Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 oC. At what temperature does benzene boil when the external pressure is 490 torrtorr?

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

wher is the rest of the question