Chapter 5, Problem 122AE

Interpretation Introduction

Interpretation:Partial pressure of the two gases is to be calculated. Mass of water formed by complete combustion is to be calculated.

Concept introduction:State of any gas is determined by some of its parameters. These parameters are pressure, amount of gas, temperature and volume. Ideal gas law helps to govern state of gas with help of relationships between these gas parameters.

Expression for ideal gas equation is as follows:

  $\text{PV}=\text{nRT}$

Where,

• $\text{P}$is pressure of gas.
• $\text{V}$is volume of gas.
• $n$is amount or moles of gas.
• $R$is universal gas constant.
• $T$is absolute temperature of gas.

Rearrange the above equation.

  $\text{n}=\frac{\text{PV}}{\text{RT}}$

Mole fraction of i-th component of a gas in a gas mixture is calculated as follows:

  ${\text{χ}}_{\text{i}}=\frac{{\text{n}}_{\text{i}}}{\text{n}}$

Where,

• ${\text{χ}}_{\text{i}}$is the mole fraction of i-th component.
• ${\text{n}}_{\text{i}}$is the mole of i-th component of gas present in the gas container.
• $\text{n}$is the total number of moles of gas present in the container.

Partial pressure of i-th component of a gas in a gas mixture enclosed in fixed volume and fixed temperature is calculated as follows:

  ${\text{P}}_{\text{i}}=\left({\text{χ}}_{\text{i}}\right)\left(\text{P}\right)$

Where,

• ${\text{χ}}_{\text{i}}$is the mole fraction of i-th component.
• ${\text{P}}_{\text{i}}$is the partial pressure of i-th component of gas present in the gas container.
• $\text{P}$is the total pressure of gas.

Combustion of a hydrocarbon $\left({\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\right)$ is described by the chemical equation as follows:

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}+\left(\text{x}+\frac{\text{y}}{\text{4}}\right){\text{O}}_{\text{2}}\to \text{x}{\text{CO}}_{\text{2}}+\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$

Answer to Problem 122AE

Partial pressure of methane is $1.3176\text{atm}$ , partial pressure of ethane is $0.1224\text{atm}$ and mass of water formed is $33.73947\text{g}$ .

Explanation of Solution

Given Information: Molefraction of methane $\left({\text{χ}}_{\text{methane}}\right)$ is 0.915. Mole fraction of ethane $\left({\text{χ}}_{\text{ethane}}\right)$ is 0.085. Volume of gas container is $\text{15}\text{.00}\text{L}$ , Total pressure of gas is $\text{1}\text{.44}\text{atm}$ , and the temperature of gas is $20°\text{C}$ or $293\text{K}$ .

Partial pressure of methane in a gas mixture enclosed in fixed volume and fixed temperature is calculated by the formula as follows:

  ${\text{P}}_{\text{methane}}=\left({\text{χ}}_{\text{methane}}\right)\left(\text{P}\right)$

Where,

• ${\text{χ}}_{\text{methane}}$is the mole fraction of methane.
• ${\text{P}}_{\text{methane}}$is the partial pressure of methane of gas present in the gas container.
• $\text{P}$is the total pressure of gas.

Value of ${\text{χ}}_{\text{methane}}$ is 0.915.

Value of $\text{P}$ is $\text{1}\text{.44}\text{atm}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{P}}_{\text{methane}}=\left(0.915\right)\left(1.44\text{atm}\right)\\ =1.3176\text{atm}\end{array}$

Partial pressure of ethane in a gas mixture enclosed in fixed volume and fixed temperature is calculated by the formula as follows:

  ${\text{P}}_{\text{ethane}}=\left({\text{χ}}_{\text{ethane}}\right)\left(\text{P}\right)$

Where,

• ${\text{χ}}_{\text{ethane}}$is the mole fraction of ethane.
• ${\text{P}}_{\text{ethane}}$is the partial pressure of ethane of gas present in the gas container.
• $\text{P}$is the total pressure of gas.

Value of ${\text{χ}}_{\text{ethane}}$ is 0.085.

Value of $\text{P}$ is $\text{1}\text{.44}\text{atm}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{P}}_{\text{ethane}}=\left(0.085\right)\left(1.44\text{atm}\right)\\ =0.1224\text{atm}\end{array}$

Conversion of $20°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =20°\text{C}+273\\ =293\text{K}\end{array}$

If all the gasses are considered as ideal gas then,mole of gas present in a close container at fixed pressure and temperature is calculated by the equation as follows:

  $\text{PV}=\text{nRT}$

Where,

• $\text{P}$is the pressure of gas.
• $\text{V}$is the volume of gas container.
• $\text{n}$is mole of gas present in container.
• $\text{R}$is universal gas constant.
• $\text{T}$is the temperature on absolute scale.

