Chapter 4, Problem 4.62E

Interpretation Introduction

(a)

Interpretation:

The equation for Gibbs energy of mixing of gases is to be derived.

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 ordered arrangement, then the entropy of the system increases. The entropy of mixing of gases is shown below.

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

Answer to Problem 4.62E

The equation for Gibbs energy of mixing of gases has been derived as shown below.

Explanation of Solution

The Gibbs free energy of mixing of gases is shown below.

${\Delta }_{\text{mix}}G=RT{\displaystyle \sum _{\text{i}=1}^{\text{no of gases}}{n}_{\text{i}}\ln x_{\text{i}}}$ …(1)

Where,

• $R$ represents the gas constant.

• $T$ represents the temperature.

• $n_{\text{i}}$ represents the number of moles of the ${\text{i}}^{\text{th}}$ gas.

• $x_{\text{i}}$ represents the mole fraction of the ${\text{i}}^{\text{th}}$ gas.

The entropy of mixing of gases is shown below.

${\Delta }_{\text{mix}}S=-R{\displaystyle \sum _{i=1}^{\text{no}\text{.}\text{of}\text{gases}}{n}_i\ln x_i}$ …(2)

The change in Gibbs energy of the system is mathematically shown below.

$\Delta G=\Delta H-T\Delta S$ …(3)

Where,

• $\Delta H$ represents the change in enthalpy of the system.

• $\Delta G$ represents the change in Gibbs energy.

• $T$ represents the temperature of the system.

• $\Delta S$ represents the change in entropy of the system.

The equation (3) can be written for the mixing process as shown below.

${\Delta }_{\text{mix}}G={\Delta }_{\text{mix}}H-T{\Delta }_{\text{mix}}S$ …(4)

Assume that ${\Delta }_{\text{mix}}H=0$.

Substitute the value of ${\Delta }_{\text{mix}}H$ in the equation (4).

$\begin{array}{rll}{\Delta }_{\text{mix}}G& =& 0-T{\Delta }_{\text{mix}}S\\ & =& -T{\Delta }_{\text{mix}}S\end{array}$

Substitute the value of ${\Delta }_{\text{mix}}S$ from equation (2) in the equation (4).

$\begin{array}{rll}{\Delta }_{\text{mix}}G& =& -T\left(-R{\displaystyle \sum _{i=1}^{\text{no}\text{.}\text{of}\text{gases}}{n}_i\ln x_i}\right)\\ & =& RT{\displaystyle \sum _{i=1}^{\text{no}\text{.}\text{of}\text{gases}}{n}_i\ln x_i}\end{array}$

Therefore, the equation for Gibbs energy of mixing of gases has been derived.

Conclusion

The equation for Gibbs energy of mixing of gases has been derived.

Interpretation Introduction

(b)

Interpretation:

The statement that the process of mixing of gases is always spontaneous is to be verified by a demonstration that the Gibbs free energy of mixing is always less than zero for a mixture of gases.

Concept introduction:

The Gibbs free energy of the system represents the maximum amount of non-expansion work achieved by a thermodynamic system at isothermal and isobaric conditions. The change in Gibbs free energy is used to predict the spontaneity of the process. The Gibbs free energy of mixing of gases is given below.

${\Delta }_{\text{mix}}G=RT{\displaystyle \sum _{\text{i}=1}^{\text{no of gases}}{n}_{\text{i}}\ln x_{\text{i}}}$

Answer to Problem 4.62E

The mole fraction of gas is always less than one. The natural logarithm of a number that is less than one is always negative. Therefore, the value of change in Gibbs free energy for mixing of gas is always negative and the process is always spontaneous.

Explanation of Solution

The Gibbs free energy of mixing of gases is shown below.

${\Delta }_{\text{mix}}G=RT{\displaystyle \sum _{\text{i}=1}^{\text{no of gases}}{n}_{\text{i}}\ln x_{\text{i}}}$ …(1)

Where,

• $R$ represents the gas constant.

• $T$ represents the temperature.

• $n_{\text{i}}$ represents the number of moles of the ${\text{i}}^{\text{th}}$ gas.

• $x_{\text{i}}$ represents the mole fraction of the ${\text{i}}^{\text{th}}$ gas.

The Gibbs free energy of mixing of two gases A and B can be given as shown below.

${\Delta }_{\text{mix}}G=RT\left(n_{\text{A}}\ln x_{\text{A}}+n_{\text{B}}\ln x_{\text{B}}\right)$

The mole fraction of both the gases is always less than one. The natural logarithm of a number that is less than one is always negative. The result of the addition of two negative values is also negative.

The right-hand side of the equation is negative for the mixing of gases.

${\Delta }_{\text{mix}}G=-\text{negative}$

Therefore, the negative value of change in Gibbs free energy indicates that the process of mixing of gases is spontaneous.

