Question

One stormy night, the temp. is 23.5^oC and atmospheric pressure is 755 mmHg. Further, the relative humidity of air was measured to be 85.6%. Determine the following:

1. Molar composition of the air in mole fractions
2. Dew point temperature (T_dew)

Answer in details. Draw a diagram if necessary.

Answer 1

Temperature = 23.5 ⁰C.

atmospheric pressure P = 755 mmHg.

Relative humidity = 85.6 % = 0.856

the vapor pressure of water at 23.5 ⁰C = 21.647 mmHg.

relative humidity is the ratio of partial pressure of water vapor present in air at that temperature to the vapor

pressure of water vapor present at that temperature.

Partial pressure of water vapor at 23.5 ⁰C = vapor pressure x relative humidity = 0.856 x 21.647 = 18.53 mmHg.

mol fraction of water vapor in air = 18.53 / 755 = 0.02454

the maximum mol fraction of water vapor in air at 23.5 ⁰C = 21.647 / 755 = 0.0286

this means the air at that temperature can bear 2.86 % of water vapor (mol %) at 23.5 ⁰C.

molar composition :

water vapor = 0.02454

air = 1 - 0.02454 = 0.97546

dew point is the temperature, where the air gets saturated and cannot hold any water vapor beyond it contains.

at this temperature, the water vapor already present in the air will be the maximum water vapor that the air can

bearable.

Thaomon - 2.45%, Air P = 755 mmity T= 23.5°C R.H=0.856 Yo = 0.0286 = 2.86% (menimum hearable mol fraction) 44,0 Cactual) = 2.45% Reduce the temperature, the manimum bearable fraction decreases from Air P=755 mmH 2.86% to 2:46% dew Point T= 20.95 °C Condition g furthur reduction of temperature below dewpoint! falater vapor present in air starts Condensing.

at this temperature, partial pressure of vapor present = vapor pressure of water.

partial pressure of water vapor present in air = 18.53 mmHg.

from antonie equation, find the temperature by substituting this as vapor pressure.

log(psat) = 8.07131 - 1730.63 T(°C) + 233.426-

log(18.53) = 8.07131 1730.63 T(°C) + 233.426

find T from the above equation

T = 20.95 ⁰C.

dew point = 20.95 ⁰C.