Chapter 1, Problem 1.2P

Interpretation Introduction

Interpretation:

The relation between the parameters of Antoine equation which is written in two different forms should be determined.

Answer to Problem 1.2P

The relation between the parameters of Antoine equation which is written in two different forms is:

A = 2.303a
B = 2.303b
C = c-273.15

And, ${\log }_{10}^{psat/kpa}=\frac{1}{0.88}{\log }_{10}^{psat/torr}$

Explanation of Solution

Given information:

Antoine equation can be expressed in following two forms-

$\log 10P^{sat}/torr=\frac{a-b}{[t/°C+c]}$ ......... (1)

And
$\ln P^{sat}/kPa=\frac{A-B}{[t/K+C]}$ .......... (2)

Where $P^{sat}$ − Liquid/vapor saturation pressure

And a, b, c, A, B, C are substance specific constants

• Antoine Equation relates the Vapor pressure of the substance to the temperature.
• The given two forms of equation differ in terms of units of pressure and temperature. Hence to find the relation b/w a, b, c and A, B, C, we must convert the equation1 so that it has the same unit as that of equation 2.

Calculation:

We have equation 1 as-

  $\log 10P^{sat}/torr=\frac{a-b}{[t/°C+c]}$ ......... (1)

• Consider Left hand side of equation (1)- First convert log₁₀into ln by multiplying equation (1) by 2.303:

  $2.303\log 10P^{sat}/torr=\ln P^{sat}/torr$

• Now convert the unit of Pressure from torr to KPa:

1 torr = 133.32 Pa = $133.32\times {10}^{-3}KPa$

  $\begin{array}{l}2.303\log 10P^{sat}/torr=\ln P^{sat}/torr\\ \text{ = ln }{P}^{sat}/(torr\times \frac{133.32\times {10}^{-3}\text{ KPa}}{1torr})\\ \text{ = ln }{P}^{sat}/133.32\times {10}^{-3}\text{ KPa }\\ \\ \end{array}$

Therefore,

  $\begin{array}{l}2.303\log 10P^{sat}/torr=\text{ ln }{P}^{sat}/133.32\times {10}^{-3}\text{ KPa }\\ \text{ = ln (7}\text{.50 }{P}^{sat})/KPa\\ \end{array}$

Now we know that, $\ln (ab)=\ln a+\ln b$

Thus,

  $\begin{array}{l}\text{ ln (7}\text{.50 }{P}^{sat}/KPa)=\ln (7.50)+\ln (P^{sat}/KPa)\\ \text{ = 0}\text{.88+ln}(P^{sat}/KPa)\\ \end{array}$

Finally,

  $\begin{array}{l}2.303\log 10P^{sat}/torr=0.88+\ln P^{sat}/KPa\\ \therefore \log 10P^{sat}/torr=\frac{1}{2.303}(0.88+\ln P^{sat}/KPa)\mathrm{......}\text{ (3)}\end{array}$

• Now consider the right hand side of the equation (1) and convert the temperature into Kelvin(K)

  $(T-273.15)/K=t/°C$

  $\begin{array}{l}\therefore \frac{a-b}{[t/°C+c]}=\frac{a-b}{[(T-273.15)/K+c]}\\ \text{ =}\frac{a-b}{[T/K+c-273.15]}\mathrm{......}\text{(4)}\end{array}$

Thus, putting equation (3) and (4) in equation (1), we get

  $\begin{array}{l}\frac{1}{2.303}(0.88+\ln P^{sat}/KPa)=\frac{a-b}{[T/K+c-273.15]}\\ (0.88+\ln P^{sat}/KPa)=2.303(\frac{a-b}{[T/K+c-273.15]})\\ \text{ =}\frac{2.303a-2.303b}{[T/K+c-273.15]}\mathrm{......}\text{(5)}\end{array}$

Hence, by comparing equation (2) and equation (5), we can relate a, b, c and A,B,C as:

A = 2.303a
B = 2.303b
C = c-273.15

And, ${\log }_{10}^{psat/kpa}=\frac{1}{0.88}{\log }_{10}^{psat/torr}$

Conclusion

The relation between the parameters of Antoine equation which is written in two different forms is:

A = 2.303a
B = 2.303b
C = c-273.15

And, ${\log }_{10}^{psat/kpa}=\frac{1}{0.88}{\log }_{10}^{psat/torr}$