Question

Consider the following data for a substance:

Critical point: 116.9 atm and 549°C

Triple point: 0.101 atm and 110°C

Normal melting point: 1.00 atm and 124°C

What is the normal boiling point of the substance?

Enter your answer in units of °C to three significant figures.

Answer 1

The normal boiling point of a substance is its boiling point at normal atmospheric pressure = 1 ATM

The normal boiling point, critical point and triple point are points in the vaporization curve of a substance.

Any point in the vaporization curve of a substance is given by the Clausius-Clapeyron equation.

It is

Here, it is given that P = 116.9 atm at T = 549 C = 822 K

And P = 0.101 atm at T = 110 C = 383 K

We ahve to find T at P = 1 atm.

Using the first two values,

Using this,