Question

A room of dimensions 10m x 8m x 5m is containing the bucket of capacity 20L full water. Calculate the water remaining in the bucket if room temperature is 250 Celsius and vapour pressure of water at 250 Celsius is 45mm Hg

Answer 1

Answer : The amount of water remaining in the bucket is 2.6 L

Explanation :

Let us find the volume of the room first.

Volume of the room = 10 m x 8 m x 5 m = 400 m³.

Let us convert the above volume from m³ to L.

The conversion factor is 1 m³ = 1000 L

$400 m^{3}\times \frac{1000L}{1m^{3}}= 400000L$

Therefore we have V = 400000L

Vapour pressure of water is given as 45 mmHg .

Let us convert this to atm unit.

The conversion factor is 1 atm = 760 mmHg.

$45 mmHg\times \frac{1atm}{760mmHg}= 0.0592 atm$

We have P = 0.0592 atm

R = 0.0821 L-atm/mol K

T = 25 + 273 = 298 K

The ideal gas equation can be written as,

$PV = nRT$

Let us rearrange this equation to solve for number of moles of vapours (n).

$n = \frac{(PV)}{(RT)}$

$n = \frac{(0.0592atm)(400000L)}{(0.0821L.atm/mol K)(298K)}$

$n = \frac{(23680)}{(24.47)}$

$n = 968 mol$

968 moles of water are present in the air.

Let us convert this to L of water.

We will use molar mass of water which is 18.02 g/mol and density of water which is 1g/mL as our conversion factors.

The set up given below.

$968mol\times \frac{18.02g}{mol}\times \frac{1mL}{1g}\times \frac{1L}{1000mL}= 17.4L$

17.4 L of water would vaporize to saturate the room at the given temperature.

The amount of water left in the bucket is 20 L - 17.4 L = 2.6 L

Answer 2

Answer:

hi

Explanation:

yes the above one is correct