Chapter 5, Problem 5.92PAE

Interpretation Introduction

Interpretation:

A mixture of Ar (0.40 mol), O₂ (0.50 mol), and CH₄ (0.30 mol) exerts a pressure of 740 mm Hg. If the methane and oxygen are ignited and complete combustion occurs, the final pressure of Ar, CO₂, H₂O, and the remainder of the excess reactant and total pressure of the system should be calculated.

Concept introduction:

An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws: -

$\text{PV = nRT}$

Here,

n = Number of moles of the gas

R = Universal gas constant

T = Temperature

P = Pressure

Ideal gases are described as the molecules which have negligible size but have an average molar kinetic energy which is dependent on the temperature. When temperature is low most of the gases behave like ideal gases and the ideal gas law might be applied to them.

Answer to Problem 5.92PAE

Solution:

Partial Pressure of each gas is as follows:

$\begin{array}{c}\text{Ar}=\text{246}.\text{7 mm Hg} \\ {\text{CO}}_{\text{2}}=\text{154}.\text{2 mm Hg}\\ {\text{H}}_{\text{2}}\text{O}=\text{3}0\text{8}.\text{3 mm Hg}\\ {\text{CH}}_{\text{4}}=\text{3}0.\text{8 mm Hg}\end{array}$

Total pressure = 740 mmHg

Given:

The number of moles of gases at initial stage are as follows:

$\begin{array}{c}\text{Ar}=0.\text{4}0\text{ mol} \\ {\text{O}}_2=0.\text{5}0\text{ mol} \\ {\text{CH}}_{\text{4}}=0.\text{3}0\text{ mol}\end{array}$

Explanation of Solution

Thus, total number of moles = $\left[0.40+0.50+0.30\right]\text{ mol = 1}\text{.20 mol}$ = N₁

As given total pressure = P = 740 mmHg = P₁ The combustion reaction is as follows:

$\text{Ar}\left(\text{g}\right)+{\text{CH}}_4\left(\text{g}\right){\text{+ 2O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{g}\right)+\text{Ar}\left(\text{g}\right)$

Argon being a noble gas does not react in the reaction.

Thus, as per the above equation limiting reagent is Oxygen

Now,

0.50 mol of O₂ forms no of mol of ${\text{CO}}_{\text{2}}$ = $\frac{1}{2}\left(0.50\right)=0.25\text{ mol}$

And,

0.30 mol of ${\text{CH}}_{\text{4}}$ forms number of moles of ${\text{CO}}_{\text{2}}$ = $1\left(0.30\right)=0.30\text{ mol}$

Thus, oxygen is limiting reagent here,

Calculate the number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced as follows:

0.50 mol of O₂ forms no of mol of ${\text{H}}_{\text{2}}\text{O}$ = $\frac{2}{2}\left(0.50\right)=0.50\text{ mol}$

Remaining CH₄will be:

${\text{n}}_{{\text{CH}}_{\text{4}}}=\left[0.30\text{ mol}-0.25\right]\text{ = 0}\text{.05 mol}$

Thus, number of moles of gas after combustion will be:

$\begin{array}{c}\text{Ar}=0.\text{4}0\text{ mol}\\ {\text{CO}}_{\text{2}}=0.\text{25 mol}\\ {\text{H}}_{\text{2}}\text{O}=0.\text{5}0\text{ mol}\\ {\text{CH}}_{\text{4}}=0.0\text{5 mol}\end{array}$

Thus, total number of moles = $\left[0.40+0.25+0.50+0.05\right]\text{ mol = 1}{\text{.20 mol = N}}_2$ = N₁

Total Pressure = P₂that needs to be calculated.

As per the ideal gas equation,

$\begin{array}{l}\frac{{\text{P}}_1}{{\text{N}}_1}\text{ = }\frac{{\text{P}}_2}{{\text{N}}_2}\\ {\text{P}}_2\text{ = }\frac{{\text{N}}_2{\text{P}}_1}{{\text{N}}_1}\\ \text{ = }\frac{1.20\text{ mol}\times 740\text{ mm Hg}}{1.20\text{ mol}}\\ \text{ = 740 mm Hg}\end{array}$

Therefore, the total pressure of the system remains the same that is 740 mmHg

Now, partial pressure is calculated as follows:

Suppose for gas A,

${\text{P}}_{\text{A}}={\text{X}}_{\text{A}}{\text{P}}_{\text{T}}$

Here, ${\text{X}}_{\text{A}}$ is mole fraction of gas A and ${\text{P}}_{\text{T}}$ is total pressure of gas.

Ar = $\frac{0.40\text{ mol}}{1.20\text{ mol}}\left(740\text{ mm Hg}\right)\text{ = 246}\text{.7 mm Hg}$

CO₂= $\frac{0.25\text{ mol}}{1.20\text{ mol}}\left(740\text{ mm Hg}\right)\text{ = 154}\text{.2 mm Hg}$

H₂O = $\frac{0.50\text{ mol}}{1.20\text{ mol}}\left(740\text{ mm Hg}\right)\text{ = 308}\text{.3 mm Hg}$

CH₄ = $\frac{0.05\text{ mol}}{1.20\text{ mol}}\left(740\text{ mm Hg}\right)\text{ = 30}\text{.8 mm Hg}$

Conclusion

An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws. Based on the ideal gas law concept we calculated that

Partial Pressure of each gas is as follows:

$\begin{array}{c}\text{Ar}=\text{246}.\text{7 mm Hg} \\ {\text{CO}}_{\text{2}}=\text{154}.\text{2 mm Hg}\\ {\text{H}}_{\text{2}}\text{O}=\text{3}0\text{8}.\text{3 mm Hg}\\ {\text{CH}}_{\text{4}}=\text{3}0.\text{8 mm Hg}\end{array}$

Total pressure = 740 mmHg