log(P) = A - B / (T + C)
where P is the vapor pressure in mmHg, T is the temperature in degrees Celsius, and A, B, and C are constants specific to the liquid. For water, the constants are A = 8.07131, B = 1730.63, and C = 233.426.
We are given that the vapor pressure of the water is 380 torr. To convert this to mmHg, we divide by 1.333, as 1 torr is equivalent to 1.333 mmHg. This gives us a vapor pressure of 380 torr / 1.333 mmHg = 285 mmHg.
We can now substitute this value into the Antoine equation and solve for T:
log(285) = 8.07131 - 1730.63 / (T + 233.426)
2.4541 = 8.07131 - 1730.63 / (T + 233.426)
-5.6172 = -1730.63 / (T + 233.426)
T + 233.426 = -1730.63 / -5.6172
T + 233.426 = 308.224
T = 308.224 - 233.426
T = 74.798 °C
Therefore, the temperature of the boiling water is approximately 74.798 °C.