9.13

[Dhwani Krishnan 1G]

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?

How can you use delta S = nRln(v2/v1) if the reaction is irreversible? Isn't this equation only applicable to reversible expansion?

[Mukil_Pari_2I]

Because you would have to calculate the entropy due to the temperature change, the reversible equation can be used for the first part. You would have to consider each portion separately. So, first you would assume the the temperature as constant (isothermal), for the compression. Then, you would assume the volume as constant when calculating the entropy for the temperature.