Chapter 13, Problem 13.4P

(a)

Interpretation Introduction

Interpretation:

Assuming Raoult’s law is valid we need to find the values of liquid mole fraction of n-pentane(1) and vapor mole fraction of n-pentane(1) for a binary system of n-pentane(1) and n-heptane(2) for temperature 65°C and when total pressure $P=\frac{1}{2}(P_1^{sat}+P_2^{sat})$ or $P=\frac{1}{2}(P_1^v+P_2^v)$ . We have to also plot the fraction of system that is vapor V vs. overall composition z₁ for the binary system of n-pentane (1) and n-heptane(2) at these same conditions.

Assuming Raoult’s law is valid we need to plot the total pressure P, liquid mole fraction of n-pentane (1), x₁, vapor fraction of n-pentane (1), y₁ vs. fraction of system that is vapor V for the binary system of n-pentane(1) and n-heptane(2) at these same conditions at temperature $T=65°C$ and z₁=0.5.

Concept introduction:

Raoult’s Law states that the partial pressure of liquid A above the solution is equal to the mole fraction of the liquid in a solution times the partial pressure of the pure liquid. This holds for ideal solutions. An ideal solution is approached by binary solutions of molecules that have similar properties (e.g. benzene and toluene). The law is mathematically expressed as-

  $p_A^{*}=x_A{P}_A^v$

Where

  $P_A^v$ = vapor pressure of A at the given temperature

  $x_A$ = mole fraction of the solute A in the liquid

  $p_A^{*}$ = equilibrium partial pressure exerted by the solute.

A solution behaves ideally when

1. The solute and the solvent molecules have similar sizes and similar intermolecular forces,
2. The excess volume of mixing is zero, and
3. The heat of mixing is zero when both the solute and the solvent are liquids

Answer to Problem 13.4P

$x_1=0.5$, $y_1=0.88015$

V is linear in z₁

  

  

  

Explanation of Solution

Find the value of $x_1$ and $y_1$at $T=65°C$and $P=\frac{1}{2}(P_1^v+P_2^v)$

The Raoult’s law expression for two components A and B to give the equilibrium partial pressure of the components in a binary mixture is written as follows-

$p_A^{*}=x_A{P}_A^v$

and

$p_B^{*}=(1-x_A)P_B^v$

The total pressure: $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$

Where

$x_A$ = mole fraction of A in liquid phase in the binary solution

$x_B$ = $(1-x_A)$ = mole fraction of B in liquid phase in the binary solution

  $P_A^v$ = vapor pressure of A at the given temperature

  $P_B^v$ = vapor pressure of B at the given temperature

  $p_A^{*}$ =equilibrium partial pressure of component A

  $p_B^{*}$ = equilibrium partial pressure of component B

The mole fraction of A in the vapor phase is given by -     $y_A^{}=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

Similarly, mole fraction of B in vapor phase is given by -     $y_B^{}=\frac{p_B^{*}}{P}=\frac{x_B{P}_B^v}{P}$

Now vapor pressures $P_A^v$ and $P_B^v$ are calculated by using the Antoine equation:

$\ln P_B^v$ or $\ln P_A^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta };$

$P_A^v$ and $P_B^v$ in mm Hg, $\theta$ in $°C$

$A\text{'}$, $B\text{'}$, $C\text{'}$ are constants

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+65}$

  $\Rightarrow \ln P_1^v=5.5116$

$\Rightarrow P_1^v=247.54711$ mm Hg at $T=65°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+65}$

  $\Rightarrow \ln P_2^v=3.51774$

$\Rightarrow P_2^v=33.70841$ mm Hg at $T=65°C$

Therefore, total pressure, $P=\frac{1}{2}(P_1^v+P_2^v)=\frac{1}{2}(247.54711+33.70841)=140.62776$ mm Hg at $T=65°C$ .

By equating total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ and $P=\frac{1}{2}(P_1^v+P_2^v)$ we find $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1{P}_1^v+(1-x_1)P_2^v=\frac{1}{2}(P_1^v+P_2^v)\\ \Rightarrow x_1=0.5\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $\begin{array}{l}{y}_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=0.88015\\ \end{array}$

For the given temperature, z₁ ranges from the liquid composition at the bubble point to the vapor composition at the dew point.

