Question

A high altitude balloon is filled with $1.41\times 10^4\mathrm{L}$ of hydrogen at a temperature of $21^{{}^{\circ }}\mathrm{C}$ and a pressure of745 torr. What is the volume of the balloon at a height of20Km, where the temperature is $-48^{{}^{\circ }}\mathrm{C}$ and the pressure is $63.1\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$?

Short answer

The final volume of balloon at a height of $20\mathrm{K}\mathrm{m}$is$12.74\times 10^4\mathrm{L}$.

Step-by-step solution

Definition of ideal gas 

The ideal gas law, or the universal gas equation, is a state equation for a hypothetical ideal gas. Despite its shortcomings, the ideal gas law approximates the behaviour of many gases in a number of settings rather well.

Explanation for ideal gas law

The ideal gas law connects pressure, volume, temperature, and the number of moles in a gas.It is provided by

$\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}$

where,

$\begin{array}{r}\mathrm{P}=\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{s}.\\ \mathrm{V}=\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{s}.\\ \mathrm{n}=\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{s}.\\ \mathrm{T}=\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{s}.\end{array}$

and

$\begin{array}{r}\mathrm{R}=\mathrm{I}\mathrm{d}\mathrm{e}\mathrm{a}\mathrm{l}\mathrm{g}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\\ =0.08206\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\end{array}$

ifthe amount of the gas $(\mathrm{n})$is in moles, temperature $(\mathrm{T})$ is in kelvin $\left(\mathrm{K}\right),$and pressure $(\mathrm{P})$is in atm.

Understanding Combined Gas Law

So, we have the Combined Gas Law with a constant number of moles of an ideal gas under two distinct circumstances.

$\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}$

$\begin{array}{r}\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\mathrm{n}\mathrm{R}\\ \frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}=\mathrm{n}\mathrm{R}\end{array}$

i.e.$\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

where,

${\mathrm{V}}_1,{\mathrm{V}}_2=$Initial and final volumes,

${\mathrm{P}}_1,{\mathrm{P}}_2=$Initial and final pressures, and

${\mathrm{T}}_1,{\mathrm{T}}_2=$Initial and final absolute temperatures.

Evaluating final volume

However, the temperatures provided are in degrees Celsius, not kelvins (absolute scale).

As a result, we must convert temperatures from degrees Celsius to kelvins. We have

$0^{\circ }\mathrm{C}=\left(0\right)+273.15K$

So,

$\begin{array}{r}21^{{}^{\circ }}=(21)+273.15\mathrm{K}\\ =294.15\mathrm{K}-48^{{}^{\circ }}\\ =(-48)+273.15\mathrm{K}\\ =225.15\mathrm{K}\end{array}$

So,

$\begin{array}{r}\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\left({\mathrm{V}}_1\right)=11.2\mathrm{L}\\ \mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{T}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\left({\mathrm{T}}_1\right)=294.15\mathrm{K}\\ \mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\left({\mathrm{P}}_1\right)=745\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\\ \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\left({\mathrm{V}}_2\right)=?\mathrm{L}\\ \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{T}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\left({\mathrm{T}}_2\right)=225.15\mathrm{K}\\ \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\left({\mathrm{P}}_2\right)=63.1\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}.\end{array}$

$\begin{array}{r}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}\\ \frac{(745)\times \left(1.41\times 10^4\right)}{294.15}=\frac{(63.1)\times \left({\mathrm{V}}_2\right)}{225.15}\\ \frac{1050.45\times 10^4}{294.15}=\frac{(63.1)\times \left({\mathrm{V}}_2\right)}{225.15}\\ 35711.37=\frac{(63.1)\times \left({\mathrm{V}}_2\right)}{225.15}\\ {\mathrm{V}}_2=\frac{(35711.37)\times (225.15)}{63.1}\\ {\mathrm{V}}_2=\frac{804.04\times 10^4}{63.1}\\ =127423.381\mathrm{L}.\\ =12.74\times 10^4\mathrm{L}.\end{array}$

Therefore, the final volume of balloon at$20\mathrm{K}\mathrm{m}$is$12.74\times 10^4\mathrm{L}$.