Chapter 13, Problem 29A

Interpretation Introduction

Interpretation:

The pressure have to be calculated.

Concept Introduction:

The ideal gas law was derived by combining three important laws of the gases. The ideal gas law can be expressed as-

  $\begin{array}{l}V\propto \frac{\mathrm{n}T}{P}\\ \mathrm{PV}=nRT\end{array}$ Where, R is universal gas constant.

The ideal gas law consists of all the variables of a gas such as pressure (P), volume (V), number of moles (n) and temperature (T).

Answer to Problem 29A

The pressure of oxygen gas is 338.72 atm.

Explanation of Solution

Data given: Mass of oxygen gas = 56.2 kg = 56200 gm

Volume = 125 L

Temperature = 21 °C = 294 K

R = 0.082 L atm/mol K

Number of moles of oxygen $‘n’=\frac{56200}{32}mol$

  $\begin{array}{l}PV=nRT\\ P=\frac{nRT}{V}\\ =\frac{56200mol\times 0.082Latm/molK\times 294K}{32\times 125L}=338.72atm\end{array}$