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow z_1=0.5\times (1-V)+0.88015\times V\\ \Rightarrow z_1=0.5-0.5V+0.88015V\\ \Rightarrow z_1=0.5+0.38015V\end{array}$

So, V is linear in z₁

  

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+65}$

  $\Rightarrow \ln P_1^v=5.5116$

$\Rightarrow P_1^v=247.54711$ mm Hg at $T=65°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+65}$

  $\Rightarrow \ln P_2^v=3.51774$

$\Rightarrow P_2^v=33.70841$ mm Hg at $T=65°C$

So, the expression for total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ in $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1\times 247.54711+(1-x_1)\times 33.70841\\ \Rightarrow P=33.70841+213.8387\times x_1\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $y_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=\frac{x_1\times 247.54711}{33.70841+213.8387\times x_1}$

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow 0.5=x_1\times (1-V)+(\frac{x_1\times 247.54711}{33.70841+213.8387\times x_1})\times V\end{array}$

  

  

(b)

Interpretation Introduction

Interpretation:

Assuming Raoult’s law is valid we need to find the values of liquid mole fraction of n-pentane(1) and vapor mole fraction of n-pentane(1) for a binary system of n-pentane(1) and n-heptane(2) for temperature 75°C and when total pressure $P=\frac{1}{2}(P_1^{sat}+P_2^{sat})$ or $P=\frac{1}{2}(P_1^v+P_2^v)$ . We have to also plot the fraction of system that is vapor V vs. overall composition z₁ for the binary system of n-pentane(1) and n-heptane(2) at these same conditions.

Assuming Raoult’s law is valid we need to plot the total pressure P, liquid mole fraction of n-pentane (1), x₁, vapor fraction of n-pentane (1), y₁ vs. fraction of system that is vapor V for the binary system of n-pentane(1) and n-heptane(2) at these same conditions at temperature $T=75°C$ and z₁=0.5.

Concept introduction:

Raoult’s Law states that the partial pressure of liquid A above the solution is equal to the mole fraction of the liquid in a solution times the partial pressure of the pure liquid. This holds for ideal solutions. An ideal solution is approached by binary solutions of molecules that have similar properties (e.g. benzene and toluene). The law is mathematically expressed as-

  $p_A^{*}=x_A{P}_A^v$

Where

  $P_A^v$ = vapor pressure of A at the given temperature

  $x_A$ = mole fraction of the solute A in the liquid

  $p_A^{*}$ = equilibrium partial pressure exerted by the solute.

A solution behaves ideally when

1. The solute and the solvent molecules have similar sizes and similar intermolecular forces,
2. The excess volume of mixing is zero, and
3. The heat of mixing is zero when both the solute and the solvent are liquids

Answer to Problem 13.4P

$x_1=0.5$, $y_1=0.87077$

V is linear in z₁

  

  

  

  

Explanation of Solution

The Raoult’s law expression for two components A and B to give the equilibrium partial pressure of the components in a binary mixture is written as follows-

$p_A^{*}=x_A{P}_A^v$

and

$p_B^{*}=(1-x_A)P_B^v$

The total pressure: $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$

Where

$x_A$ = mole fraction of A in liquid phase in the binary solution

$x_B$ = $(1-x_A)$ = mole fraction of B in liquid phase in the binary solution

  $P_A^v$ = vapor pressure of A at the given temperature

  $P_B^v$ = vapor pressure of B at the given temperature

  $p_A^{*}$ =equilibrium partial pressure of component A

  $p_B^{*}$ = equilibrium partial pressure of component B

The mole fraction of A in the vapor phase is given by -     $y_A^{}=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

Similarly, mole fraction of B in vapor phase is given by -     $y_B^{}=\frac{p_B^{*}}{P}=\frac{x_B{P}_B^v}{P}$

Now vapor pressures $P_A^v$ and $P_B^v$ are calculated by using the Antoine equation:

$\ln P_B^v$ or $\ln P_A^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta };$

$P_A^v$ and $P_B^v$ in mm Hg, $\theta$ in $°C$

$A\text{'}$, $B\text{'}$, $C\text{'}$ are constants

Now, For benzene (1) $A\text{'}$ =15.9037, $B\text{'}$ =2789.01, $C\text{'}$ =220.79

