Question

An apartment has the dimensions 20 m by 10 m by 7 m. The temperature is 30°C, and the relative humidity is 52 percent. What is the total mass (in kg) of water vapor in the air in the apartment?

Answer 1

Answer:

The mass of the water vapor in the apartment would be 22.1 kg

Explanation:

The ideal gas equation would be used to obtain the total mass and is expressed as;

P V = n R T  .............1

but n = m/M

Substituting into equation 1 we have;

P V = (m/M)RT

Making m the subject formula we have;

m = M P V / R T  ..............................3

Where M is the molar mass of water =  18 g/ mol

NB the pressure of water at $30^{0}$C  from the saturation table is 4250 Pa

P is the pressure at 52% relative humidity = 0.52 x 4250 =2210 Pa

V is the volume = 20 x 10 x 7 = 1400 $m^{2}$

T is the temperature, T = $30^{0}$C   + 273.15 = 303.15 K

R is the ideal gas constant = 8.3245 J/mol/K

m = (18 g/ mol x  2210 Pa x 1400 $m^{2}$ )/ (8314 J x 303.15 K)

m = 55692000 / 2520.3891

m =22096.58 g

m = 22096.58/1000

m = 22.1 kg

Therefore the mass of the water vapor in the apartment would be 22.1 kg

Answer 2

Answer:

Mass will be 1627.776 kg

Explanation:

We have given that dimension of apartment 20 m by 10 m by 7 m

So volume $V=20\times 10\times 7=1400m^3$

Temperature T = 30°C = 273+30 = 303 K

Molar mass of air = 29 g/mol

Form ideal gas we know that $pv=nRT$ .here p is pressure, v is volume , n is number of moles, R is gas constant and T is temperature

So number of moles $n=\frac{pv}{RT}=\frac{1.01\times 10^5\times 1400}{8.314\times 303}=56130.22moles$

We know that $number\ of\ moles=\frac{mass\ in\ gram}{molar\ mass}$

So mass in gram = 56130.22×29 = $1627776.38$ gram = 1627.776 kg