Discussion Forum



1)

For the process, $H_{2}O (l)\rightarrow H_{2}O(g)$ at   T= 100° C and 1 atmosphere  pressure, the correct choice is 


A) $\triangle S_{system}$ > 0 and $\triangle S_{surrounding}$ > 0

B) $\triangle S_{system}$ > 0 and $\triangle S_{surrounding}$ < 0

C) $\triangle S_{system}$ < 0 and $\triangle S_{surrounding}$ > 0

D) $\triangle S_{system}$ < 0 and $\triangle S_{surrounding}$ < 0

Answer:

Option B

Explanation:

Plan  This problem is based on assumption that total entropy change of universe is zero

 At 100°  C and 1-atmosphere pressure

   $H_{0}O(l)\rightleftharpoons H_{2}O(g)$ is at equilibrium

 For equilibrium ,   $\triangle S_{total}=0$  and 

    $\triangle S_{system}$+  $\triangle S_{surrounding}$  =0

 As we know during conversion of liquid to gas entropy of system increases, in a similar manner entropy of surrounding decreases.

   $\therefore$      $ \triangle S_{system} $  > 0  and  $ \triangle S_{surrounding} <$ 0

 Hence, (b) is the correct choice