The vapor pressure of ethanol is 400. mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is the vapor pressure of ethanol, in mmhg, at 34.9°c? (r = 8.314 j/k • mol)

The vapor pressure at the new temperature is 352 mmHg.

We can obtain the vapor pressure at different temperatures using the formula;  ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)).

P1 = initial pressure

P2 = final pressure

ΔHvap = enthalpy of vaporization

R = gas constant

T1 = initial temperature

T2= final temperature

Substituting values;

ln(400/P2) = (39.3 × 10^3 J/mol/ 8.314 J/K.mol) (1/308 - 1/337)

ln(400/P2) = 0.128

(400/P2) = e^0.128

P2 = 400/e^0.128

P2 = 352 mmHg

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100 mm Hg

Explanation: -

P 1 =400 mm Hg

T 1 = 63.5 C + 273 = 336.5 K

T 2 = 34.9 C + 273 = 307.9 K

ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹

R = 8.314 J ⁻¹K mol⁻¹

Now using the Clausius Clapeyron equation

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

Plugging in the values

ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x ( -

= 1.38

P₂ = 100 mm Hg

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