Chapter 13, Problem 40A

Interpretation Introduction

Interpretation: -Volume of Oxygen gas is calculated using given variables.

Concept Introduction: - For calculating volume of Oxygen, the formula for ideal gas equation is applied.

It is represented as follows:

  $\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}$

Or,

  $\mathrm{n}=\frac{\mathrm{P}\mathrm{V}}{\text{R}\mathrm{T}}$

Where

  $\mathrm{V}$ is volume of the gas

  $\mathrm{n}$ is the number of moles of the gas

  $\mathrm{P}$ is pressure of the gas

  $\mathrm{T}$ is temperature of the gas

  $\text{R}$ is gas constant

Answer to Problem 40A

Volume of oxygen at $25°\mathrm{C}$ is $14.7\mathrm{L}$ .

Explanation of Solution

There is a relation between volume of a gas and number of molecules of gas present in it, this relation is expressed by Avogadro’s law

According to Avogadro’s law 1 mole of every substance occupies $22.4\mathrm{L}$ at STP weigh’s equal to the molecular mass and contains Avogadro’s number $6.023\times {10}^{23}$ of particles.

  $\begin{array}{c}\text{no}\text{. of moles = }\frac{\text{given mass}}{\text{molar mass}}\\ =\frac{5.53\text{ g}}{44.1\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}\\ =0.125\mathrm{moles}\end{array}$

The combustion reaction is represented as follows:

  ${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3\mathrm{C}{\mathrm{O}}_2+4{\mathrm{H}}_2\mathrm{O}$

According to stoichiometry, $1\text{ mole}$ of propane requires $\text{5 moles}$ of Oxygen.

So, $0.125\text{ moles}$ will require number of oxygen atoms as follows:

  $\begin{array}{c}{\mathrm{n}}_{{\text{O}}_{\text{2}}}=0.125\times 5\\ =0.625\text{ moles}\end{array}$

According to the Ideal gas equation

  $\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}$

P = pressure of the gas = 1.04atm

V = volume of the gas =?

T = Temperature of the gas

The Temperature in Kelvin will be:

  $\begin{array}{c}\mathrm{T}={25}^0\text{C}+273\\ =29\text{8 K}\end{array}$

R = gas constant = 0.0821 $\frac{\text{atm L}}{\text{k mol}}$

n = moles of gas = 0.625 mol

Thus, volume for 0.125 moles of propane will be:

  $\begin{array}{c}\mathrm{T}=24+273\text{ K}\\ =297\mathrm{K}\end{array}$

R is 0.0821 L atm/K mol

Putting the values,

  $\begin{array}{c}\mathrm{V}=\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{P}}\\ =\frac{0.625\times 0.0821\times 298}{1.04}\\ \mathrm{V}=\frac{15.2911}{1.04}\\ =14.7\mathrm{L}\end{array}$

Volume of oxygen at $25°\mathrm{C}$ is $14.7\mathrm{L}$ .

Conclusion

Volume of oxygen at $25°\mathrm{C}$ is $14.7\mathrm{L}$ .