Chapter 3, Problem 3.24E

${\text{SO}}_{\text{2}}$ in a piston chamber kept in a constant-temperature bath at $25.0°\text{C}$ expands from $25.0\text{mL}$ to $75.0\text{mL}$ very, very slowly. Assume ${\text{SO}}_{\text{2}}$ behaves as a van der Waals gas, and its van der Waals parameters are $a=6.714\text{atm}\cdot {\text{L}}^{\text{2}}/{\text{mol}}^2$ and $b=0.05636\text{L}/\text{mol}$. If there is $0.00100$ mole of ideal gas in the chamber, calculate $\Delta S_{\text{sys}},\Delta S_{\text{surr}},$ and $\Delta S_{\text{univ}}$ for the process.

Interpretation Introduction

Interpretation:

The values of $\Delta S_{\text{sys}},\Delta S_{\text{surr}},$ and $\Delta S_{\text{univ}}$ for the process of expansion of ${\text{SO}}_{\text{2}}$ gas are to be calculated.

Concept introduction:

The change in entropy of the system is given by the equation shown below.

$\Delta S_{\text{sys}}=\frac{q_{\text{rev}}}{T}$

This equation is in the terms of heat for the reversible processes. The change in entropy of the system is given by the equation as shown below.

$\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}$

The change in entropy of the universe is the sum of the change in entropy of system and surrounding.

Answer to Problem 3.24E

The values of $\Delta S_{\text{sys}},\Delta S_{\text{surr}},$ and $\Delta S_{\text{univ}}$ for the process of expansion of ${\text{SO}}_{\text{2}}$ gas are $9.12\times {10}^{-3}\text{J}{\text{K}}^{-1}$, $0.0164\text{J}{\text{K}}^{-1}$ and $0.0255\text{J}{\text{K}}^{-1}$ respectively.

Explanation of Solution

The given process occurs at constant temperature, therefore $\Delta \text{U}=\text{0}$ for the process. Thus, the $q=-w$.

The work done is calculated using the formula given below.

$w=-p_{\text{ex}}\Delta V$…(1)

Where,

• $p_{\text{ex}}$ is the external pressure.

• $\Delta V$ is the change in volume.

• $w$ is the work done.

The pressure of ${\text{SO}}_{\text{2}}$ gas can be calculated using the van der Waals gas equation as shown below.

$p=\frac{nRT}{V-nb}-a{\left(\frac{n}{V}\right)}^2$

Where,

• $p$ is the pressure.

• $V$ is the volume.

• $R$ is the gas constant.

• $T$ is the temperature.

• $a,b$ are the van der Waals coefficients.

• $n$ is the number of moles.

The number of moles, volume and temperature given for ${\text{SO}}_{\text{2}}$ gas are $0.00100\text{mol}$, $25.0\text{mL}$ and $25.0°\text{C}$ respectively. The values of $a$ and $b$ for ${\text{SO}}_{\text{2}}$ gas are given as $6.714\text{atm}\cdot {\text{L}}^{\text{2}}/{\text{mol}}^2$ and $0.05636\text{L}/\text{mol}$ respectively.

Substitute the values of number of moles, volume and temperature in the above equation as shown below.

$\begin{array}{rll}p& =& \frac{nRT}{V-nb}-a{\left(\frac{n}{V}\right)}^2\\ & =& \left(\begin{array}{l}\frac{\left(0.00100\text{mol}\right)\left(0.0821\text{L}\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(25+273.15\text{K}\right)}{\left(25.0\text{mL}\times \frac{\text{1}\text{L}}{1000\text{mL}}\right)-\left(0.00100\text{mol}\times 0.05636\text{L}/\text{mol}\right)}\\ -\left(6.714\text{atm}\cdot {\text{L}}^{\text{2}}/{\text{mol}}^2\right){\left(\frac{0.00100\text{mol}}{25.0\text{mL}\times \frac{\text{1}\text{L}}{1000\text{mL}}}\right)}^2\end{array}\right)\\ & =& \left(\frac{0.0244}{0.025-5.636\times {10}^{-5}}\right)-0.01074\\ & =& 0.9676\text{atm}\end{array}$

The initial and final volume given is $25.00\text{mL}$ and $75.00\text{mL}$ respectively. Substitute the values in the equation (1) as shown below.

