Chapter 5, Problem 5.103P

According to government standards, the 8h threshold limit value is 5000 ppmv for CO2 and 0.1 ppmv for ${\text{Br}}_{\text{2}}$ (1 ppmv is 1 part by volume in 106 parts by volume). Exposure to either gas for 8 h above these limits is unsafe. At STP, which of the following would be unsafe for 8 h of exposure?

(a) Air with a partial pressure of 0.2 torr of ${\text{Br}}_{\text{2}}$ (b) Air with a partial pressure of 0.2 torr of ${\text{CO}}_{\text{2}}$ (c) 1000 L of air containing 0.0004 g of ${\text{Br}}_{\text{2}}$ gas

(d) 1000 L of air containing $\text{2}\text{.8}\times {\text{10}}^{\text{22}}$ molecules of ${\text{CO}}_{\text{2}}$

Interpretation Introduction

Interpretation:

Among the given options, the correct one should be determined which is unsafe for 8 h of exposure.

1. Air having partial pressure of 0.2 torr of ${\text{Br}}_{\text{2}}$ .
2. Air having partial pressure of 0.2 torr of ${\text{CO}}_2$ .
3. 1000 L of air comprises 0.0004 g of ${\text{Br}}_{\text{2}}$ gas.
4. 1000 L of air comprises $\text{2}\text{.8}\times {\text{10}}^{22}$ molecules of ${\text{CO}}_2$ .

Concept Introduction:

The expression for mole fraction:

  $X_A=\frac{P_A}{P_{Total}}$

Where, $X_A$ = Mole fraction of gas A

  $P_A$ = Partial pressure of gas A

  $P_{Total}$ = Total pressure

Answer to Problem 5.103P

Correct option: Option (a) is unsafe for8 h of exposure.

Explanation of Solution

Given information:

8-h threshold limit value = 5000 ppmv for carbon dioxide,0.1ppmv for bromine.

1 ppmv is equal to 1 part by volume in 10⁶ parts by volume.

Reason for correct option:

(a)

Change the unit of pressure in torr to atm.

Since, 1 atm = 760 torr

Thus, pressure in atm = $0.2\text{ torr}\times \frac{1\text{ atm}}{760\text{ torr}}=2.63\times {10}^{-4}\text{ atm}$

Mole fraction of bromine gas is calculated as:

  $X_{B{r}_2}=\frac{P_{B{r}_2}}{P_{Total}}$

  $P_{Total}$ = 1 atm

Put the values,

  $X_{B{r}_2}=\frac{2.63\times {10}^{-4}\text{ atm}}{1.00\text{ atm}}$

  $X_{B{r}_2}=2.63\times {10}^{-4}\text{ atm}$

Convert the above value of bromine gas in ppmv:

Since, 1 ppmv = $\frac{1\text{ part of volume}}{{10}^6\text{ part of volume}}$

Thus, value of bromine gas in ppmv = $2.63\times {10}^{-4}\times \frac{{10}^6\text{ part of volume}}{1\text{ part of volume}}$

   = $263\text{ ppmv}$

The above pressure of bromine gas is more than0.1ppmv and the pressure is beyond the 8 h limit of threshold.

Thus, the air having partial pressure of bromine gas of 0.2 torr is unsafe.

Conclusion

Reason for incorrect option:

(b)

Change the unit of pressure in torr to atm.

Since, 1 atm = 760 torr

Thus, pressure in atm = $0.2\text{ torr}\times \frac{1\text{ atm}}{760\text{ torr}}=2.63\times {10}^{-4}\text{ atm}$

Mole fraction of carbon dioxide is calculated as:

  $X_{C{O}_2}=\frac{P_{C{O}_2}}{P_{Total}}$

  $P_{Total}$ = 1 atm

Put the values,

  $X_{C{O}_2}=\frac{2.63\times {10}^{-4}\text{ atm}}{1.00\text{ atm}}$

  $X_{C{O}_2}=2.63\times {10}^{-4}\text{ atm}$

Convert the above value of carbon dioxide in ppmv:

Since, 1 ppmv = $\frac{1\text{ part of volume}}{{10}^6\text{ part of volume}}$

Thus, value of carbon dioxide in ppmv = $2.63\times {10}^{-4}\times \frac{{10}^6\text{ part of volume}}{1\text{ part of volume}}$

   = $263\text{ ppmv}$

The above pressure of carbon dioxide is less than 5000 ppmv and the pressure doesn’t get beyond the 8 h limit of threshold.

Thus, the air having partial pressure of carbon dioxide of 0.2 torr is safe.

(c)

Molar mass of bromine gas = 159.9 g/mole

Mass of bromine gas = 0.0004 g

Number of moles of bromine gas = $\frac{\text{mass}}{\text{Molar mass}}$

Put the values,

Number of moles of bromine gas = $\frac{0.0004\text{ g}}{159.9\text{ g/mole}}$

   = $\text{2}\text{.50}\times {\text{10}}^{-6}$ mole

Since, volume of one mole of bromine gas at STP = 22.4 L

Thus, volume of bromine gas at STP = $\text{2}\text{.50}\times {\text{10}}^{-6}{\text{ mol Br}}_2\times \frac{22.4{\text{ L Br}}_2}{1{\text{ mol Br}}_2}$

   = $5.6\times {10}^{-5}\text{ L }$

The ppmv of bromine gas in 1000 L air is calculated as:

  $\frac{5.6\times {10}^{-5}\text{ L}}{1000\text{ L}}\times \frac{{10}^6\text{ part of volume}}{1\text{ part of volume}}$ = $0.056\text{ ppmv}$

The above value is less than 0.1 ppmv, therefore, 1000 L of air comprises 0.0004 g of bromine gas is safe.

(d)

Number of molecules = $2.8\times {10}^{22}\text{ molecules}$

Since, 1 mole = $6.023\times {10}^{23}\text{ molecules}$

Thus,

Number of moles of carbon dioxide gas = $2.8\times {10}^{22}\text{ molecules}\times \frac{1{\text{ mole of CO}}_2}{6.023\times {10}^{23}\text{ molecules}}$

   = 0.046 mole

Since, volume of one mole of carbon dioxide gas at STP = 22.4 L

Thus, volume of carbon dioxide gas at STP = $\text{0}\text{.046 mol}{\mathrm{CO}}_2\times \frac{22.4\text{ L }{\mathrm{CO}}_2}{1\text{ mol }{\mathrm{CO}}_2}$

   = $1.0304\text{ L }$

The ppmv of carbon dioxide gas in 1000 L air is calculated as:

  $\frac{1.0304\text{ L }}{1000\text{ L}}\times \frac{{10}^6\text{ part of volume}}{1\text{ part of volume}}$ = $1030.4\text{ ppmv}$

The above value is less than 5000 ppmv, therefore, 1000 L of air comprises $2.8\times {10}^{22}\text{ molecules}$ of carbon dioxide gas is safe.