Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. when the flask contents are at 58.3°c, the height of the mercury in the arm of the manometer connected to the flask is 747 mm, and that in the arm open to the atmosphere is 52 mm. at 110°c, the mercury level is 577 mm in the arm connected to the flask, and 222 mm in the other arm. atmospheric pressure is 755 mmhg. (a) using the data given, find δhv and b in the clausius-clapeyron equation. (b) air saturated with chlorobenzene at 130°c and 101.3kpa (absolute) is cooled to 58.3°c at constant pressure. estimate the percentage of the chlorobenzene originally in the vapor that condenses. (hint: draw a flowchart with separate vapor and liquid streams)

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

= 58.3°C

= 747mmHg

= 52mmHg

= 110°C

= 577mmHg

= 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

= 58.3°C + 273.2

= 331.5k

= 110°C + 273.2

= 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

=

The vapor pressure under the first set of conditions is:

= 755mmHg - (747mmHg - 52mmHg)

= 60mmHg

The vapor pressure under the second set of conditions is:

= 755mmHg - (577mmHg = 222mmHg)

= 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

where  (delta) H = ΔH

The slope of the

Calculating the slope; we have:

=

= -4661.4k

= 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B =

B =

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P =

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature  of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

=

= 6.595

=

= 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

= 731.44mmHg

= 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

At inlet conditions, the mole fraction of chlorobenzene is:

At outlet conditions , the mole fraction of chlorobenzene is:

Since there is no reaction, the total balance around the condensation is :

Let assume, that 100 moles of liquid chlorobenzene (CB) is condensed, therefore the equation becomes:

+ 100mol

The chlorobenzene balance using the mole fractions calculated above is :

substituting equation (1) into equation above; we have:

we can solve for n1, i.e ;

n1 = 104.3mol total air

Therefore the moles of chlorobenzene that will produce 100 moles of CB liquid is:

= 100.34mol CB

Now, calculating the percentage of chlorobenzene that condenses: we have;

% CB Condensation =  

% CB Condensation = 99.7%

The percentage of chlorobenzene originally in the vapor that condenses is 99.7%

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

the answer is option a.