Chapter 8, Problem 26QRT

(a)

Interpretation Introduction

Interpretation:

Amount of Helium gas in a $\text{1}\text{.05}\times {\text{10}}^3\text{ L}$ balloon has to be calculated when the balloon is on the ground.

Concept Introduction:

Ideal gas Equation:

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Pressure can be expressed in different units.  But it is normally reported in

$\text{mm}\text{Hg}\left(\text{millimeters}\text{of}\text{mercury}\right)$

General unit conversions:

  $\begin{array}{l}\text{1}\text{Pa}\text{=}{\text{1Nm}}^{-2}{\text{ = 1 kgm}}^{-1}{s}^{-2};{\text{ 1 kPa = 10}}^{\text{3}}\text{ Pa}\\ \\ \text{1}\text{bar}\text{=}{\text{10}}^{\text{5}}{\text{Pa = 10}}^{\text{3}}{\text{ kPa; mbar = 10}}^{\text{2}}\text{ Pa}\\ \\ \text{1}\text{atm}\text{=}1.01325\times {10}^5\text{ Pa = 101}\text{.325 kPa}\\ \\ 1\text{ atm = 760 Torr = 760}\text{mm}\text{Hg}\\ \\ 1\text{ atm = 14}{\text{.7 lb/in}}^{\text{2}}\left(psi\right)\text{ = 1}\text{.01325 bar}\end{array}$

Answer to Problem 26QRT

Amount of Helium (in moles) is $42.8\text{ mol}\text{.}$

Explanation of Solution

Given data is shown below,

  $\begin{array}{c}V\text{ = 1}\text{.05}\times {\text{10}}^3\text{ L}\\ \\ \text{P = 745 mmHg = 745 mmHg}\times \frac{1\text{ atm}}{\text{760 mmHg}}\\ \\ =\text{ 0}\text{.980 atm}\\ \\ \text{T = 20}°\text{C = }\left(20+273\right)K=\text{293 K}\end{array}$

Amount of Helium (in moles) can be calculated using Ideal gas equation as shown below,

  $\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{0}\text{.980 atm}\right)\left(1.05\times {10}^3\text{ L}\right)\text{= n}\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{293 K}\right)\\ \\ n=\frac{\left(\text{0}\text{.980 atm}\right)\left(1.05\times {10}^3\text{ L}\right)}{\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{293 K}\right)}\\ \\ =42.8\text{ mol He}\end{array}$

Amount of Helium (in moles) is $42.8mol.$

(b)

Interpretation Introduction

Interpretation:

Volume of Helium gas in a balloon has to be determined when the balloon ascends to a height of $2\text{ mil}\text{.}$

Concept Introduction:

General Gas Law:

Combine Charles’s law and Boyle’s law to get the General gas law or combined gas law given below,

  $\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

Answer to Problem 26QRT

Volume of Helium gas is $\text{1}\text{.07}\times {\text{10}}^3\text{ L}\text{.}$

Explanation of Solution

Given data is shown below,

  $\begin{array}{c}{\text{P}}_{\text{1}}\text{= 745 mmHg }\\ \\ {\text{P}}_2\text{= 600 mmHg }\\ \\ {\text{T}}_{\text{1}}\text{=}\text{ 20}°\text{C}=\left(\text{20}+273\right)K\text{ = 293 K}\\ \\ {\text{T}}_2\text{=}-\text{33}°\text{C}=\left(-\text{33}+273\right)K\text{ = 240 K}\\ \\ {\text{V}}_1\text{ = 1}\text{.05}\times {\text{10}}^3\text{ L}\end{array}$

Volume of Helium gas can be calculated using combined gas law that is proposed by combining the Boyle’s law and Charles’ law as follows,

  $\begin{array}{c}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ \\ \frac{\left(\text{745 mmHg}\right)\left(\text{1}\text{.05}\times {\text{10}}^3\text{ L}\right)}{\left(\text{293 K}\right)}\text{=}\frac{\left(\text{600 mmHg }\right){\text{V}}_{\text{2}}}{\left(\text{240 K}\right)}\\ \\ {\text{V}}_{\text{2}}\text{ = }\frac{\left(\text{745 mmHg}\right)\left(\text{1}\text{.05}\times {\text{10}}^3\text{ L}\right)\left(\text{240 K}\right)}{\left(\text{293 K}\right)\left(\text{600 mmHg }\right)}\\ \\ =\text{ 1}\text{.07}\times {\text{10}}^3\text{ L}\end{array}$

Volume of Helium gas is $1.07\times 10^{3}L.$