Chapter 13, Problem 61A

Interpretation Introduction

Interpretation:

On the basis of given conditions, the balloon should be bust or not is to be explained.

Concept introduction:

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

  $\text{PV}\text{= nRT}$

Where, P = pressure of the gas

V = Volume of the gas

R = Gas constant

T = temperature of the gas

Answer to Problem 61A

Yes, the balloon will bust because the volume at high altitude at a particular pressure is $\text{3L}$ which exceeds the manufacture’s design restraints.

Explanation of Solution

Given information:

The Pressure at sea level, ${\text{P}}_{\text{1}}\text{=}\text{1atm}\text{=}\text{760}\text{mmHg}$

The Volume of the gas in the balloon at this pressure, ${\text{V}}_{\text{1}}\text{=}\text{2}\text{.0L}$

The Pressure at a particular aptitude, ${\text{P}}_{\text{2}}\text{=}\text{500}\text{mmHg}$

Or, ${\text{P}}_{\text{2}}\text{=}\text{0}\text{.638}\text{atm}$

Now according to the ideal gas equation for two conditions at a constant temperature

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{nRT}}_{\text{1}}\\ \text{And,}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\text{=}{\text{nRT}}_{}\end{array}$

So, on comparing these equations to each other, we get

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\begin{array}{l}\text{2}\\ \end{array}}\\ {\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}\\ \text{=}\frac{\text{1atm}\text{×2}}{\text{0}\text{.658}}\\ \text{=}\text{3L}\end{array}$

This volume is greater than the balloon’s capacity. So, the balloon will burst.