Chapter 5, Problem 5.147P

(a)

Interpretation Introduction

Interpretation:

At STP condition the exposure to ${\text{Br}}_{\text{2}}$ gas at a partial pressure of $0.2\text{torr}$ for $8\text{ h}$ is safe or unsafe is to be determined.

Concept introduction:

The expression to calculate the partial pressure of the gas is as follows:

  $P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

  $X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

The ideal gas equation can be expressed as follows,

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.147P

The exposure to ${\text{Br}}_{\text{2}}$ at partial pressure of $0.2\text{torr}$ for $8\text{ h}$ would be unsafe.

Explanation of Solution

The expression to calculate the mole fraction of ${\text{Br}}_{\text{2}}$ is as follows:

  $X_{{\text{Br}}_{\text{2}}}=\frac{P_{{\text{Br}}_{\text{2}}}}{P_{\text{total}}}$        (1)

Here, $X_{{\text{Br}}_{\text{2}}}$ is the mole fraction of ${\text{Br}}_{\text{2}}$, $P_{{\text{Br}}_{\text{2}}}$ is the partial pressure of ${\text{Br}}_{\text{2}}$ and $P_{\text{total}}$ is the total pressure.

Substitute $0.2\text{ torr}$ for $P_{{\text{Br}}_{\text{2}}}$ and $760\text{ torr}$ for $P_{\text{total}}$ in the equation (1).

  $\begin{array}{c}{X}_{{\text{Br}}_{\text{2}}}=\left(\frac{0.2\text{ torr}}{760\text{ torr}}\right)\\ =\left(2.63\times {10}^{-4}\right)\left({10}^6\right)\\ =2.63\times {10}^2\text{ppmv}\text{.}\end{array}$

According to the government standard, the threshold limit for ${\text{Br}}_{\text{2}}$ is $0.1\text{ ppmv}$ for eight hours. If the $\text{ppmv}$ is greater than the threshold limit for eight hours then the exposure of the gas is unsafe. The $\text{ppmv}$ value of ${\text{Br}}_{\text{2}}$ is more than the threshold limit value. Thus, the exposure of ${\text{Br}}_{\text{2}}$ would be unsafe.

Conclusion

The exposure of ${\text{Br}}_{\text{2}}$ would be unsafe because $\text{ppmv}$ is greater than the threshold limit for eight hours.

(b)

Interpretation Introduction

Interpretation:

The exposure of ${\text{CO}}_{\text{2}}$ gas at a partial pressure of $0.2\text{torr}$ is safe or unsafe is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows:

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

  $P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows:

  $X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

Answer to Problem 5.147P

The exposure to ${\text{CO}}_{\text{2}}$ at partial pressure of $0.2\text{torr}$ for 8 hours would be safe.

Explanation of Solution

The expression to calculate the mole fraction of ${\text{CO}}_{\text{2}}$ is as follows:

  $X_{{\text{CO}}_{\text{2}}}=\frac{P_{{\text{CO}}_{\text{2}}}}{P_{\text{total}}}$        (2)

Here, $X_{{\text{CO}}_{\text{2}}}$ is the mole fraction of ${\text{CO}}_{\text{2}}$, $P_{{\text{CO}}_{\text{2}}}$ is the partial pressure of ${\text{CO}}_{\text{2}}$ and $P_{\text{total}}$ is the total pressure.

Substitute $0.2\text{ torr}$ for $P_{{\text{CO}}_{\text{2}}}$ and $760\text{ torr}$ for $P_{\text{total}}$ in the equation (2).

  $\begin{array}{c}{X}_{{\text{CO}}_{\text{2}}}=\left(\frac{0.2\text{ torr}}{760\text{ torr}}\right)\\ =\left(2.63\times {10}^{-4}\right)\left({10}^6\right)\\ =2.63\times {10}^2\text{ppmv}\text{.}\end{array}$

According to the government standard, the threshold limit for ${\text{CO}}_{\text{2}}$ is $\text{5000 ppmv}$ for eight hours. If the $\text{ppmv}$ is greater than the threshold limit for eight hours then the exposure of the gas is unsafe. The $\text{ppmv}$ value of ${\text{CO}}_{\text{2}}$ is less than the threshold limit value. Thus, the exposure of ${\text{CO}}_{\text{2}}$ would be safe.

Conclusion

The exposure of ${\text{CO}}_{\text{2}}$ would be safe because $\text{ppmv}$ is less than the threshold limit for eight hours.

(c)

Interpretation Introduction

Interpretation:

The exposure to $1000\text{ L}$ of air that contains $0.0004\text{ g}$ of ${\text{Br}}_{\text{2}}$ gas is safe or unsafe is to be determined.

Concept introduction:

The expression to calculate the moles of solute is as follows:

  $\text{Moles of solute}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}}$

The ideal gas equation can be expressed as follows,

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.147P

The exposure to $1000\text{ L}$ of air that contains $0.0004\text{ g}$ of ${\text{Br}}_{\text{2}}$ would be safe.

Explanation of Solution

The expression to calculate the moles of ${\text{Br}}_{\text{2}}$ is as follows:

  ${\text{Moles of Br}}_{\text{2}}=\frac{{\text{Given mass of Br}}_{\text{2}}}{{\text{Molar mass of Br}}_{\text{2}}}$        (3)

Substitute the value $\text{0}\text{.004 g}$ for the given mass of ${\text{Br}}_{\text{2}}$ and $\text{159}\text{.80 g/mol}$ for molar mass of ${\text{Br}}_{\text{2}}$ in the equation (3).

