Question

Calculate the changes in free energy, enthalpy, and entropy when 1.00 mole of $\mathbf{Ar}\left(g\right)$at ${\mathbf{278}}^{\mathbf{o}}\mathbf{C}$ is compressed isothermally from 100.0 L to 1.00 L.

Short answer

The changes in free energy, enthalpy, and entropy are $\mathbf{0}$, $\mathbf{-}\mathbf{38}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$, and $\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kJ}$.

Step-by-step solution

Step 1: Introduction to the Concept

Isothermal process refers to a thermodynamic process in which there is no change in temperature, i.e. the temperature remains constant.

Step 2: Determination of ∆H,∆Gand ∆S

Because there is no change in temperature in an isothermal operation, the change in enthalpy for an ideal gas will be zero.

$\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{0}$

The entropy change,$∆\mathrm{S}$, is determined using the formula:

$\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{nRln}\left(\frac{V_2}{V_1}\right)$

Where ndenotes thenumber of moles, R is the universal Gas Constant, and ${\mathbf{V}}_{\mathbf{1}}$ denotes theinitial volume and ${\mathbf{V}}_{\mathbf{2}}$denotes thefinal volume.

Substituting the values is as follows:

$$\begin{gathered} ∆\mathrm{S}=\left(1.00\mathrm{mol}\right)\times \left(8.3145\mathrm{J}/\mathrm{Kmol}\right)\times \ln \left(\frac{1.00\mathrm{L}}{100.0\mathrm{L}}\right) \\ ∆\mathrm{S}=8.3145\mathrm{J}/\mathrm{K}\times \left(-4.61\right) \\ ∆\mathrm{S}=-38.33\mathrm{J}/\mathrm{K} \end{gathered}$$

Using the formula,

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{H}\mathbf{-}\mathbf{T}\mathbf{∆}\mathbf{S}$,

$\mathbf{∆}\mathbf{S}$is determined.

Where $\mathbf{∆}\mathbf{H}$ denotes enthalpy change,$\mathbf{∆}\mathbf{S}$ denotes entropy change, and Tdenotes temperature.

Substituting the values is as follows:

$$\begin{gathered} ∆\mathrm{G}=\left(0\mathrm{kJ}\right)-\left(300\mathrm{K}\right)\times \left(-38.33\mathrm{J}/\mathrm{K}\right) \\ ∆\mathrm{G}=11499\mathrm{J}\times \left(\frac{1\mathrm{kJ}}{{10}^3\mathrm{J}}\right) \\ ∆\mathrm{G}=11.5\mathrm{kJ} \end{gathered}$$

Therefore the $\mathbf{∆}\mathbf{H}$is $\mathbf{0}$, $\mathbf{∆}\mathbf{S}$is $\mathbf{-}\mathbf{38}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$and $\mathbf{∆}\mathbf{G}$is $\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kJ}$.