For the process H2O(l)(1bar, 373 K) → H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is

Question

For the process H2O(l)(1bar, 373 K) → H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is (A) Δ G =0, Δ S = +ve (B) Δ G =0, Δ S = -ve (C) Δ G =+ve Δ S = 0 (D) Δ G =-ve Δ S = +ve

Tardigrade Admin · Accepted Answer

Since, liquid is passing into gaseous phase so entropy will increase and at 373 K during the phase transformation it remains at equilibrium. So, Δ G = 0

Q. For the process $H_{2} O (l) (1$ bar, $373 K) \to H_{2} O_{(g)} (1$ bar, $373 K)$ , the correct set of thermodynamic parameters is

$Δ G = 0, Δ S = + v e$

$Δ G = 0, Δ S = - v e$

$Δ G = + v e Δ S = 0$

$Δ G = - v e Δ S = + v e$

Since, liquid is passing into gaseous phase so entropy will increase and at $373 K$ during the phase transformation it remains at equilibrium. So, $Δ G = 0$

Q. For the process H2​O(l)(1bar, 373K)→H2​O(g)​(1 bar, 373K), the correct set of thermodynamic parameters is

ΔG=0,ΔS=+ve

ΔG=0,ΔS=−ve

ΔG=+veΔS=0

ΔG=−veΔS=+ve