Answer :
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!