Chapter 5, Problem 5.92P

Interpretation Introduction

Interpretation:

The number in metric tons (t) should be calculated for each gas emitted per year.

Concept Introduction:

Ideal gas law equation is shown as:

  $\text{PV= nRT}$

Where,

P = Pressure

V = Volume

n = Number of moles

R = Universal gas constant

T = Temperature

Rearrange the above formula in terms of number of moles:

  $n=\frac{PV}{RT}$

Number of moles is defined as the ratio of mass to the molar mass.

The mathematical expression is given as:

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

Mole fraction is defined as the ratio of number of moles of a gas to the total number of moles of gas.

The mathematical expression is given as:

  $\text{Mole fraction = }\frac{\text{Number of moles of gas}}{\text{Total number of moles of gases}}$

Answer to Problem 5.92P

Mass of ${\text{CO}}_2$ per year = 481.8 tons per year

Mass of $\text{CO}$ per year = 9.136 tons per year

Mass of ${\text{H}}_{\text{2}}\text{O}$ per year = 149.285 tons per year

Mass of ${\text{SO}}_2$ per year = 169.85 tons per year

Mass of ${\text{S}}_2$ per year = 0.43 tons per year

Mass of $\text{HCl}$ per year = 0.652 tons per year

Mass of ${\text{H}}_{\text{2}}$ per year = 0.212 tons per year

Mass of ${\text{H}}_{\text{2}}\text{S}$ per year = 0.229 tons per year

Explanation of Solution

Given information:

Hawaiian volcano Kilauea emits an average of $1.5\times {10}^3{\text{ m}}^3$ ofa gas each day.

Temperature = 298 K

Pressure = 1.00 atm

Mole fractions: 0.4896 ${\text{CO}}_2$ , $0.0146\text{ CO}$ , $0.3710{\text{ H}}_{\text{2}}\text{O}$ , $0.1185{\text{ SO}}_2$ , $0.0003\text{ S}$ , $0.0047{\text{ H}}_2$ , $0.0008\text{ HCl}$ , $0.0003{\text{ H}}_{\text{2}}\text{S}$

The number of moles is calculated as:

  $n=\frac{PV}{RT}$

Volume in L = $1.5\times {10}^6\text{L}$

Put the values,

  $n=\frac{(1\text{ atm)}\times (1.5\times {10}^6\text{L)}}{0.0821\text{ atm L/mol K}\times 298\text{ K}}$

   = $6.13\times {10}^4\text{ mol}$

Now, number of moles of carbon dioxide is calculated as:

  ${\text{Number of moles of CO}}_2\text{ = Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.4896}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $3.00\times {10}^4\text{ moles}$

Molar mass of ${\text{CO}}_{\text{2}}$ = 44 g/mole

Molar mass of ${\text{CO}}_{\text{2}}$ in kg = $44\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{CO}}_{\text{2}}$ = ${\text{Moles of CO}}_2\times {\text{molar mass of CO}}_{\text{2}}$

   = $3.00\times {10}^4\text{ moles}\times 44\times {10}^{-3}\text{ kg/mole}$

   = $1320\text{ kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{CO}}_{\text{2}}$ = $1320\text{ kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $1.32\text{ tons each day}$

Mass of ${\text{CO}}_{\text{2}}$ per year = $1.32\times 365\text{ per year}$

   = 481.8 tons per year

Number of moles of $\text{CO}$ is calculated as:

  $\text{Number of moles of CO = Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.0146}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.0894\times {10}^4\text{ mole}$

Molar mass of $\text{CO}$ = 28 g/mole

Molar mass of $\text{CO}$ in kg = $28\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of $\text{CO}$ = $\text{Moles of CO}\times \text{molar mass of CO}$

   = $0.0894\times {10}^4\text{ mole}\times 28\times {10}^{-3}\text{ kg/mole}$

   = $25.032\text{ kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of $\text{CO}$ = $25.032\text{ kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.025032\text{ tons each day}$

Mass of $\text{CO}$ per year = $0.025032\times 365\text{ per year}$

   = 9.136 tons per year

Number of moles of ${\text{H}}_{\text{2}}\text{O}$ is calculated as:

  ${\text{Number of moles of H}}_{\text{2}}\text{O = Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.3710}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $2.274\times {10}^4\text{ mole}$

Molar mass of ${\text{H}}_{\text{2}}\text{O}$ = 18 g/mole

Molar mass of ${\text{H}}_{\text{2}}\text{O}$ in kg = $18\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{H}}_{\text{2}}\text{O}$ = ${\text{Moles of H}}_{\text{2}}\text{O}\times {\text{molar mass of H}}_{\text{2}}\text{O}$

   = $2.274\times {10}^4\text{ mole}\times 18\times {10}^{-3}\text{ kg/mole}$

   = $\text{409 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{H}}_{\text{2}}\text{O}$ = $\text{409 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.409\text{ tons each day}$