$\begin{array}{rll}w& =& -p_{\text{ex}}\Delta V\\ & =& -0.9676\text{atm}\times \left(75.00\text{mL}-25.00\text{mL}\right)\times \frac{\text{1}\text{L}}{1000\text{mL}}\times \frac{101.32\text{J}}{\text{L}\text{atm}}\\ & =& -0.9676\text{atm}\times \left(50.0\text{mL}\right)\times \frac{\text{1}\text{L}}{1000\text{mL}}\times \frac{101.32\text{J}}{\text{L}\text{atm}}\\ & =& -4.90\text{J}\end{array}$

Thus, $q=4.90\text{J}$. Since this heat is given to the surrounding, therefore the $\Delta S_{\text{surr}}$ is calculated as shown below.

$\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}$

Substitute the values in above equation as shown below.

$\begin{array}{rll}\Delta S_{\text{surr}}& =& \frac{q_{\text{surr}}}{T}\\ & =& \frac{4.90\text{J}}{298.15\text{K}}\\ & =& 0.0164\text{J}{\text{K}}^{-1}\end{array}$

The work done for the reversible process is calculated using the formula given below.

$w=-nRT\ln \frac{V_{\text{f}}}{V_{\text{i}}}$

Where,

• $n$ is the number of moles.

• $R$ is the gas constant.

• $T$ is the temperature.

• $V_{\text{f}}$ is the final volume.

• $V_{\text{i}}$ is the initial volume.

Substitute the values of $R$, $T$, $V_{\text{f}}$ and $V_{\text{i}}$ in the above equation as shown below.

$\begin{array}{rll}w& =& -nRT\ln \frac{V_{\text{f}}}{V_{\text{i}}}\\ & =& -0.00100\text{mol}\times 8.314\text{J}{\text{mol}}^{-\text{1}}{\text{K}}^{-1}\times 298.15\text{K}\times \ln \frac{75.00\text{mL}}{25.00\text{mL}}\\ & =& -2.4788\times \left(1.0986\right)\\ & =& -2.72\text{J}\end{array}$

Thus, $q_{\text{rev}}=2.72\text{J}$. The $\Delta S_{\text{sys}}$ is calculated using the equation given below as shown below.

$\Delta S_{\text{sys}}=\frac{q_{\text{rev}}}{T}$

Substitute the values in above equation as shown below.

$\begin{array}{rll}\Delta S_{\text{sys}}& =& \frac{q_{\text{rev}}}{T}\\ & =& \frac{2.72\text{J}}{298.15\text{K}}\\ & =& 9.12\times {10}^{-3}\text{J}{\text{K}}^{-1}\end{array}$

The value of $\Delta S_{\text{univ}}$ is calculated using the equation given below as shown below.

$\Delta S_{\text{univ}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$

Substitute the values in above equation as shown below.

$\begin{array}{rll}\Delta S_{\text{univ}}& =& \Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ & =& 9.12\times {10}^{-3}\text{J}{\text{K}}^{-1}+0.0164\text{J}{\text{K}}^{-1}\\ & =& 0.0255\text{J}{\text{K}}^{-1}\end{array}$

Conclusion

The values of $\Delta S_{\text{sys}},\Delta S_{\text{surr}},$ and $\Delta S_{\text{univ}}$ for the process of expansion of ${\text{SO}}_{\text{2}}$ gas are $9.12\times {10}^{-3}\text{J}{\text{K}}^{-1}$, $0.0164\text{J}{\text{K}}^{-1}$ and $0.0255\text{J}{\text{K}}^{-1}$ respectively.