Question

What is the amount of entropy produced or destroyed in the process when one kilogram of propane, initially at 8 and 50 °c, undergoes a rapid expansion in a piston-cylinder assembly to 3 , 20 °c? The heat transfer between the propane and its surroundings occurs at an average temperature of 35 °c. Given that the measured work done by the propane is 42.4 , and kinetic and potential energy effects can be ignored, also, determine whether it is possible for the work measurement to be correct.

Answer 1

The amount of entropy produced or destroyed in the process is 0.1377 J/K. It is not possible for the work measurement to be correct as the change in internal energy is not zero.

Step-by-step explanation:

To calculate the amount of entropy produced or destroyed in the process, we can use the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature. In this case, the heat transfer is given as 42.4 J and the average temperature is given as 35 °C (308 K). So, the change in entropy can be calculated as ΔS = 42.4 J / 308 K = 0.1377 J/K.

Now, to determine the possibility of the work measurement being correct, we need to consider the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat transfer (Q) minus the work done (W). If the work measurement is correct, then the change in internal energy should be zero since the initial and final temperatures are the same. Therefore, we can calculate the change in internal energy as ΔU = Q - W = 42.4 J - 0 J = 42.4 J. Since the change in internal energy is not zero, it suggests that the work measurement is not correct.