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

Answer: Option B

Solution(By Examveda Team)

Since here heat is taken from the system \[Q = - ve\]
From second law of thermodynamics \[TdS \ge \delta Q\]
Since given reversible process \[TdS = \delta Q\]
$$ \Rightarrow dS = - ve$$
From first law of thermodynamics
$$\eqalign{ & \delta Q = dU \cr & \Rightarrow dU = - ve \cr} $$