Question

The normal boiling point of ethanol is 78.3 ∘C, and the heat of vaporization is ΔHvap = 38.6 kJ/mol.

Part A

What is the boiling point of ethanol in ∘C on top of Pikes Peak in Colorado, where P = 401 mmHg?

Express your answer with the appropriate units.

Answer 1

Part A:

Given data ,

P2 = 401 mmHg

T1 = 78.3⁰ C + 273 = 351.3 K

P1 = 1 atm = 760 mmHg

H_vap = 38.6 kJ / mol = 38600 J/ mol

Let us consider a Clausius - Claperyon equation

ln(P2 / P1) = (-H_vap / R) x (1 / T2 - 1 / T1)

ln (401 / 760) =( -38600 / 8.314) x (1 / T2 - 1 / 351.3 )

- 0.6462 = - 4642.77 x (1 / T2 - 1 / 351.3  )

0.0001391 = (1 / T2 - 1 / 351.3  )

0.0001391 = 1 / T2 - 0.0028465

0.0029856 = 1 / T2

T2 = 335 K - 273

= 62⁰ C

Therefore the new boiling point is 62⁰ C