did i do this right?

Homework Question 03 P,= 1 atm = 760 torr T. - 1860K A Antimony has a normal boiling point of 1860 K and AH vap = 77.14 kJ/mol. What happens to the vapor pressure of antimony when the temperature is increased to 1950 K? P2 = ? T2 = 1950 K (A) The vapor pressure exceeds 760 torr. B. The vapor pressure is 760 torr. C. The vapor pressure falls between 760 and 380 torr. D. The vapor pressure falls below 380 torr. In(o atm 77.14 kJ/mol 8.314 x 10-3 1950 K In Pz - lip (1) = 0.230231407 COD + InxD) + In(1) In P2 = 0.230231407 0.2302 31407 e = 1.26atm = 958 torr - ( 1860k ) )