And For toluene (2) $A\text{'}$ =16.00531, $B\text{'}$ =3090.78, $C\text{'}$ =219.14

Find the value of $x_1$ and $y_1$at $T=75°C$and $P=\frac{1}{2}(P_1^v+P_2^v)$

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+75}$

  $\Rightarrow \ln P_1^v=5.78048$

$\Rightarrow P_1^v=323.91605$ mm Hg at $T=75°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+75}$

  $\Rightarrow \ln P_2^v=3.87269$

$\Rightarrow P_2^v=48.072$ mm Hg at $T=75°C$

Therefore, total pressure, $P=\frac{1}{2}(P_1^v+P_2^v)=\frac{1}{2}(323.91605+48.072)=185.994$ mm Hg at $T=75°C$ .

By equating total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ and $P=\frac{1}{2}(P_1^v+P_2^v)$ we find $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1{P}_1^v+(1-x_1)P_2^v=\frac{1}{2}(P_1^v+P_2^v)\\ \Rightarrow x_1=0.5\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $y_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=0.87077$

For the given temperature, z₁ ranges from the liquid composition at the bubble point to the vapor composition at the dew point.

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow z_1=0.5\times (1-V)+0.87077\times V\\ \Rightarrow z_1=0.5-0.5V+0.87077V\\ \Rightarrow z_1=0.5+0.37077V\end{array}$

So, V is linear in z₁

  

  

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+75}$

  $\Rightarrow \ln P_1^v=5.78048$

$\Rightarrow P_1^v=323.91605$ mm Hg at $T=75°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+75}$

  $\Rightarrow \ln P_2^v=3.87269$

$\Rightarrow P_2^v=48.072$ mm Hg at $T=75°C$

So, the expression for total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ in $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1\times 323.91605+(1-x_1)\times 48.072\\ \Rightarrow P=48.072+275.84405\times x_1\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $y_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=\frac{x_1\times 323.91605}{48.072+275.84405\times x_1}$

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow 0.5=x_1\times (1-V)+(\frac{x_1\times 323.91605}{48.072+275.84405\times x_1})\times V\end{array}$

  

  

(c)

Interpretation Introduction

Interpretation:

Assuming Raoult’s law is valid we need to find the values of liquid mole fraction of n-pentane(1) and vapor mole fraction of n-pentane(1) for a binary system of n-pentane(1) and n-heptane(2) for temperature 85°C and when total pressure $P=\frac{1}{2}(P_1^{sat}+P_2^{sat})$ or $P=\frac{1}{2}(P_1^v+P_2^v)$ . We have to also plot the fraction of system that is vapor V vs. overall composition z₁ for the binary system of n-pentane(1) and n-heptane(2) at these same conditions.

Assuming Raoult’s law is valid we need to plot the total pressure P, liquid mole fraction of n-pentane (1), x₁, vapor fraction of n-pentane (1), y₁ vs. fraction of system that is vapor V for the binary system of n-pentane(1) and n-heptane(2) at these same conditions at temperature $T=85°C$ and z₁=0.5.

Concept introduction:

Raoult’s Law states that the partial pressure of liquid A above the solution is equal to the mole fraction of the liquid in a solution times the partial pressure of the pure liquid. This holds for ideal solutions. An ideal solution is approached by binary solutions of molecules that have similar properties (e.g. benzene and toluene). The law is mathematically expressed as-

  $p_A^{*}=x_A{P}_A^v$

Where

  $P_A^v$ = vapor pressure of A at the given temperature

  $x_A$ = mole fraction of the solute A in the liquid

  $p_A^{*}$ = equilibrium partial pressure exerted by the solute.