  $\begin{array}{c}{\text{Moles of Br}}_{\text{2}}=\frac{0.0004\text{ g}}{159.44\text{ g/mol}}\\ =2.50\times {10}^{-6}\text{ mol}\end{array}$

The formula to convert $°\text{C}$ to Kelvin is:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (4)

Substitute $0°\text{C}$ for $\text{T }°\text{C}$ in equation (4).

  $\begin{array}{c}T\left(\text{K}\right)=0°\text{C}+273.15\\ =273.15\text{ K}\end{array}$

The expression to calculate the moles of ${\text{Br}}_{\text{2}}$ is as follows:

  $PV=nRT$        (5)

Rearrange equation (5) for $n$ as follows:

  $n=\frac{PV}{RT}$        (6)

Substitute $\text{1 atm}$ for $P$, $\text{273 K}$ for $T$, $\text{1000 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (6).

  $\begin{array}{c}n=\frac{\left(\text{1 atm}\right)\left(\text{32}\text{.4 L}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{273 K}\right)}\\ =1.44557\text{ mol}\end{array}$

The expression to calculate the concentration of ${\text{Br}}_{\text{2}}$ as follows:

  ${\text{Concentration of Br}}_{\text{2}}=\left(\frac{{\text{Mole of Br}}_{\text{2}}}{\text{Mole of air}}\right)\left({10}^6\right)$        (7)

Substitute $2.503\times {10}^{-6}\text{ mol}$ for moles of ${\text{Br}}_{\text{2}}$ and $44.616\text{ mol}$ for moles of air in the equation (7).

  $\begin{array}{c}{\text{Concentration of Br}}_{\text{2}}=\left(\frac{2.503\times {10}^{-6}\text{ mol}}{44.616\text{ mol}}\right)\left({10}^6\right)\\ =0.056103\text{ ppmv}\\ =0.06\text{ ppmv}\text{.}\end{array}$

According to the government standard, the threshold limit for ${\text{Br}}_{\text{2}}$ is $0.1\text{ ppmv}$ for eight hours. If the $\text{ppmv}$ is greater than the threshold limit for eight hours then the exposure of the gas is unsafe. The $\text{ppmv}$ value of ${\text{Br}}_{\text{2}}$ is less than the threshold limit value. Thus, the exposure of ${\text{Br}}_{\text{2}}$ would be safe.

Conclusion

The exposure of ${\text{Br}}_{\text{2}}$ would be safe because $\text{ppmv}$ is less than the threshold limit for eight hours.

(d)

Interpretation Introduction

Interpretation:

The exposure to $1000\text{ L}$ to of air that contains $2.8\times {10}^{22}\text{ molecule}$ of ${\text{CO}}_{\text{2}}$ gas is safe or unsafe is to be determined.

Concept introduction:

The expression to calculate the moles of solute is as follows:

  $\text{Moles of solute}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}}$

The ideal gas equation can be expressed as follows,

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant

Answer to Problem 5.147P

The exposure to $1000\text{ L}$ to of air that contains $2.8\times {10}^{22}\text{ molecule}$ of ${\text{CO}}_{\text{2}}$ gas is safe

Explanation of Solution

The expression to calculate the moles of ${\text{CO}}_{\text{2}}$ is as follows:

  ${\text{Moles of CO}}_{\text{2}}=\frac{{\text{Molecule of CO}}_{\text{2}}}{\text{Avogadro's number}}$        (8)

Substitute $2.8\times {10}^{22}\text{ molecules}$ for moles of ${\text{CO}}_{\text{2}}$ and $\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{ molecules}$ for Avogadro’s number in the equation (8).

  $\begin{array}{c}{\text{Moles of CO}}_{\text{2}}=\frac{2.8\times {10}^{22}\text{ molecules}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{ molecules}}\\ =0.04696\text{ mol}\end{array}$

The expression to calculate the concentration of ${\text{CO}}_{\text{2}}$ as follows:

  ${\text{Concentration of CO}}_{\text{2}}=\left(\frac{{\text{Mole of CO}}_{\text{2}}}{\text{Mole of air}}\right)\left({10}^6\right)$        (9)

Substitute $\text{0}\text{.046496 mol}$ for moles of ${\text{CO}}_{\text{2}}$ and $44.616\text{ mol}$ for moles of air in the equation (9).

  $\begin{array}{c}{\text{Concentration of Br}}_{\text{2}}=\left(\frac{\text{0}\text{.046496 mol}}{44.616\text{ mol}}\right)\left({10}^6\right)\\ =1042.1\text{ ppmv}\\ =1.0\times {10}^3\text{ ppmv}\text{.}\end{array}$

According to the government standard, the threshold limit for ${\text{CO}}_{\text{2}}$ is $\text{5000 ppmv}$ for eight hours. If the $\text{ppmv}$ is greater than the threshold limit for eight hours then the exposure of the gas is unsafe. The $\text{ppmv}$ value of ${\text{CO}}_{\text{2}}$ is less than the threshold limit value. Thus, the exposure of ${\text{CO}}_{\text{2}}$ would be safe.

Conclusion

The exposure of ${\text{CO}}_{\text{2}}$ would be safe because $\text{ppmv}$ is less than the threshold limit for eight hours.