Chapter 9.8, Problem 17P

A)

Interpretation Introduction

Interpretation:

The concentration of the resulting mixture has to be determined.

Concept introduction:

The molar flow rate $\left({\dot{n}}_i\right)$ of component is given as follows:

${\dot{n}}_i=\frac{{\dot{m}}_i}{M{W}_i}$

Here, mass flow rate of component is ${\dot{m}}_i$, and molecular weight of benzene is $M{W}_i$.

The molecular weight of benzene as $\text{78}\text{.11}\text{g}/\text{mol}$.

The molecular weight of 2- propanol as $\text{60}\text{.1}\text{g}/\text{mol}$.

the mole fraction of component $\left(y_i\right)$ is given as follows:

$y_i=\frac{{\dot{n}}_i}{{\dot{n}}_{i_1}+{\dot{n}}_{i_2}}$

Here, molar flow rate of the component 2 is $n_{i_2}$.

B)

Interpretation Introduction

Interpretation:

The heat effect $\left(\dot{Q}\right)$ of mixing has to be determined.

Concept introduction:

Expression for the heat effect $\left(\dot{Q}\right)$ of mixing using the first law of thermodynamics is given as follows:

${\dot{n}}_{benzene}{H}_1+{\dot{n}}_{2-propanol}{H}_2-{\dot{n}}_{benzene+propanol}{H}_3+\dot{Q}=0$

Here, enthalpy of benzene is $H_1$, enthalpy of 2-propanol is $H_2$, mole fraction of benzene and 2-propanol is ${\dot{n}}_{benzene+propanol}$, and enthalpy of the resulting mixture (benzene and 2-propanol) is $H_3$.

The mole fraction of benzene and 2-propanol $\left({\dot{n}}_{benzene+propanol}\right)$ is given as follows;

${\dot{n}}_{benzene+propanol}={\dot{n}}_{benzene}+{\dot{n}}_{2-propanol}$

C)

Interpretation Introduction

Interpretation:

The temperature of the resulting solution has to be determined.

Concept introduction:

The energy balance for th given system is given as follows.

${\dot{n}}_{benzene}{H}_1+{\dot{n}}_{2-propanol}{H}_2-{\dot{n}}_{benzene+propanol}{H}_3=0$