Chapter 12, Problem 12.136P

(a)

Interpretation Introduction

Interpretation:

The mass of water that must be removed every time the inside air is completely replaced with outside air is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

  $PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

The conversion factor to convert ${\text{m}}^3$ to $\text{L}$ is as follows:

  $1{\text{m}}^3=1000\text{L}$

Answer to Problem 12.136P

The mass of water that must be removed is $49.3\text{tons}$.

Explanation of Solution

Substitute $22°\text{C}$ in equation (2) to calculate the temperature $T$ in Kelvin as follows:

  $\begin{array}{c}T\left(\text{K}\right)=22°\text{C}+273.2\\ =295.2\text{ K}\end{array}$

Rearrange the equation (1) to calculate the number of moles at $31\text{torr}$.

  $n_1=\frac{PV}{RT}$        (3)

Substitute the value $31\text{torr}$ for $P$, $295.2\text{ K}$ for $T$, $2.4\times {10}^6{\text{m}}^3$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (3).

  $\begin{array}{c}{n}_1=\frac{\left(31\text{torr}\right)\left(\frac{\text{1}\text{atm}}{760\text{torr}}\right)\left(2.4\times {10}^6{\text{m}}^3\right)\left(\frac{1\text{L}}{{10}^{-3}{\text{m}}^3}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(295.2\text{ K}\right)}\\ =4,039,244.229\text{mol}\end{array}$

The formula to calculate mass at $31\text{torr}$ is as follows:

  $m_1=\left(n_1\right)\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$        (4)

Substitute the value $4,039,244.229\text{mol}$ for $n_1$ and $18.02\text{ g/mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the equation (4).

  $\begin{array}{c}{m}_1=\left(4,039,244.229\text{mol}\right)\left(18.02\text{ g/mol}\right)\\ =72787181\text{ g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\left(\frac{\text{1}\text{ton}}{\text{1000}\text{kg}}\right)\\ =72.7872\text{tons}\end{array}$

Rearrange the equation (1) to calculate the number of moles at $10\text{torr}$.

  $n_2=\frac{PV}{RT}$        (5)

Substitute the value $10\text{torr}$ for $P$, $295.2\text{ K}$ for $T$, $2.4\times {10}^6{\text{m}}^3$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (5).

  $\begin{array}{c}{n}_2=\frac{\left(10\text{torr}\right)\left(\frac{\text{1}\text{atm}}{760\text{torr}}\right)\left(2.4\times {10}^6{\text{m}}^3\right)\left(\frac{1\text{L}}{{10}^{-3}{\text{m}}^3}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(295.2\text{ K}\right)}\\ =1,302,980.7\text{mol}\end{array}$

The formula to calculate mass at $10\text{torr}$ is as follows:

  $m_2=\left(n_2\right)\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$        (6)

Substitute the value $1,302,980.7\text{mol}$ for $n_2$ and $18.02\text{ g/mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the equation (6).

  $\begin{array}{c}{m}_2=\left(1,302,980.7\text{mol}\right)\left(18.02\text{ g/mol}\right)\\ =23479712.2\text{ g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\left(\frac{\text{1}\text{ton}}{\text{1000}\text{kg}}\right)\\ =23.4797\text{tons}\end{array}$

The formula to calculate the mass of ${\text{H}}_{\text{2}}\text{O}$ removed is as follows:

  ${\text{Mass of H}}_{\text{2}}\text{O removed}=m_1-m_2$        (7)

Substitute the value $72.7872\text{tons}$ for $m_1$ and $23.4797\text{tons}$ for $m_2$ in the equation (7).

  $\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O removed}=72.7872\text{tons}-23.4797\text{tons}\\ =49.3075\text{tons}\\ \approx 49.3\text{tons}\text{.}\end{array}$

Conclusion

49 metric tons of water must be removed every time the inside air is completely replaced with outside air.

(b)

Interpretation Introduction

Interpretation:

The heat released when $49.3\text{tons}$ of water condenses is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Answer to Problem 12.136P

The heat released when $49.3\text{tons}$ of water condenses is $-1.11\times {10}^8\text{kJ}$.

Explanation of Solution

The heat of condensation for water is $-40.7\text{kJ/mol}$. The formula to calculate the heat released when $49.3\text{tons}$ mass of water condenses is as follows:

  $\text{Heat released}=\left(\frac{\text{given mass of water}}{\text{molar mass of water}}\right)\left(\text{heat of condensation for water}\right)$        (8)

Substitute the value $49.3\text{tons}$ for given mass of water, $18.02\text{ g/mol}$ for molar mass of water and $-40.7\text{kJ/mol}$ for heat of condensation for water in the equation (8).

  $\begin{array}{c}\text{Heat released}=\left(\frac{49.3\text{tons}}{18.02\text{ g/mol}}\right)\left(-40.7\text{kJ/mol}\right)\\ =-1.113660\times {10}^8\text{kJ}\\ \approx -1.11\times {10}^8\text{kJ}\text{.}\end{array}$

Conclusion

Condensation is an exothermic reaction that is heat is released when water is condensed. The heat released when $49.3\text{tons}$ of water condenses is $-1.11\times {10}^8\text{kJ}$.