A solution behaves ideally when

1. The solute and the solvent molecules have similar sizes and similar intermolecular forces,
2. The excess volume of mixing is zero, and
3. The heat of mixing is zero when both the solute and the solvent are liquids

Answer to Problem 13.4P

$x_1=0.5$, $y_1=0.96984$

V is linear in z₁

  

  

  

  

Explanation of Solution

The Raoult’s law expression for two components A and B to give the equilibrium partial pressure of the components in a binary mixture is written as follows-

$p_A^{*}=x_A{P}_A^v$

and

$p_B^{*}=(1-x_A)P_B^v$

The total pressure: $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$

Where

$x_A$ = mole fraction of A in liquid phase in the binary solution

$x_B$ = $(1-x_A)$ = mole fraction of B in liquid phase in the binary solution

  $P_A^v$ = vapor pressure of A at the given temperature

  $P_B^v$ = vapor pressure of B at the given temperature

  $p_A^{*}$ =equilibrium partial pressure of component A

  $p_B^{*}$ = equilibrium partial pressure of component B

The mole fraction of A in the vapor phase is given by -     $y_A^{}=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

Similarly, mole fraction of B in vapor phase is given by -     $y_B^{}=\frac{p_B^{*}}{P}=\frac{x_B{P}_B^v}{P}$

Now vapor pressures $P_A^v$ and $P_B^v$ are calculated by using the Antoine equation:

$\ln P_B^v$ or $\ln P_A^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta };$

$P_A^v$ and $P_B^v$ in mm Hg, $\theta$ in $°C$

$A\text{'}$, $B\text{'}$, $C\text{'}$ are constants

Now, For benzene (1) $A\text{'}$ =15.9037, $B\text{'}$ =2789.01, $C\text{'}$ =220.79

And For toluene (2) $A\text{'}$ =16.00531, $B\text{'}$ =3090.78, $C\text{'}$ =219.14

Find the value of $x_1$ and $y_1$at $T=85°C$and $P=\frac{1}{2}(P_1^v+P_2^v)$

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+85}$

  $\Rightarrow \ln P_1^v=6.0324$

$\Rightarrow P_1^v=416.71575$ mm Hg at $T=85°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+85}$

  $\Rightarrow \ln P_2^v=4.2041$

$\Rightarrow P_2^v=66.96039$ mm Hg at $T=85°C$

Therefore, total pressure, $P=\frac{1}{2}(P_1^v+P_2^v)=\frac{1}{2}(416.71575+66.96039)=214.83807$ mm Hg at $T=85°C$ .

By equating total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ and $P=\frac{1}{2}(P_1^v+P_2^v)$ we find $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1{P}_1^v+(1-x_1)P_2^v=\frac{1}{2}(P_1^v+P_2^v)\\ \Rightarrow x_1=0.5\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $\begin{array}{l}{y}_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=0.96984\\ \end{array}$

For the given temperature, z₁ ranges from the liquid composition at the bubble point to the vapor composition at the dew point.

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow z_1=0.5\times (1-V)+0.96984\times V\\ \Rightarrow z_1=0.5-0.5V+0.96984V\\ \Rightarrow z_1=0.5+0.46984V\end{array}$

So, V is linear in z₁

  

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+85}$

  $\Rightarrow \ln P_1^v=6.0324$

$\Rightarrow P_1^v=416.71575$ mm Hg at $T=85°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+85}$

  $\Rightarrow \ln P_2^v=4.2041$

$\Rightarrow P_2^v=66.96039$ mm Hg at $T=85°C$

So, the expression for total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ in $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1\times 416.71575+(1-x_1)\times 66.96039\\ \Rightarrow P=66.96039+349.75536\times x_1\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $y_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=\frac{x_1\times 416.71575}{66.96039+349.75536\times x_1}$

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow 0.5=x_1\times (1-V)+(\frac{x_1\times 416.71575}{66.96039+349.75536\times x_1})\times V\end{array}$

  

  

(d)

Interpretation Introduction

Interpretation:

Assuming Raoult’s law is valid we need to find the values of liquid mole fraction of n-pentane(1) and vapor mole fraction of n-pentane(1) for a binary system of n-pentane(1) and n-heptane(2) for temperature 95°C and when total pressure $P=\frac{1}{2}(P_1^{sat}+P_2^{sat})$ or $P=\frac{1}{2}(P_1^v+P_2^v)$ . We have to also plot the fraction of system that is vapor V vs. overall composition z₁ for the binary system of n-pentane(1) and n-heptane(2) at these same conditions.

Assuming Raoult’s law is valid we need to plot the total pressure P, liquid mole fraction of n-pentane (1), x₁, vapor fraction of n-pentane (1), y₁ vs. fraction of system that is vapor V for the binary system of n-pentane(1) and n-heptane(2) at these same conditions at temperature $T=95°C$ and z₁=0.5.

