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The vapor pressure of ethanol is [tex]1.00 \times 10^2 \, \text{mmHg}[/tex] at [tex]34.90^\circ\text{C}[/tex]. What is its vapor pressure at [tex]54.81^\circ\text{C}[/tex]? (The enthalpy of vaporization for ethanol is [tex]39.3 \, \text{kJ/mol}[/tex].)

mmHg

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Answer :

jufsanabriasa jufsanabriasa · Answer 1 · 2024-02-16T21:53:20-05:00

Answer:

2,54x10² mmHg

Explanation:

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

[tex]Ln(\frac{P_{2}}{P_{1}} ) =\frac{ deltaH_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )[/tex]

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:

[tex]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} )[/tex]

Thus, P2 is 2,54x10² mmHg

I hope it helps!