Question

Enter your answer in the provided box. The vapor pressure of ethanol is 1.00 x 10² mmHg at 34.90°C. What is its vapor pressure at 54.81°C? ( AH yap for ethanol is 39.3 kJ/mol.) mmHg

Answer 1

Answer:

2,54x10² mmHg

Explanation:

To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:

$Ln(\frac{P_{2}}{P_{1}} ) =\frac{ deltaH_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )$

Where:

P1 is 1,00x10² mmHg

ΔHvap is 39,3 kJ/mol

R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹

T1 is 34,90°C + 273,15 = 308,05 K

T2 is 54,81°C + 273,15 = 327,96 K

Thus:

$Ln(\frac{P_{2}}{1,0x10^{2}mmHg}) =\frac{39,3kJ/mol}{8,314x10^{-3}kJ/molK} (\frac{1}{308,05K}-\frac{1}{327,96K} )$

Thus, P2 is 2,54x10² mmHg

I hope it helps!