Chapter 13, Problem 67A

Interpretation Introduction

Interpretation:

The number of moles of gases present in the cylinder at a particular temperature is to be determined and the number of oxygen atoms also to be determined.

Concept introduction:

The ideal gases are those gases which obey the ideal gas equation and the ideal gas equation is written as

  $\text{PV}\text{= nRT}$

Answer to Problem 67A

The number of moles of gas in the cylinder is $\text{22}{\text{.8×10}}^{\text{-3}}\text{moles}$ and the number of molecules of oxygen gas is $\text{3}{\text{.8×10}}^{\text{23}}\text{atoms}$ .

Explanation of Solution

Given information:

Volume = ${\text{618cm}}^{\text{3}}$

Volume in liter = ${\text{618×10}}^{\text{-3}}\text{L}$

Temperature = $\text{75°C}\text{=}\text{75}\text{+}\text{273=}\text{348K}$

Pressure = $\text{802}\text{mmHg}$

Pressure in atmospheric pressure ₌$\frac{\text{802}}{\text{760}}\text{=}\text{1}\text{.1}\text{atm}$

Presence of oxygen in air by volume % = $\text{20\%}$

Calculation:

According to the ideal gas equation,

  $\begin{array}{l}\text{n=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{1}{\text{.055×618×10}}^{\text{-3}}}{\text{0}\text{.0821×348}}\\ \text{=}\text{22}{\text{.8×10}}^{\text{-3}}\text{moles}\end{array}$

Number of molecules of the oxygen atom

Number of molecules = $\begin{array}{l}\frac{\text{20}}{\text{32}}{\text{×N}}_{\text{A}}\\ \frac{\text{20}}{\text{32}}\text{×}\text{6}{\text{.023×10}}^{\text{23}}\\ \text{3}{\text{.8×10}}^{\text{23}}\text{atoms}\end{array}$

The oxygen contains $\text{3}{\text{.8×10}}^{\text{23}}\text{atoms}$ .

Conclusion

The number of moles of the gas is $\text{22}{\text{.8×10}}^{\text{-3}}\text{moles}$ and air contains $\text{3}{\text{.8×10}}^{\text{23}}\text{atoms}$ oxygen atoms.