Answer:
ΔHvap = 1.796 x 10⁵ J = 179.6 kJ
Tb = 1737 ºC
Explanation:
The expression we should use to solve this equation is given by
ln p = - ΔHvap/R x (1/T) +c
where ln p is the natural log of the pressues, ΔHvap is the enthalpy of vaporization, R is the gas constant (8.314 J/Kmol), T is the temperature and c is a constant. This is called the Clausius-Clayperon equation for the temperature dependence of pressure.
Note this equation has a form
y= mx + b
where m the slope is equal to - ΔHvap/R.
We have the values of T and P, therefore, we can determine the slope of these data to answer our question.
T(K) 1/T (K⁻¹) P (torr) lnp
1500 6.67 x 10⁻⁴ 19.72 2.98
1600 6.25 x 10⁻⁴ 48.48 3.88
1700 5.88 x 10⁻⁴ 107.2 4.67
1800 5.56 x 10⁻⁴ 217.7 5.38
1900 5.26 x 10⁻⁴ 408.2 6.01
Using a comercial software or calculating by hand the best slope, one is able to determine the desired values. In excel we get the equation of line:
y = -21598x + 17.38
with R²= 1
Therefore
- ( 21598 ) = -ΔHvap/ 8.314
⇒ ΔHvap = 1.796 x 10⁵ J = 179.6 kJ
and to calculate the boiling point we will use the equation for the best line just determined with p = 760 torr
ln p = -21598x + 17.38
where p is the pressure , and Tb is the normal boiling point:
ln ( 760 ) = -21598 ( 1/Tb ) + 17.38
Tb = 2009 K = 1737 ºC
