8. Compare the values obtained for the pressure(atm) of 2 mols O2 at 298.15K held in a8.25 dm bulb usinga) Redlick-Kwong equation: For O2: Te = 154.6 K; Pc = 50.43 barb) Beattie - Bridgeman equationFor O2, the Beattie - Bridgeman constants:Ao = 0.14911 Pa mé mol? Bo = 46.24 x 10 m molc = 4.80 x 10' m Kmol a= 25.62 x 104m mol b= 4.208 x 10 m molPa-mmole kUse R=8.314IVERSI

Moles of O2, n = 2 mol Temperature of O2, T = 298.15 K Volume of the bulb, V = 8.25 dm3 = 0.00825 m3

Answered: 8. Compare the values obtained for the…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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8. Compare the values obtained for the pressure(atm) of 2 mols O2 at 298.15K held in a
8.25 dm bulb using
a) Redlick-Kwong equation: For O2: Te = 154.6 K; Pc = 50.43 bar
b) Beattie - Bridgeman equation
For O2, the Beattie - Bridgeman constants:
Ao = 0.14911 Pa mé mol? Bo = 46.24 x 10 m mol
c = 4.80 x 10' m Kmol a= 25.62 x 104m mol b= 4.208 x 10 m mol
Pa-m
mole k
Use R=8.314
IVERSI

Expert Solution

Step 1

Moles of O₂, n = 2 mol

Temperature of O₂, T = 298.15 K

Volume of the bulb, V = 8.25 dm³ = 0.00825 m³

Step 2

(a)

For O₂, critical temperature and critical pressure are given as:

$\begin{matrix} T_{C} = 154.6 K \\ P_{C} = 50.43 bar \end{matrix}$

The Redlich-Kwong equation to be used is:

$\begin{matrix} P^{R K} = \frac{R T}{V_{m} - b} - \frac{a (T)}{V_{m} (V_{m} + b)} ...... (1) \\ Here, \\ a (T) = 0.42748 \frac{T_{r}^{- 1 / 2} R^{2} T_{C}^{2}}{P_{C}} ...... (2) \\ b = 0.08664 \frac{R T_{C}}{P_{C}} ...... (3) \end{matrix}$

Step 3

Calculate reduced temperature and molar volume as:

$\begin{matrix} T_{r} = \frac{T}{T_{C}} \\ = \frac{298.15 K}{154.6 K} \\ = 1.929 \\ V_{m} = \frac{V}{n} \\ = \frac{0.00825 m^{3}}{2 mol} \\ = 0.004125 \frac{m^{3}}{mol} \end{matrix}$

Now, use equations (2) and (3) to calculate the parameters a(T) and b respectively.

$\begin{matrix} a (T) = 0.42748 \frac{T_{r}^{- 1 / 2} R^{2} T_{C}^{2}}{P_{C}} \\ = 0.42748 \frac{{(1.929)}^{- 1 / 2} {(8.314 \frac{Pa \cdot m^{3}}{mol \cdot K})}^{2} {(154.6 K)}^{2}}{50.43 bar \times \frac{10^{5} Pa}{bar}} \\ = 0.1008 \frac{Pa \cdot m^{6}}{{mol}^{2}} \\ b = 0.08664 \frac{R T_{C}}{P_{C}} \\ = 0.08664 \frac{(8.314 \frac{Pa \cdot m^{3}}{mol \cdot K}) (154.6 K)}{50.43 bar \times \frac{10^{5} Pa}{bar}} \\ = 2.2083 \times 10^{- 5} \frac{m^{3}}{mol} \end{matrix}$

Step 4

Now, use these values in equation (1) and calculate the pressure of the gas in the bulb as:

$\begin{matrix} P^{R K} = \frac{R T}{V_{m} - b} - \frac{a (T)}{V_{m} (V_{m} + b)} \\ P^{R K} = \frac{(8.314 \frac{Pa \cdot m^{3}}{mol \cdot K}) (298.15 K)}{(0.004125 \frac{m^{3}}{mol}) - (2.2083 \times 10^{- 5} \frac{m^{3}}{mol})} - \frac{0.1008 \frac{Pa \cdot m^{6}}{{mol}^{2}}}{(0.004125 \frac{m^{3}}{mol}) ((0.004125 \frac{m^{3}}{mol}) + (2.2083 \times 10^{- 5} \frac{m^{3}}{mol}))} \\ P^{R K} = 598267.76 Pa \\ P^{R K} = 5.98 bar \end{matrix}$

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