Question

One mole of steam is condensed at ${100}^{o}C$ the water is cooled to $0^{o}C$ and frozen to ice. Which of the following statements are correct, given heat of vapourization and fusion are $540cal/gm$ and $80cal/gm$ (average heat capacity of liquid water $=1cal$ ${gm}^{-1}$ ${degree}^{-1}$?

• A. Entropy change during the condensation of steam is $-26.06cal{/}^{o}C$
• B. Entropy change during cooling of water from ${100}^{o}C$ to $0^{o}C$ is $-5.62cal{/}^{o}C$
• C. Entropy change during freezing of water at $0^{o}C$ is $-5.27cal{/}^{o}C$
• D. Total entropy change is $-36.95cal{/}^{o}C$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Entropy change during the condensation of steam is $-26.06cal{/}^{o}C$
B Entropy change during cooling of water from ${100}^{o}C$ to $0^{o}C$ is $-5.62cal{/}^{o}C$
C Entropy change during freezing of water at $0^{o}C$ is $-5.27cal{/}^{o}C$
D Total entropy change is $-36.95cal{/}^{o}C$

1 mol $H_{2}O$=18 g

Given transformations are:

1. $H_{2}O\left(g\right)\to H_{2}O\left(l\right)$; $\Delta H=-18\times 540cal$ and $T=373K$

Entropy change during condensation of steam is:

$\Delta S_1=\frac{\Delta H}{T}$

$\Delta S_1=\frac{-9720}{373}=-26.06cal/K$

2. $H_{2}O\left(l\right)at373K\to H_{2}O\left(l\right)at273K$; $C_p=18\times 1cal/K$

Entropy change during cooling of water is:

$\Delta S_2={\int }_{T_1}^{T_2}{C}_p\frac{dT}{T}=C_p\ln \left(\frac{T_2}{T_1}\right)={\int }_{373}^{273}{C}_p\frac{dT}{T}$

$\Delta S_2==-18\ln \left(\frac{273}{373}\right)=-5.62cal/K$

3. $H_{2}O\left(l\right)\to H_{2}O\left(s\right)$; $\Delta H=-18\times 80cal/g$ and $T=273K$

Entropy change during freezing of water is:

using $\Delta S_3=\frac{\Delta H}{T}$

$\Delta S_3=\frac{-1440}{273}=-5.27cal/K$

Total entropy change $\Delta S=\Delta S_1+\Delta S_2+\Delta S_3$

$\Delta S=(-26.06)+(-5.62)+(-5.27)=-36.95cal/K=-36.95cal{/}^{\circ }C$

Therefore all options A,B,C,D are true.