Answer:
THE TEMPERATURE AT THE EXTERNAL PRESSURE OF 490 TORR IS 550.72 C
Explanation:
Heat of vaporization = 30.72 kj/mol
Normal boiling point = 80.1 C
Pressure = 490 torr
So therefore, with the above variables given, the relationship between heat of vaporization, pressure and temperature is given by this equation:
ln (P2/P1) = - ^H vap/ R ( 1/ T2 - 1/T1)
H vap = 30.72 kj/mol
P1 = 760 torr ( this is so because the normal boiling point occurs at the vapor pressure equal to the atmospheric pressure which value is 760 torr)
T1 = 80.1 + 273 = 353.1 K
P2 = 490 torr
R = 8.314 J/mol K
T2 = unknown
Re-arranging the formula bu=y making T2 the subject of the equation, we have;
T2 = 1 / (1/T1) - (R / Hvap) (ln (P2/P1))
Introducing the variables into the equation, we have:
T2 = 1 / (1 / 353.1 ) - (8.341 / 30.72 * 10 ^3) ln (490 / 760)
T2 = 1/ (0.00283 - 0.27* 10^-3 ln 0.6447)
T2 = 1 / 0.00283 - 0.27*10^-3 (- 0.43897)
T2 = 1 / 0.00283 + 0.1185 * 10 ^-3
T2 = 1 / 0.1214 *10^-3
T2 = 1 / 0.001214
T2 = 823.72
T2 = 823.72 - 273 = 550.72 C
The temperature when the external pressure is 490 torr is 550.72 C
