Chapter 4, Problem 4.10P

Interpretation Introduction

(a)

Interpretation:

The latent heat of vaporization for one of the compounds namely, Acetone in this case should be calculated using the equation 4.13 and the calculated value should be compared with the values in table B.2.

Concept Introduction:

The latent heat of vaporization can be calculated as follows:

  $\frac{\Delta H_n}{R{T}_n}=\frac{1.092\times (ln{P}_c-1.013)}{0.93-T_{rn}}$

Here, $\Delta H_n$ is latent heat of vaporization, R is Universal gas constant, $P$ is pressure and T is temperature.

Interpretation Introduction

(b)

Interpretation:

To find the heat of vaporization for one of the compounds of the four, take n-pentane

Concept Introduction:

Heat of vaporisation $\Delta H_{2calc}$ is given as

  $\frac{\Delta H_{2calc}}{\Delta H_1}=\left({\left.\frac{1-T_{r2}}{1-T_{r1}}\right)}^{0.38}\right.$

Where,

$T_{r2}=\frac{T_2}{T_c}$

$T_{r1}=\frac{T_1}{T_c}$

Latent heat of vaporisation at normal boiling point, 25.79 kJ/mol =

  $\Delta H_{2calc}$ = Latent heat of vaporisation at 25^o C (in J/mol)

  $\Delta H_2=m\times \Delta H_{25}$

where, m = molecular weight = 72.15 g/mol

  $\Delta H_{25}$ = Latent heat of vaporisation at 25 ^oC (in kJ/g) = 366.3 J/g = 0.3663 kJ/g

  $\Delta H_2$ = Latent heat of vaporisation at 25 ^oC (in kJ/mol)