Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The change in entropy should be determined for CO₂ and N₂ blending together to form a gas mixture at same conditions assuming them to be ideal gases.

Concept introduction:

The internal energy change for any reversible process is written as:

$dU=TdS-PdV$

Where, dU is change in internal energy, dS is change in entropy and dV is change in volume.

Here, T is temperature and P is pressure.

Ideal gas equation is given by:

PV = nRT

Where, P = Pressure

V = volume

n = number of moles

R = gas constant

T = temperature

And, total entropy change is identified by adding the change in entropy for individual gases.

That is, ΔS_total =ΔS₁ +ΔS₂

Explanation of Solution

Given Information:

Initial volumes of the gases are:

$\begin{array}{c}C{O}_2=0.7m^3\\ N_2=0.3m^3\end{array}$

Final Volume $=0.7m^3+0.3m^3=1.0m^3$

Pressure of each gas is $\text{1 bar}$

Temperature is ${\text{25}}^{\text{0}}\text{C=298 K}$ .

Step1:

Establish equation for entropy change.

Since, after mixing there has been no change in internal energy. So, the dU term will become zero such as:

$\begin{array}{c}dU=TdS-PdV\\ 0=TdS-PdV\\ TdS=PdV\end{array}$

Hence, change in entropy can be written as:

$dS=\frac{PdV}{T}$

Further, the gases are considered to be ideal and so as per ideal gas equation:

$\begin{array}{c}PV=nRT\\ \frac{P}{T}=\frac{nR}{V}\end{array}$

Putting the value of P/T in change in entropy equation we get:

$dS=\frac{PdV}{T}=\frac{nRdV}{V}$

The equation can be simplified as:

$\Delta S=nR\ln \left(\frac{V_2}{V_1}\right)$

Step2:

Entropy change for ${\text{CO}}_{\text{2}}$ gas is calculated as:

Moles of ${\text{CO}}_{\text{2}}$ ;

$nR=\frac{PV}{T}$

Adding the values we get:

$n=\frac{PV}{RT}=\frac{1bar\times 0.7m^3}{8.314\times {10}^{-5}{m}^{3}bar/Kmol\times 298K}=28.25mol$

Hence,

$\begin{array}{c}\Delta S=nR\ln \left(\frac{V_2}{V_1}\right)\\ =28.25mol\times 8.314J/K.mol\ln \left(\frac{1}{0.7}\right)\\ =83.75J/K\end{array}$

Step3:

Entropy change for N₂ gas is calculated as.

Moles of N₂;
$n=\frac{PV}{RT}$

Adding the values we get:

$n=\frac{PV}{RT}=\frac{1bar\times 0.3m^3}{8.314\times {10}^{-5}{m}^{3}bar/Kmol\times 298K}=12.11mol$

Hence,

$\begin{array}{c}\Delta S=nR\ln \left(\frac{V_2}{V_1}\right)\\ =12.11mol\times 8.314J/K.mol\ln \left(\frac{1}{0.3}\right)\\ =121.22J/K\end{array}$

Step4:

Total entropy change is calculated by adding the change in entropy for individual gases as follows:

$\begin{array}{c}\Delta S_{Total}=\Delta S_{C{O}_2}+\Delta S_{N_2}\\ =83.75J/K+121.22J/K\\ =204.97J/K\end{array}$