Answer to Question #125254 in General Chemistry for Jay Shendurnikar

Question

The relative humidity of the humid air is 51.2% at constant pressure of 760 mm of Hg and temperature of 50 °C. Calculate the dew point of the mixture and the mole fraction of water vapor. Antoine’s equation for vapour pressure is as follows: ln (vp in mm of hg) = 18.3036 - 38.1644/T (ln k) - 46.13

Answer 1

An equation that is frequently employed to determine dew point according to the T and RH, is as follows:

T_dew = (237.3 × [ln(RH/100) + ((17.27×T) / (237.3+T) )]) / (17.27 - [ln(RH/100) + ((17.27×T)/(237.3+T) )])

Where:

T_dew = dew point temperature in degrees Celsius (°C),

T = air temperature in degrees Celsius (°C),

RH = relative humidity (%),

ln = natural logarithm.

T_dew = (237.3 × [ln(51.2/100) + ((17.27×50) / (237.3+50))]) / (17.27 – [ln(51.2/100) + ((17.27×50) / (237.3+50))])

T_dew = (237.3 × [-0.6694 + (863.5/287.3)]) / (17.27 – [-0.6694 + (863.5/287.3)])

T_dew = (237.3 × [-0.6694 + 3.0055]) / (17.27 – [-0.6694 + 3.0055])

T_dew = (237.3 × 2.336) / (17.27 – 2.3361)

T_dew = 554.35 / 14.9339 = 37.1 °C