AR$17$ A precise table or diagram is essential. Deliver a perfect solution urgently. Strict warning: AI tools are prohibited. "Molecular Thermodynamics Time: $65$ minutes

Open book, open notes; $3$ problems.

Name:

$1.$ Vapor pressures of benzene $(1)$ and toluene $(2)$ are given by: In p$.$m $($k$*$p$\_\{$a$\})=13\mathrm{.}7819-2726\mathrm{.}81/($t $+217\mathrm{.}572)$

benzene:

toluene:

i$\_\{$n$\}*$p$,$ wt$($k$*$p$\_\{$a$\})=13\mathrm{.}932-3056\mathrm{.}96(1+217\mathrm{.}625)$

where t is in centigrade. Calculate:

a$.$ P and y$\_\{1\}$ for x$\_\{1\}=0\mathrm{.}33$ and t$-100$C b$.$ t and xi for y$\_\{1\}=0\mathrm{.}33$ and P $=120$kPa

$2.$ A liquid mixture of species $1$ and $2$ for which x$\_\{1\}=0\mathrm{.}6$ is in equilibrium with its vapor at $144$ deg $*$ C Determine numerical values for the equilibrium pressure P and vapor composition y$\_\{1\}$ using the following information for this system:

In y$\_\{1\}=$ A $*$ x$\_\{2\}^2$ and In gamma$\_\{2\}=$ A $*$ x$\_\{1\}^2$ represent the vapor$-$liquid

constant.

At $144$ deg $*$ C P$=75\mathrm{.}20$ kPa and Pat $=31\mathrm{.}66$ kPa

equilibrium well. Here, A is a

The system forms an azeotrope at $144$C for which the azeotropic mole fraction of species

$1,$ is x$\_\{1\}=$ y$\_\{1\}=0\mathrm{.}294$ For a binary system in VLE, azeotrope means the vapor and liquid phase mole fractions are the same, as given to be $0\mathrm{.}294.$

$3.$ For the ammonia synthesis reaction written:

$1/9*$ N$\_\{2\}($g$)+3/2*$ H$\_\{2\}($g$)->$ N$*$H$\_\{3\}($g$)$

with $0\mathrm{.}5$ mol N$\_\{2\}$ and $1\mathrm{.}5$ mol H$2$ as the initial amounts of reactants and with the assumption that the equilibrium mixture is an ideal gas, show that:

epsilon $=1-(1+1\mathrm{.}299$K $*$ p$/(1$bar$))^(-1/2)"$