Question

In a dew point apparatus, a metal beaker is cooled by gradually adding ice water to the water initially at room temperature. The moisture from the room air begins to condense on the beaker when its temperature is ${12.8}^{\circ }C.$ If the room temperature is ${21}^{\circ }C$ and the barometric pressure is $1.01325$ bar, determine the parts by mass of water vapor in the room air?
$(at{12.8}^{\circ }C{P}_{sat}=1.479kN/m^2)$

• A. $0.00913$
• B. $0.007214$
• C. $0.00863$
• D. $0.008714$

Answer 1

The correct option is A $0.00913$

$\text{Partial pressure of water vapour at DPT of}{12.8}^{\circ }C=P_v=1.479kN/m^2\text{Partial pressure of dry air}{P}_a=101.325-1.479=99.846kN/m^2\text{Specific humidity}\omega =\frac{m_v}{m_a}=0.622\times \frac{P_v}{P_a}=0.622\times \frac{1.479}{99.846}=0.009214\frac{kgw.v}{kgd.a}\text{Parts by mass of water vapour in room air is}=\frac{m_v}{m}=\frac{\omega }{1+\omega }\text{where, m = mass of mixture}\because (m=m_a+m_v)\frac{m_v}{m}=\frac{0.009214}{1.009214}=0.00913\frac{kgw.v}{kgmixture}$