**1. The problem statement, all variables and given/known data**

A vessel of 0.3m

^{3}capacity contains a mixture of air and steam which is 0.75 dry. If the pressure is 7 bar and temperature is 116.9 degree Celsius, find: Mass of water present, mass of dry saturated vapour, mass of air.

Answer key: 0.102kg, 0.307kg, 1.394kg

**2. Relevant equations**

Relative humidity = mv/mg = Pv/Pg (pressure of vapour/ pressure of dry saturated vapour)

humidity ratio = mv/ma

Moist Volume *V* = total volume/mass of dry air

**3. The attempt at a solution**

I was able to determine the mass of dry saturated vapour. The moist volume (*V*) of dry saturated vapour from the steam table is 0.9774m^{3}/kg and hence:

0.9974 = volume/mass of dry saturated vapour

0.9974 = 0.3/mg

mg = 0.3069kg

I have absolutely no idea however, on how to find the other two masses. I don’t understand the term 0.75 dry, I don’t know how to find the mass of water (is this mass of water vapour???) without relative humidity. I’d appreciate any help. Thank you very much.

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