Question

Calculate the change in entropy for the process : \( H _{2} O _{\left(\text {liq, } 20^{\circ} C , 1 atm \right)} \) to \( H _{2} O _{\left.\text {(gas, } 250^{\circ} C , 1 atm \right)} \) You are given, heat capacity of water \( ( \) liq \( )=75.6 JK ^{-1} \cdot mol ^{-1} \) Heat capacity of water \( ( \) vap \( )=36.2 JK ^{-1} mol ^{-1} \) Heat of vaporization of \( H _{2} O \) at \( 100^{\circ} C \) and \( 1 atm 40.85 JK mol ^{-1} \)

Answer 1

Given transformation break into three parts —

(1) H₂O (l) (20° C) → H₂O (l) (100° C) \((c_{p_{liq}}\) = 75.6 JK^-1 mol^-1)

(2) H₂O (l) (100° C) → H₂O (g) (100° C) \((\triangle H_{tras}\) = 40.85 JK mol^-1)

(3) H₂O (g) (100° C) → H₂O (g) (250° C) \((c_{p_{vap}}\) = 36.2 JK^-1 mol^-1)

Pressure remain constant = 1 atm (in the whole process)

We have use formula —

ΔS \(=\displaystyle\int\limits^{T_f}_{T_i}\,\frac{c_p\,dT}T\) [q = c_p dT]

ΔS = c_p ln \(\frac{T_f}{T_i}\)  .....(IV)

and ΔS \(=\cfrac{\Delta_{Tras}H}{T_{Trans\,-}}\) ....(V)

For equation (1) —

H₂O (l) (293 K) → H₂O (l) (373 K)

ΔS₁ = 75.6 x ln 373/293

ΔS₁ = 18.25 JK^-1 mol^-1

For equation (2) —

H₂O (l) (373 K) \(\overset{\text{Phase transition}}{\longrightarrow}\) H₂O (g) (373 K)

Using equation (V) —

ΔS₂ \(=\frac{40.85\,J\,mol^{-1}}{373\,K}\)

ΔS₂ = 0.109 JK^-1 mol^-1

For equation (3) —

H₂O (g) (373 K) → H₂O (g) (523 K)

Using equation (IV) —

ΔS₃ = 36.2 x ln 523/373

ΔS₃ = 12.24 JK^-1 mol^-1

Therefore, total change in entropy for whole process

= ΔS₁ + ΔS₂ + ΔS₃

= (18.25 + 0.11 + 12.24) JK^-1 mol^-1

ΔS = 30.60 JK^-1 mol^-1