Chemical Engineering Thermodynamics

Q1:

A closed system is cooled reversibly from 100°C to 50°C. If no work is done on the system

A its internal energy (U) decreases and its entropy (S) increases.

B U and S both decreases

C U decreases but S is constant.

D U is constant but S decreases.

ANS:B - U and S both decreases

In a closed system undergoing a reversible cooling process, the internal energy (𝑈U) and entropy (𝑆S) change. However, the change in each depends on the specifics of the process. In this scenario, the system is being cooled, which means heat is being removed from the system. Since no work is being done on the system, the cooling process is likely to be adiabatic (no heat transfer). Consequently, the change in internal energy is primarily due to the heat transfer, and the change in entropy is due to the cooling process. Given that the system is cooled reversibly from 100°C to 50°C:

  • The internal energy (𝑈U) decreases because the system loses heat.
  • The entropy (𝑆S) increases because the cooling process is reversible and heat is being removed from the system.
So, the correct statement is: "its internal energy (U) decreases and its entropy (S) increases."



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