Rearrange the above equation as follows:

  $\text{n}=\frac{\text{PV}}{\text{RT}}$

Total pressure of gas is $\text{1}\text{.44}\text{atm}$ .

Volume of gas container is $\text{15}\text{.00}\text{L}$ .

Temperature of gas on absolute scale is $293\text{K}$ .

Universal gas constant is $0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{n}=\frac{\left(1.44\text{atm}\right)\left(15.00\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{-1}\right)\left(293\text{K}\right)}\\ =0.899\text{mol}\end{array}$

Mole of methane in a gas mixture is calculated by the formula as follows:

  ${\text{n}}_{\text{methane}}=\left({\text{χ}}_{\text{methane}}\right)\left(\text{n}\right)$

Where,

• ${\text{χ}}_{\text{methane}}$is the mole fraction of methane.
• ${\text{n}}_{\text{methane}}$is the mole of methane present in the gas container.
• $\text{n}$is the total number of moles of gas present in the container.

Value of ${\text{χ}}_{\text{methane}}$ is 0.915.

Value of $\text{n}$ is $0.899\text{mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{n}}_{\text{methane}}=\left(0.915\right)\left(0.899\text{mol}\right)\\ =0.822585\text{mol}\end{array}$

Mole of ethane in a gas mixture is calculated by the formula as follows:

  ${\text{n}}_{\text{ethane}}=\left({\text{χ}}_{\text{ethane}}\right)\left(\text{n}\right)$

Where,

• ${\text{χ}}_{\text{ethane}}$is the mole fraction of ethane.
• ${\text{n}}_{\text{ethane}}$is the mole of ethane present in the gas container.
• $\text{n}$is the total number of moles of gas present in the container.

Value of ${\text{χ}}_{\text{ethane}}$ is is 0.085.

Value of $\text{n}$ is $0.899\text{mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{n}}_{\text{ethane}}=\left(0.085\right)\left(0.899\text{mol}\right)\\ =0.076415\text{mol}\end{array}$

Combustion of methane $\left({\text{CH}}_{\text{4}}\right)$ is described by the chemical equation as follows:

  ${\text{CH}}_4+2{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Hence for combustion of $1\text{mol}$ methane $\left({\text{CH}}_{\text{4}}\right)$ , $2\text{mol}$ water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is formed.For combustion of $0.822585\text{mol}$ methane,moles of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ formed are calculated as follows.

  $\begin{array}{c}\text{moles of water formed}=\left(0.822585\text{mol}\right)\left(2\right)\\ =1.64517\text{mol}\end{array}$

Combustion of ethane $\left({\text{C}}_2{\text{H}}_6\right)$ is described by the chemical equation as follows:

  ${\text{C}}_2{\text{H}}_6+\frac{7}{2}{\text{O}}_{\text{2}}\to 2{\text{CO}}_{\text{2}}+3{\text{H}}_{\text{2}}\text{O}$

Hence for combustion of $1\text{mol}$ ethane $\left({\text{C}}_2{\text{H}}_6\right)$ , $3\text{mol}$ water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is formed.For combustion of $0.076415\text{mol}$ ethane, moles of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ formed is calculated as follows.

  $\begin{array}{c}\text{moles of water formed}=\left(0.076415\text{mol}\right)\left(3\right)\\ =0.229245\text{mol}\end{array}$

Total moles of water, that is formed by the complete combustion of both gases in the natural gas sample is calculated as follows:

  $\begin{array}{c}\text{Total moles of water}=\left(1.64517\text{mol}+0.229245\text{mol}\right)\\ =1.874415\text{mol}\end{array}$

Molar mass of water is $18\text{g}{\text{mol}}^{\text{-1}}$ .

Total mass of water, that is formed by the complete combustion of both gases in the natural gas sample is calculated as follows:

  $\begin{array}{c}\text{Total mass of water}=\left(1.874415\text{mol}\right)\left(18\text{g}{\text{mol}}^{-1}\right)\\ =33.73947\text{g}\end{array}$

Conclusion

Partial pressure of methane is $1.3176\text{atm}$ , partial pressure of ethane is $0.1224\text{atm}$ and mass of water formed is $33.73947\text{g}$ .