Chapter 6, Problem 6.95QP

(a)

Interpretation Introduction

Interpretation:

The mass of Helium in the balloon has to be calculated.

Concept Introduction:

Ideal gas law is applicable to those gases which obey Boyle’s law and Charles’s law. The ideal gas equation can be obtained by combining the equations of Boyle’s law and Charles’s law.

At constant temperature, (Boyle’s law)

$\text{Pα}\frac{\text{1}}{\text{V}}$

At constant volume, (Charles’s law)

$\text{PαT}$

By combining the above equations,

$\begin{array}{l}\text{Pα}\frac{\text{T}}{\text{V}}\\ \\ \text{PVαT}\\ \\ \text{PV=RT}\end{array}$

Where R= proportionality constant called as gas constant.

The general equation for ideal gas law is written as,

$\text{PV=nRT}$

Where n= number of moles

Answer to Problem 6.95QP

The mass of Helium in the balloon is $\text{3}{\text{.4×10}}^{\text{5}}\text{g}\text{.}$

Explanation of Solution

Given,

Diameter of balloon is $\text{16}\text{m}$

Temperature of inflation is $\text{18°C}$

The volume of balloon is calculated as,

$\begin{array}{l}\text{Volume}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \\ \text{Volume}\text{=}\frac{\text{4}}{\text{3}}\text{π}{\left(\text{8}\text{m}\right)}^{\text{3}}\\ \\ \text{Volume}\text{=}{\left(\text{2}{\text{.1×10}}^{\text{3}}\text{m}\right)}^{\text{3}}\text{×}\frac{\text{1000}\text{L}}{\text{1}{\text{m}}^{\text{3}}}\\ \\ \text{Volume}\text{=}\text{2}{\text{.1×10}}^{\text{6}}\text{L}\end{array}$

The mass of Helium is calculated as,

$\begin{array}{l}{\text{n}}_{\text{He}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \\ {\text{n}}_{\text{He}}\text{=}\frac{\left(\text{98}\text{.7}\text{kPa×}\frac{\text{1atm}}{\text{1}{\text{.01325×10}}^{\text{2}}\text{kPa}}\right)\left(\text{2}{\text{.1×10}}^{\text{6}}\text{L}\right)}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{mole}\text{K}}\right)\left(\text{273+18}\right)\text{K}}\\ \\ {\text{n}}_{\text{He}}\text{=}\text{8}{\text{.6×10}}^{\text{4}}\text{mole}\text{He}\\ \\ \text{Mass}\text{of}\text{Helium}\text{=}\left(\text{8}{\text{.6×10}}^{\text{4}}\text{mole}\right)\text{×}\frac{\text{4}\text{.003g}\text{He}}{\text{1}\text{mole}\text{He}}\\ \\ \text{Mass}\text{of}\text{Helium}\text{=}\text{3}{\text{.4×10}}^{\text{5}}\text{g}\end{array}$

The mass of Helium in the balloon is $\text{3}{\text{.4×10}}^{\text{5}}\text{g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The work done in joules has to be calculated.

Concept Introduction:

The work done on the system or by the system can be calculated from the equation

$\text{w=-PΔV}$

Where w=work done

P=Pressure

$\text{ΔV}$=Change in volume

Answer to Problem 6.95QP

The work done in joules is $\text{-2}{\text{.0×10}}^{\text{8}}\text{J}\text{.}$

Explanation of Solution

Given,

Diameter of balloon is $\text{16}\text{m}$

Temperature of inflation is $\text{18°C}$

Pressure is $\text{98}\text{.7}\text{kPa}$

The volume of balloon is calculated as,

$\begin{array}{l}\text{Volume}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \\ \text{Volume}\text{=}\frac{\text{4}}{\text{3}}\text{π}{\left(\text{8}\text{m}\right)}^{\text{3}}\\ \\ \text{Volume}\text{=}{\left(\text{2}{\text{.1×10}}^{\text{3}}\text{m}\right)}^{\text{3}}\text{×}\frac{\text{1000}\text{L}}{\text{1}{\text{m}}^{\text{3}}}\\ \\ \text{Volume}\text{=}\text{2}{\text{.1×10}}^{\text{6}}\text{L}\end{array}$

The work done is calculated as,

$\begin{array}{l}\text{w}\text{=}\text{-PΔV}\\ \\ \text{w}\text{=}\text{-}\left(\text{98}\text{.7kPa×}\frac{\text{1atm}}{\text{1}{\text{.01325×10}}^{\text{2}}\text{kPa}}\right)\left(\text{2}{\text{.1×10}}^{\text{6}}\text{L}\right)\left(\frac{\text{101}\text{.3J}}{\text{1L}\text{.atm}}\right)\\ \\ \text{w}\text{=}\text{-2}{\text{.0×10}}^{\text{8}}\text{J}\end{array}$

The work done in joules is $\text{-2}{\text{.0×10}}^{\text{8}}\text{J}\text{.}$