Conclusion

The mole fraction of gas is always less than one. The natural logarithm of a number that is less than one is always negative. Therefore, the value of change in Gibbs free energy for mixing of gas is always negative and the process is always spontaneous.

Interpretation Introduction

(c)

Interpretation:

The value of ${\Delta }_{\text{mix}}G$ for mixing $1.0\text{mol}\text{Ne}$, $2.0\text{mol}\text{He}$, and $3.0\text{mol}\text{Ar}$ at $35.0°\text{C}$ is to be calculated.

Concept introduction:

The Gibbs free energy of the system represents the maximum amount of non-expansion work achieved by a thermodynamic system at isothermal and isobaric conditions. The change in Gibbs free energy is used to predict the spontaneity of the process. The Gibbs free energy of mixing of gases is shown below.

${\Delta }_{\text{mix}}G=RT{\displaystyle \sum _{\text{i}=1}^{\text{no of gases}}{n}_{\text{i}}\ln x_{\text{i}}}$

Answer to Problem 4.62E

The value of ${\Delta }_{\text{mix}}G$ for mixing $1.0\text{mol}\text{Ne}$, $2.0\text{mol}\text{He}$, and $3.0\text{mol}\text{Ar}$ at $35.0°\text{C}$ is $-15538.400\text{J}$.

Explanation of Solution

The number of moles of neon gas is $1.0\text{mol}$.

The number of moles of helium gas is $2.0\text{mol}$.

The number of moles of argon gas is $3.0\text{mol}$.

The temperature of mixing is $35.0°\text{C}$.

The temperature of mixing in Kelvin is shown below.

$\begin{array}{rll}T& =& \left(273+35.0°\text{C}\right)\text{K}\\ & =& \text{308}\text{K}\end{array}$

The total number of moles of gases is calculated as,

$n_{\text{t}}=n_{\text{Ne}}+n_{\text{He}}+n_{\text{Ar}}$

Where,

• $n_{\text{Ne}}$ represents the number of moles of neon gas.

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

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

Substitute the value of $n_{\text{Ne}}$, $n_{\text{He}}$, and $n_{\text{Ar}}$ in the above equation.

$\begin{array}{rll}{n}_{\text{t}}& =& 1.0\text{mol}+2.0\text{mol}+3.0\text{mol}\\ & =& 6.0\text{mol}\end{array}$

The mole fraction of a substance present in a system is shown below.

$x_{\text{a}}=\frac{n_{\text{a}}}{n_{\text{t}}}$ …(5)

Where,

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

• $n_{\text{t}}$ represents the total number of moles.

Substitute the value of the number of moles of neon gas and $n_{\text{t}}$ in the equation (5).

$\begin{array}{rll}{x}_{\text{Ne}}& =& \frac{1.0\text{mol}}{6.0\text{mol}}\\ & =& 0.1667\end{array}$

Substitute the value of the number of moles of helium gas and $n_{\text{t}}$ in the equation (5).

$\begin{array}{rll}{x}_{\text{He}}& =& \frac{2.0\text{mol}}{6.0\text{mol}}\\ & =& 0.3333\end{array}$

Substitute the value of the number of moles of argon gas and $n_{\text{t}}$ in the equation (5).

$\begin{array}{rll}{x}_{\text{Ar}}& =& \frac{3.0\text{mol}}{6.0\text{mol}}\\ & =& 0.50\end{array}$

The Gibbs free energy of mixing of gases is given as shown below.

${\Delta }_{\text{mix}}G=RT\left(n_{\text{Ne}}\ln x_{\text{Ne}}+n_{\text{He}}\ln x_{\text{He}}+n_{\text{Ar}}\ln x_{\text{Ar}}\right)$ …(6)

Where,

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

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

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

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

Substitute the value of $R$, temperature, number of moles and mole fractions of gases in the equation (6).

$\begin{array}{rll}{\Delta }_{\text{mix}}G& =& \left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{308}\text{K}\right)\left(\left(1.0\text{mol}\right)\ln \left(0.1667\right)+\left(2.0\text{mol}\right)\ln \left(0.3333\right)+\left(3.0\text{mol}\right)\ln \left(0.50\right)\right)\\ & =& \left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{308}\text{K}\right)\left(-6.068\text{mol}\right)\\ & =& -15538.400\text{J}\end{array}$

Therefore, the value of ${\Delta }_{\text{mix}}G$ for mixing $1.0\text{mol}\text{Ne}$, $2.0\text{mol}\text{He}$, and $3.0\text{mol}\text{Ar}$ at $35.0°\text{C}$ is $-15538.400\text{J}$.

Conclusion

The value of ${\Delta }_{\text{mix}}G$ for mixing $1.0\text{mol}\text{Ne}$, $2.0\text{mol}\text{He}$, and $3.0\text{mol}\text{Ar}$ at $35.0°\text{C}$ is $-15538.400\text{J}$.