91Ó°ÊÓ

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.

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${\mathrm{L}}_{\mathrm{v}}=2256×{10}^3\mathrm{J}/\mathrm{kg}$

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

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

Put value of heat in formula for change in entropy

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

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

b) When compare the result of $∆\mathrm{S}=6.05×{10}^3\mathrm{J}/\mathrm{K}$for vaporisation with the result of $∆\mathrm{S}=6.05×{10}^3\mathrm{J}/\mathrm{K}$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.