INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
Question
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Chapter 3, Problem 3.60P
Interpretation Introduction

Interpretation:

To find the mass of n -butane vapors in the tank which are in equilibrium with 14m3 of its n -butane liquid in 30m3 tank at 25oC . The vapor pressure of n -butane at given temperature is 2.43bar .

Concept Introduction:

The molar volume of butane vapor in vessel is given as:

  V=Vvaporn

The mass of n -butanevapours present in tank is:

  m=n×Mm=VvaporV×M

Where n is number of moles of butane vapor present in tank, M is molecular weight of butane.

The V is molar volume and Vvapor is total volume of vapors in tank.

The molar volume of butane vapor in vessel is also calculated from generalized correlations for gases given by Pitzer at given conditions. The expression is:

Z=1+B0PrTr+ωB1PrTr where PV=ZRT........(1)

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Chapter 3 Solutions

INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<

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