Chapter 3, Problem 3.32E

$4.00\text{L}$ of $\text{Ar}$ and $2.50\text{L}$ of $\text{He}$, each at $298\text{K}$ and $1.50\text{atm}$, were mixed isothermally and isobarically. The mixture was then expanded to a final volume of $20.0\text{L}$ at $298\text{K}$. Write chemical processes for each step, and determine the change in entropy for the complete process.

Interpretation Introduction

Interpretation:

The chemical processes for each step of the given process are to be stated. The change in entropy for the complete process is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases. The entropy of mixing of gases is given as,

${\Delta }_{\text{mix}}S=-R{\displaystyle \sum _{i=1}^{\text{no}\text{.}\text{of}\text{gases}}{n}_i\ln x_i}$

Answer to Problem 3.32E

The first process is the mixing of gas and the second process is the expansion of mixture. The entropy of mixing for the given mixture is $2.2081\text{J}/\text{K}$.The change in entropy by the volume change of the given system is $3.7246\text{J}/\text{K}$. The change in entropy for the complete process is $5.9327\text{J}/\text{K}$.

Explanation of Solution

The initial volume of the $\text{Ar}$ is $4.00\text{L}$.

The initial volume of the $\text{He}$ is $2.50\text{L}$.

The temperature is $298\text{K}$.

The pressure is $1.50\text{atm}$.

The ideal gas equation is given as,

$PV=nRT$ …(1)

Where,

• $V$ represents the volume occupied by the ideal gas.

• $P$ represents the pressure of the ideal gas.

• $n$ represents the number of moles of the ideal gas.

• $T$ represents the temperature of the ideal gas.

• $R$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}$.

Rearrange the above equation for the value of $n$.

$n=\frac{PV}{RT}$ …(2)

Substitute the values of the volume of argon gas $P$, $T$ and $R$ in the equation (2).

$\begin{array}{rll}n& =& \frac{\left(4.00\text{L}\right)\left(1.50\text{atm}\right)}{\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(298\text{K}\right)}\\ & =& 0.2453\text{mol}\end{array}$

Therefore, the number of moles of the $\text{Ar}$ is $0.2453\text{mol}$.

Substitute the values of the volume of helium gas $P$, $T$ and $R$ in the equation (2).

$\begin{array}{rll}n& =& \frac{\left(2.50\text{L}\right)\left(1.50\text{atm}\right)}{\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(298\text{K}\right)}\\ & =& 0.1533\text{mol}\end{array}$

Therefore, the number of moles of the $\text{He}$ is $0.1533\text{mol}$.

The mole fraction of a substance present in the two-component system is given as,

$x_{\text{a}}=\frac{n_{\text{a}}}{n_{\text{a}}+n_{\text{b}}}$ …(3)

Where,

• $n_{\text{a}}$ represents the number of moles of substance $\text{a}$.

• $n_{\text{b}}$ represents the number of moles of substance $\text{b}$.

Substitute the number of moles of argon gas and helium gas to calculate the mole fraction of argon gas in the equation (3).

$\begin{array}{rll}{x}_{\text{Ar}}& =& \frac{0.2453\text{mol}}{0.2453\text{mol}+0.1533\text{mol}}\\ & =& \frac{0.2453\text{mol}}{0.3986\text{mol}}\\ & =& 0.6154\end{array}$

Substitute the number of moles of argon gas and helium gas to calculate the mole fraction of helium gas in the equation (3).

$\begin{array}{rll}{x}_{\text{Ar}}& =& \frac{0.1533\text{mol}}{0.2453\text{mol}+0.1533\text{mol}}\\ & =& \frac{0.1533\text{mol}}{0.3986\text{mol}}\\ & =& 0.3846\end{array}$

The entropy of mixing of gases to form given mixture is given as,

${\Delta }_{\text{mix}}S=-R\left(n_{\text{Ar}}\ln x_{\text{Ar}}+n_{\text{He}}\ln x_{\text{He}}\right)$ …(4)

Where,

• $n_{\text{Ar}}$ represents the number of mole of $\text{Ar}$ gas.