Concept introduction:

Raoult’s Law states that the partial pressure of liquid A above the solution is equal to the mole fraction of the liquid in a solution times the partial pressure of the pure liquid. This holds for ideal solutions. An ideal solution is approached by binary solutions of molecules that have similar properties (e.g. benzene and toluene). The law is mathematically expressed as-

  $p_A^{*}=x_A{P}_A^v$

Where

  $P_A^v$ = vapor pressure of A at the given temperature

  $x_A$ = mole fraction of the solute A in the liquid

  $p_A^{*}$ = equilibrium partial pressure exerted by the solute.

A solution behaves ideally when

1. The solute and the solvent molecules have similar sizes and similar intermolecular forces,
2. The excess volume of mixing is zero, and
3. The heat of mixing is zero when both the solute and the solvent are liquids

Answer to Problem 13.4P

$x_1=0.5$, $y_1=0.852544$

V is linear in z₁

  

  

  

  

Explanation of Solution

Find the value of $x_1$ and $y_1$at $T=95°C$and $P=\frac{1}{2}(P_1^v+P_2^v)$

Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+95}$

  $\Rightarrow \ln P_1^v=6.0324$

$\Rightarrow P_1^v=527.90536$ mm Hg at $T=95°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+95}$

  $\Rightarrow \ln P_2^v=4.51422$

$\Rightarrow P_2^v=91.30635$ mm Hg at $T=95°C$

Therefore, total pressure, $P=\frac{1}{2}(P_1^v+P_2^v)=\frac{1}{2}(527.90536+91.30635)=309.60586$ mm Hg at $T=95°C$ .

By equating total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ and $P=\frac{1}{2}(P_1^v+P_2^v)$ we find $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1{P}_1^v+(1-x_1)P_2^v=\frac{1}{2}(P_1^v+P_2^v)\\ \Rightarrow x_1=0.5\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $\begin{array}{l}{y}_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=0.852544\\ \end{array}$

For the given temperature, z₁ ranges from the liquid composition at the bubble point to the vapor composition at the dew point.

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow z_1=0.5\times (1-V)+0.852544\times V\\ \Rightarrow z_1=0.5-0.5V+0.852544V\\ \Rightarrow z_1=0.5+0.35254V\end{array}$

So, V is linear in z₁

   Now, For n-pentane (1) $A\text{'}$ =13.7667, $B\text{'}$ =2451.88, $C\text{'}$ =232.014

And For n-heptane (2) $A\text{'}$ =13.8622, $B\text{'}$ =2911.26, $C\text{'}$ =216.432

Now vapor pressure for n-pentane (1) can be calculated using Antoine equation as follows: -

  $\ln P_1^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_1^v=13.7667-\frac{2451.88}{232.014+95}$

  $\Rightarrow \ln P_1^v=6.0324$

$\Rightarrow P_1^v=527.90536$ mm Hg at $T=95°C$

Now vapor pressure for n-heptane (2) can be calculated using Antoine equation as follows:

  $\ln P_2^v=A\text{'}-\frac{B\text{'}}{C\text{'}+\theta }$

  $\Rightarrow \ln P_2^v=13.8622-\frac{2911.26}{216.432+95}$

  $\Rightarrow \ln P_2^v=4.51422$

$\Rightarrow P_2^v=91.30635$ mm Hg at $T=95°C$

So, the expression for total pressure $P=p_A^{*}+p_B^{*}=x_A{P}_A^v+(1-x_A)P_B^v$ in $x_1$

  $\begin{array}{l}P=p_1^{*}+p_2^{*}=x_1\times 527.90536+(1-x_1)\times 91.30635\\ \Rightarrow P=91.30635+436.59901\times x_1\end{array}$

Using the formula $y_A=\frac{p_A^{*}}{P}=\frac{x_A{P}_A^v}{P}$

we find

  $y_1=\frac{p_1^{*}}{P}=\frac{x_1{P}_1^v}{P}=\frac{x_1\times 527.90536}{91.30635+436.59901\times x_1}$

Now, $\begin{array}{l}{z}_1=x_1\times (1-V)+y_1\times V\\ \Rightarrow 0.5=x_1\times (1-V)+(\frac{x_1\times 527.90536}{91.30635+436.59901\times x_1})\times V\end{array}$