At 50.14 k a substance has a vapor pressure of 258.9 torr. its heat of vaporization in kJ/mol it if has a vapor pressure of 161.2 torr at 277.5 k is
given that :
temperature T1 = 50.14 K
pressure P1 = 258.9 torr
temperature T2 = 277.5 K
pressure P2 = 161.2 torr
using the Clausius Clapeyron , we get :
ln ( P1 / P2 ) = - ΔH / R (1/ T1 - 1/T2)
ln ( 258.9 / 161.2 ) = - ΔH / 8.314 ( 1/ 50.14 - 1 / 277.5 )
ΔH = 41.83 kJ/ mol
The heat of vaporization ΔH = 41.83 kJ/ mol.
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