Question

20.26 What is the change in entropy of 0.130 kg of helium gas at the normal boiling point of helium when it all condenses isothermally to 1.00 L of liquid helium? (Hint: See Table 17.4 in Section 17.6.)

Short answer

The change in Entropy in the process of condensation of Helium is -644.44 J/K.

Step-by-step solution

Formula to calculate the change in entropy

Heat transport divided by temperature equals the change in entropy.

Entropy change in isothermal process is given by

$$\begin{gathered} \mathbf{∆}\mathbf{S}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{T}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{mL}}_{\mathbf{v}}}{\mathbf{T}} \end{gathered}$$

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

Given data

Mass of Helium m=0.130 kg

Latent heat of vaporisation of helium ${\mathrm{L}}_{\mathrm{v}}=20.9\times {10}^3\mathrm{J}/\mathrm{kg}$

Boiling temperature of Helium (Temperature at which Helium condenses)

$\mathrm{T}=4.216\mathrm{K}$

Determine the change in entropy

Since this is the process of condensation of Helium, heat is released from Helium in the process phase transition, so we use negative for the heat release, thus the formula becomes

$∆\mathrm{S}=\frac{-{\mathrm{mL}}_{\mathrm{v}}}{\mathrm{T}}$
Put the values in the above formula

$$\begin{gathered} ∆\mathrm{S}=\frac{-0.130\mathrm{kg}\times 20.9\times {10}^3\mathrm{J}/\mathrm{kg}}{4.216\mathrm{K}} \\ =-644.44\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the change in Entropy in the process of condensation of Helium is $-644.44\mathrm{J}/\mathrm{K}$.