Question

What is the entropy change when 2 g of water changes from 0 °C to steam at 234 °C if (The thermal capacity is pressure) Cp = 1 J / kg.K and a latent heat of vaporization is 54 J/Kg?​

Answer 1

Answer:

$\boxed{ \sf \: \Delta S = 0.219J /K}$

Explanation:

Given:

mass of substance (water) = 2 g

initial temperature= 0°C = 273.15 K

final temperature= 234°C= 507.15

thermal capacity = 1 J/kg.M

latent heat of vaporization = 54 J/Kg

To find:

Entropy change ∆S = ?

Solution:

Entropy is measure of randomness of a system.

The change in entropy can be calculated in many ways depending upon the nature of substance & process like solid or liquid in isothermal process and when the process is not isothermal, gases are real or ideal etc.

If the process is isothermal (Temperature doesn't change) then the general equation to find change in entropy is,

$\sf \Delta S = \frac{ q_{rev}}{T}= state \: function$

If the temperature of reaction is changing then change in entropy can be calculated by

$\sf \: \Delta S = nC_p ln \frac{T_f}{T_i}$

Where n is number of moles of substance,

Cp is thermal capacity,

& T f and Ti are final and initial temperature respectively.

In the above reaction the water is being heated from 0°C (liquid) to 234°C (steam), the number of processes happening is

1. Entropy change (∆S1) for physical process where, water being heated from 0°C to 100°C
2. Entropy change (∆S2) for phase transformation of isothermal process at 100°C where liquid H2O is turning into vapours
3. Entropy change (∆S3)for physical process, where vapours heated upto 234°C

Calculating the number of moles of substance,

Given mass of water= 2g

Molecular mass of water = 18 g

$\sf n = \frac{given \: mass \:}{molar \: mass} = \frac{2}{18} = 0.11moles$

Change in entropy for physical process 1,

The initial temperature of water is 0°C(273.15 K), and the water turns into vapour at 100°C (373.15 K), this is a purely heating of liquid at constant pressure, hence the formula that would be used is,

$\sf \: \Delta S_1 = nC_p ln \frac{T_f}{T_i} \\ \sf \: \Delta S_1 = 0.11 \times 1 \times ln \frac{373.15}{273.15} \\ \sf \: \Delta S_1 = 0.11 \times ln 1.366 \\ \sf \: \Delta S_1 = 0.11 \times 0.311 \\ \sf \: \Delta S_1 = 0.0343 J/K$

At boiling point of water the change in entropy for phase transformation of liquid water to gaseous vapours at 100°C can be calculated,

Now the latent heat of vaporization (∆H) is given,

(Latent heat is the heat which is required to change the phase of any substance)

Substituting the data,

$\sf \: \Delta S_2 = \frac{\Delta H}{T} \\ \sf \: \Delta S_2 = \frac{54}{373.15} \\ \sf \: \Delta S_2 = 0.145 J/K$

Now the third and final physical process of heating vapours from 100°C to 234°C, at constant pressure,

$\sf \: \Delta S_3 = nC_p ln \frac{T_f}{T_i} \\ \sf \: \Delta S_3 = 0.11 \times 1 \times ln \frac{507.15}{373.15} \\ \sf \: \Delta S_3 = 0.11 \times ln 1.359 \\ \sf \: \Delta S_3 = 0.11 \times 0.306 \\ \sf \: \Delta S_3 = 0.0337J/K$

The total entropy change ∆S during the process is,

$\sf \: \Delta S = \Delta S_1+ \Delta S_2 + \Delta S_3 \\ \sf \: \Delta S = 0.0343+ 0.145 + 0.0337 \\ \sf \: \Delta S = 0.219J/K$

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