Mass of ${\text{H}}_{\text{2}}\text{O}$ per year = $0.409\times 365\text{ per year}$

   = 149.285 tons per year

Number of moles of ${\text{SO}}_2$ is calculated as:

  ${\text{Number of moles of SO}}_2\text{ = Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.1185}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.7264\times {10}^4\text{ mole}$

Molar mass of ${\text{SO}}_2$ = 64.06 g/mole

Molar mass of ${\text{SO}}_2$ in kg = $64.06\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{SO}}_2$ = ${\text{Moles of SO}}_2\times {\text{molar mass of SO}}_2$

   = $0.7264\times {10}^4\text{ mole}\times 64.06\times {10}^{-3}\text{ kg/mole}$

   = $\text{465}\text{.3 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{SO}}_2$ = $\text{465}\text{.3 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.465\text{ tons each day}$

Mass of ${\text{SO}}_2$ per year = $0.465\times 365\text{ per year}$

   = 169.85 tons per year

Number of moles of ${\text{S}}_2$ is calculated as:

  ${\text{Number of moles of S}}_2\text{ = Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.0003}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.00184\times {10}^4\text{ mole}$

Molar mass of ${\text{S}}_2$ = 64.12 g/mole

Molar mass of ${\text{S}}_2$ in kg = $64.12\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{S}}_2$ = ${\text{Moles of S}}_2\times {\text{molar mass of S}}_2$

   = $0.00184\times {10}^4\text{ mole}\times 64.12\times {10}^{-3}\text{ kg/mole}$

   = $\text{1}\text{.18 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{S}}_2$ = $\text{1}\text{.18 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.00118\text{ tons each day}$

Mass of ${\text{S}}_2$ per year = $0.00118\times 365\text{ per year}$

   = 0.43 tons per year

Number of moles of ${\text{H}}_{\text{2}}$ is calculated as:

  ${\text{Number of moles of H}}_{\text{2}}\text{= Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.0047}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.0288\times {10}^4\text{ mole}$

Molar mass of ${\text{H}}_{\text{2}}$ = 2.016 g/mole

Molar mass of ${\text{H}}_{\text{2}}$ in kg = $2.016\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{H}}_{\text{2}}$ = ${\text{Moles of H}}_{\text{2}}\times {\text{molar mass of H}}_{\text{2}}$

   = $0.0288\times {10}^4\text{ mole}\times 2.016\times {10}^{-3}\text{ kg/mole}$

   = $\text{0}\text{.58 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{H}}_{\text{2}}$ = $\text{0}\text{.58 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.00058\text{ tons each day}$

Mass of ${\text{H}}_{\text{2}}$ per year = $0.00058\times 365\text{ per year}$

   = 0.212 tons per year

Number of moles of $\text{HCl}$ is calculated as:

  $\text{Number of moles of HCl= Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.0008}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.0049\times {10}^4\text{ mole}$

Molar mass of $\text{HCl}$ = 36.458 g/mole

Molar mass of $\text{HCl}$ in kg = $36.458\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of $\text{HCl}$ = $\text{Moles of HCl}\times \text{molar mass of HCl}$

   = $0.0049\times {10}^4\text{ mole}\times 36.458\times {10}^{-3}\text{ kg/mole}$

   = $\text{1}\text{.786 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of $\text{HCl}$ = $\text{1}\text{.786 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.001786\text{ tons each day}$

Mass of $\text{HCl}$ per year = $0.001786\times 365\text{ per year}$

   = 0.652 tons per year

Number of moles of ${\text{H}}_{\text{2}}\text{S}$ is calculated as:

  ${\text{Number of moles of H}}_{\text{2}}\text{S= Mole fraction}\times \text{total moles of gases}$

  $\text{= 0}\text{.0003}\times \text{6}\text{.13}\times {\text{10}}^4\text{ mol}$

   = $0.00184\times {10}^4\text{ mole}$

Molar mass of ${\text{H}}_{\text{2}}\text{S}$ = 34.076 g/mole

Molar mass of ${\text{H}}_{\text{2}}\text{S}$ in kg = $34.076\times {10}^{-3}\text{ kg/mole}$ (1 kg = 1000 g)

Mass of ${\text{H}}_{\text{2}}\text{S}$ = ${\text{Moles of H}}_{\text{2}}\text{S}\times {\text{molar mass of H}}_{\text{2}}\text{S}$

   = $0.00184\times {10}^4\text{ mole}\times 34.076\times {10}^{-3}\text{ kg/mole}$

   = $\text{0}\text{.627 kg}$

Since, 1 kg = 0.001 ton

Thus,

Mass of ${\text{H}}_{\text{2}}\text{S}$ = $\text{0}\text{.627 kg}\times \frac{1\text{ ton}}{1000\text{ kg}}$

   = $0.000627\text{ tons each day}$

Mass of ${\text{H}}_{\text{2}}\text{S}$ per year = $0.000627\times 365\text{ per year}$

   = 0.229 tons per year