Question

Consider$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{o}\mathit{f}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{a}\mathit{t}\mathbf{300}\mathit{K}\mathit{a}\mathit{n}\mathit{d}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{a}\mathit{t}\mathit{m}\mathbf{.}$The gas expands until the final pressure is$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{a}\mathit{t}\mathit{m}$. For each of the following conditions describing the expansion, calculate$\mathit{\Delta }\mathit{S}\mathbf{,}\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{r}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathit{\Delta }{\mathit{S}}_{\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{v}}\mathbf{.}{\mathit{C}}_{\mathbf{p}}\mathit{f}\mathit{o}\mathit{r}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{i}\mathit{s}\mathbf{37}\mathbf{.}\mathbf{1}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\mathbf{,}$ and assume that the gas behaves ideally.

1. The expansion occurs isothermally and reversibly.
2. The expansion occurs isothermally against a constant external pressure of$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{a}\mathit{t}\mathit{m}$
3. The expansion occurs adiabatically and reversibly

Short answer

a.$\begin{array}{l}{\text{ΔS}}_{\text{univ}}=0\\ {\text{ΔS}}_{\text{sys}}=13.4\text{\hspace{0.17em}J/K}\\ {\text{ΔS}}_{\text{surr}}=-13.4\text{\hspace{0.17em}J/K}\end{array}$

b.$\begin{array}{l}{\text{ΔS}}_{\text{univ}}=6.77\text{\hspace{0.17em}\hspace{0.17em}J/K}\\ {\text{ΔS}}_{\text{sys}}=13.4\text{\hspace{0.17em}J/K}\\ {\text{ΔS}}_{\text{surr}}=-6.63\text{\hspace{0.17em}J/K}\end{array}$

c.

$\begin{array}{l}{\text{ΔS}}_{\text{univ}}=0\\ {\text{ΔS}}_{\text{sys}}=0\\ {\text{ΔS}}_{\text{surr}}=0\end{array}$

Step-by-step solution

Definition of isothermal expansion

Isothermal expansion is defined as keeping the gas at a constant temperature.

Calculations

a.${\text{ΔS}}_{\text{univ}}=0\left(\text{reversible}\right)$

role="math" localid="1663759404643" $\begin{array}{l}{\text{ΔS}}_{\text{sys}}=\text{nRln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\\ {\text{ΔS}}_{\text{sys}}=\text{1.00mol}\times \text{8.314\hspace{0.17em}J/mol}\times \text{ln}\left(\frac{\text{5.00}}{\text{1.00}}\right)\\ {\text{ΔS}}_{\text{sys}}=13.4\text{\hspace{0.17em}J/K}\\ {\text{ΔS}}_{\text{univ}}={\text{ΔS}}_{\text{sys}}+{\text{ΔS}}_{\text{surr}}\\ {\text{ΔS}}_{\text{surr}}=-{\text{ΔS}}_{\text{sys}}\\ {\text{ΔS}}_{\text{surr}}=-13.4\text{\hspace{0.17em}J/K}\end{array}$

b.$\begin{array}{l}{\text{ΔS}}_{\text{sys}}=13.4\text{\hspace{0.17em}J/K}\\ {\text{V}}_{\text{i}}=4.92\text{\hspace{0.17em}L}\\ {\text{V}}_{\text{f}}=24.6\text{\hspace{0.17em}L}\end{array}$

$\begin{array}{l}\text{q}={\text{P}}_{\text{ex}}\text{ΔV}\\ \text{q}=1990\text{\hspace{0.17em}J}\\ {\text{ΔS}}_{\text{surr}}=-6.63\text{\hspace{0.17em}J/K}\\ {\text{ΔS}}_{\text{univ}}={\text{ΔS}}_{\text{sys}}+{\text{ΔS}}_{\text{surr}}\\ =13.4-6.63\\ {\text{ΔS}}_{\text{univ}}=6.77\text{\hspace{0.17em}J/K}\end{array}$

c.

$\begin{array}{l}{\text{ΔS}}_{\text{univ}}=0\\ {\text{q}}_{\text{rev}}=0\\ {\text{ΔS}}_{\text{sys}}=0\\ {\text{ΔS}}_{\text{surr}}=\frac{{\text{-q}}_{\text{actual}}}{\text{T}}\\ {\text{ΔS}}_{\text{surr}}=0\end{array}$