Question

20.27 (a) Calculate the change in entropy when 1.00 kg of water at $\mathbf{100}\mathbf{°}\mathbf{C}$is vaporized and converted to steam at $\mathbf{100}\mathbf{°}\mathbf{C}$(see Table 17.4). (b) Compare your answer to the change in entropy when 1.00 kg of ice is melted at $\mathbf{0}\mathbf{°}\mathbf{C}$, calculated in Example 20.5 (Section 20.7). Is the change in entropy greater for melting or for vaporization? Interpret your answer using the idea that entropy is a measure of the randomness of a system

Short answer

a) The change in entropy when $1.00\mathrm{kg}$of water at $100°\mathrm{C}$is vaporized and converted to steam at is $100°\mathrm{C}\mathrm{is}∆\mathrm{S}=6.05\times {10}^3\mathrm{J}/\mathrm{K}$.

b) The change in entropy $∆\mathrm{S}$is about five times greater for vaporisation than in melting.

This is because the system always tends to randomness, wherein the vaporisation process, the molecules will be in a greater state of randomness.

Step-by-step solution

Formula for Entropy change and Heat

Heat transport divided by temperature equals the change in entropy.

Entropy change in isothermal process is given by

$$\begin{gathered} \mathbf{∆}\mathbf{S}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{T}} \\ \mathbf{∆}\mathbf{S}\mathbf{=}\frac{{\mathbf{mL}}_{\mathbf{v}}}{\mathbf{T}} \end{gathered}$$ ….. (1)

The heat,

$\mathbf{Q}\mathbf{=}{\mathbf{mL}}_{\mathbf{v}}$ ….. (2)

Where, $∆\mathrm{S}$is the change in entropy, is mass in kg, ${\mathrm{L}}_{\mathrm{v}}$is Latent heat of vaporisation in $\mathrm{J}/\mathrm{kg}$, $\mathrm{T}$is boiling temperature in Kelvin.

Given data

Mass of water$\mathrm{m}=1.00\mathrm{kg}$

Temperature at which phase change occurs$\mathrm{T}=100.0°\mathrm{C}=373\mathrm{K}$

Heat of vaporisation for water${\mathrm{L}}_{\mathrm{v}}=2256\times {10}^3\mathrm{J}/\mathrm{kg}$

(a) Determine the change in entropy for water to vapour conversion at T=100.0°C:

To determine the change in entropy, first calculate the heat bu substituting known values into equation (2).

$$\begin{gathered} \mathrm{Q}={\mathrm{mL}}_{\mathrm{v}} \\ =1.00\mathrm{kg}\times 2256\times {10}^3\mathrm{J}/\mathrm{K} \\ =2256\times {10}^3\mathrm{J} \end{gathered}$$

Put value of heat in formula for change in entropy

$$\begin{gathered} ∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}} \\ =\frac{2256\times {10}^3\mathrm{J}}{373\mathrm{K}} \\ =6.05\times {10}^3\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the change in entropy when water is vaporised and converted to steam at $100.0°\mathrm{C}$is $∆\mathrm{S}=6.05\times {10}^3\mathrm{J}/\mathrm{K}$.

Compare with change in entropy when ice is converted to water at

b) When compare the result of $∆\mathrm{S}=6.05\times {10}^3\mathrm{J}/\mathrm{K}$for vaporisation with the result of $∆\mathrm{S}=6.05\times {10}^3\mathrm{J}/\mathrm{K}$in example 20.5 for melting, we find that the change in entropy is about five times greater for vaporisation. This is expected because the system always tends to random, wherein the vaporisation process, the molecules will be in a greater state of random.