Chapter 10, Problem 10.64SP

Interpretation Introduction

Interpretation:

Mass of ${\text{O}}_{\text{2}}$ in grams contained in room that measure $4.0\text{ m}$ by $5.0\text{ m}$ by $\text{2}\text{.5 m}$ is to be calculated.

Concept introduction:

Any gas obeys the assumption laid down in kinetic gas theory is said to be an ideal gas. The combination of all the gas laws namely Boyle’s law, Charles law and Avogadro’s leads to the ideal gas equation that can be used to relate all the four properties such as given below:

$PV=nRT$

Where,

$\text{R}$ is gas constant.

$V$ denotes the volume.

$n$ denotes the number of moles.

$T$ denotes temperature.

$P$ denotes the pressure.

The expression to determine pressure as per ideal gas equation is given as follows:

$P=\frac{m}{M}\frac{RT}{V}$

Rearrange the above expression to obtain the expression of mass.

$m=\frac{PVM}{RT}$

Where,

• $\text{R}$ is gas constant.
• $M$ demotes molar mass.
• $m$ represents mass.
• $V$ denotes the volume.
• $n$ denotes the number of moles.
• $T$ denotes temperature.
• $P$ denotes the pressure.