For the process:
$H_{2} O (l) (1 bar, 373 K) ⇌ H_{2} O (g) (1 bar, 373 K)$
the correct set of thermodynamic parameters is:

Δ G > 0, Δ S = + v e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ G = 0, Δ S = + v e

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ G = + v e, Δ S = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ G = - v e, Δ S = + v e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $Δ G = 0, Δ S = + v e$
This is a phase transformation and hence, $Δ G = 0$ . Since, $H_{2} O (l)$ is changing to $H_{2} O (g)$ at the same temperature and pressure, $Δ S = + v e$ .

Suggest Corrections

Similar questions

H_{2} O_{(l)} (1 bar, 373 K) ⇌ H_{2} O_{(g)} (1 bar, 373 K)

For $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ .The correct set of thermodynamic parameter is

H_{2} O (l) (1 b a r, 373 K) ⇌ H_{2} O (g) (1 b a r, 373 K)

For the process $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ , the correct set of thermodynamic parameters is

H_{2} O (l) ⇌ H_{2} O (g) (1 b a r, 373 K),

Join BYJU'S Learning Program