecules will also be constant. The volume decreases so that the movement of molecules increases as a result, entropy of molecules will also increase.

This means that $\Delta S = 2.490 J/K > 0$