Question

The relative humidity of air equals the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature, times 100%.% and its temperature is 68 oF , how many moles of water are present in a room measuring 12 ft x 10 ft x 8 ft?

Answer 1

Question:

The relative humidity of air equals the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature times 100%. If the relative humidity of the air is 58% and its temperature is 68 oF , how many moles of water are present in a room measuring 12 ft x 10 ft x 8 ft?

Answer:

15.1379 moles  of water are present in a room measuring 12 ft x 10 ft x 8 ft.

Explanation:

Given:

Relative humidity of the air = 58%

Temperature = $68^{\circ}F$

To Find:

Number of moles of water are present in a room measuring 12 ft x 10 ft x 8 ft = ?

Solution:

Converting temperature Fahrenheit to Celsius

$T_({\circ}_C) = (T_({\circ}_F) - 32) \times \frac{5}{9}$

$T_({\circ}_C) = (68 - 32) \times \frac{5}{9}$

$T_({\circ}_C) = (36) \times \frac{5}{9}$

$T_({\circ}_C) = \frac{180}{9}$

$T_({\circ}_C) = 20 ^{\circ}$

The vapour pressure of $H_2O$ is equal to 17.54 torr

vapour pressure of air

= $\frac{58}{100} \times 17.54$ torr

= 10.17132 torr

Volume of the room = $Length \times Breadth\times Height$

=> $12 \times 10 \times 8$

=> $960 ft^3$

Using the ideal gas equation

PV = nRT

$1 ft^3 = 28.3168 L$

Volume in Litres = 27184.2 L

Pressure in atm

= $\frac{10.17132}{760}$

= 0.0133945 atm

Now substituting the values we get

Pv = nRT

$(0.0133945)(27184.2) = n \times 0.0821 \times (273+20)$

$364.118767 = n \times 0.0821 \times (293)$

$364.118767 = n \times 24.0533$

$n= \frac{364.118767}{24.0533}$

n = 15.1379 moles