Chapter 16, Problem 148CP

Interpretation Introduction

Interpretation:

The minimum volume of the container needs to be determined.

Concept introduction:

The density is calculated as follows:

  $d=\frac{m}{V}$

Here, m is mass and V is volume.

The ideal gas equation is given below:

  $PV=nRT$

Here,

P = pressure

V = volume

n = number of moles

R = gas constant

T = Temperature

Answer to Problem 148CP

Minimum volume of the container = 5779 feet³

The cube dimensions are as follows:

a = 17.95 feet

b = 17.95 feet

c = 17.95 feet

Explanation of Solution

The volume of water can be converted into gallons as follows:

  $\begin{array}{l}Volumeofwater=1gallon\times \left(\frac{3785c{m}^3}{1gallon}\right)\\ =3785c{m}^3\end{array}$

Density of water = 0.998 g/cm³

Temperature of water = 25ºC

Now, calculate the moles of water in 1 gallon as:

  $\begin{array}{l}Density=\frac{mass}{volume}\hfill \\ \begin{array}{l}massofwater=density\times volume\\ =0.998g/c{m}^3\times 3785c{m}^3\\ =3777.4gofwater\end{array}\hfill \\ \hfill \\ \begin{array}{l}molesofwater=\frac{massofwater}{molarmassofwater}\\ =\frac{3777.4g}{18.02g/mol}\\ =209.62molofwater\end{array}\hfill \\ \hfill \end{array}$

Next step is to determine the minimum volume of the container.

The volume of the water vapour is equal to the volume of the container

  $PV=nRT$

P = vapour pressure of water at 25ºC

n = 209.62 mol of water

R = 0.082 atm L/K mol

T = 25ºC + 273 = 298 K

  $\begin{array}{l}vapourpressureofwaterat25ºC=23.8torr\times \left(\frac{1atm}{760torr}\right)\\ =0.0313atm\end{array}$

Now rearrange the above equation to get,

  $\begin{array}{l}V=nRT/P\\ =\frac{\left(209.62mol\right)\left(0.082atmL/Kmol\right)\left(298K\right)}{0.0313atm}\\ =163651L\times \left(\frac{1fee{t}^3}{28.3168L}\right)\\ =5779fee{t}^3\end{array}$

Now we determine the cube dimensions.

Volume of the cube = a x b x c

Since each side of a cube is the same: volume of the cube = $a\times a\times a=a^3$

  $\begin{array}{l}a=\sqrt[3]{\left(volumeofthecube\right)}\\ =\sqrt[3]{\left(5779fee{t}^3\right)}\\ =17.95feet\end{array}$

Cube dimensions are as follows:

a = 17.95 feet

b = 17.95 feet

c = 17.95 feet