A certain substance has a heat of vaporization of 41.25 kJ/mol.41.25 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.505.50times higher than it was at 309 K?
T1 = 309 K
P2 = 5.50 x atm
P1 = x
Hvap= 41.25
kJ/mol
R = 8.314 x 10^-3 kJ/K mol
from clausius -clayperon equation
ln(P2/P1) = ( Hvap/ R )
[1/T1 - 1/T2]
ln (5.50 / 1) = 41.25 / 8.314 x 10^-3 [1 / 309 - 1/ T2]
T2 = 345.7 K
temperature = 346 K
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