• $x_{\text{Ar}}$ represents the mole fraction of $\text{Ar}$ gas.

• $n_{\text{He}}$ represents the number of mole of $\text{He}$ gas.

• $x_{\text{He}}$ represents the mole fraction of $\text{He}$ gas.

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\text{K}$.

Substitute the value of $R$, number of moles and mole fractions of gases in the equation (4).

$\begin{array}{rll}{\Delta }_{\text{mix}}S& =& -\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(\left(0.2453\text{mol}\right)\ln \left(0.6154\right)+\left(0.1533\text{mol}\right)\ln \left(0.3846\right)\right)\\ & =& -\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(-0.2656\text{mol}\right)\\ & =& 2.2081\text{J}/\text{K}\end{array}$

Therefore, the entropy of mixing for the given mixture is $2.2081\text{J}/\text{K}$.

The total volume of the given mixture is given as,

$V_{\text{t}}=V_{\text{1}}+V_{\text{2}}$

Where,

• $V_{\text{e}}$ represents the volume of argon.

• $V_{\text{w}}$ represents the volume of helium.

Substitute the value of the volume of argon and the volume of helium in the above equation.

$\begin{array}{rll}{V}_{\text{t}}& =& 4.00\text{L}+2.50\text{L}\\ & =& 6.50\text{L}\end{array}$

Therefore, the initial volume of the given mixture is $6.50\text{L}$.

The final volume of thegiven mixture is $20.0\text{L}$.

The change entropy bythe expansion of the given system is given as,

$\Delta S=\left(n_{\text{Ar}}+n_{\text{He}}\right)R\ln \frac{V_{\text{f}}}{V_{\text{i}}}$ …(5)

Where,

• $V_{\text{f}}$ represents the initial volume of the mixture.

• $V_{\text{i}}$ represents the final volume of the mixture.

Substitute the value of $V_{\text{f}}$, $V_{\text{i}}$, $R$, $n_{\text{He}}$, and $n_{\text{Ar}}$ in the equation (5).

$\begin{array}{rll}\Delta S_{\text{expansion}}& =& \left(0.2453\text{mol}+0.1533\text{mol}\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\ln \left(\frac{20.0\text{L}}{6.50\text{L}}\right)\\ & =& \left(0.2453\text{mol}+0.1533\text{mol}\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\ln \left(\frac{20.0\text{L}}{6.50\text{L}}\right)\\ & =& \left(0.3986\text{mol}\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\left(1.1239\right)\\ & =& 3.7246\text{J}/\text{K}\end{array}$

Therefore, the change entropy by the volume change of the given system is $3.7246\text{J}/\text{K}$.

The total entropy change of the given mixture is given as,

$\Delta S_{\text{t}}=\Delta S_{\text{expansion}}+{\Delta }_{\text{mix}}S$

Where,

• ${\Delta }_{\text{mix}}S$ represents the entropy change by mixing.

• $\Delta S_{\text{expansion}}$ represents the entropy change by expansion.

Substitute the value of the entropy change by mixingand the entropy change by expansionin the above equation.

$\begin{array}{rll}\Delta S_{\text{t}}& =& 3.7246\text{J}/\text{K}+2.2081\text{J}/\text{K}\\ & =& 5.9327\text{J}/\text{K}\end{array}$

Therefore, the change in entropy for the complete process is $5.9327\text{J}/\text{K}$.

Conclusion

The first process is the mixing of gas and the second process is the expansion of mixture. The entropy of mixing for the given mixture is $2.2081\text{J}/\text{K}$.The change entropy by the volume change of the given system is $3.7246\text{J}/\text{K}$. The change in entropy for the complete process is $5.9327\text{J}/\text{K}$.