Question

A balloon that is $100.21\mathrm{L}$ at ${21}^{\circ }\mathrm{C}$and $0.981$$\mathbf{a}\mathbf{t}\mathbf{m}$ is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is $144.53\mathrm{L}$ at a temperature of ${5.24}^{\circ }\mathrm{C}$, what is the pressure experienced by the balloon as it clears Mount Crumpet?

Short answer

Pressure experienced by the balloon as it clears Mount Crumpet is $0.643\mathrm{a}\mathrm{t}\mathrm{m}.$

Step-by-step solution

Definition of volume

A substance's volume is the amount of space it occupies.

Use ideal gas law

We know that Ideal gas law relates pressure, volume, temperature and number of moles of a gas.

It is given by

$PV=nRT$

where,

$P$= pressure of the gas.

$V$= volume of the gas.

$n$= number of moles of the gas.

$T$= temperature of the gas.

$R=$Ideal gas constant$=0.08206L\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{K}^{-1}$.

If, amount of the gas$(n)$is in moles , temperature$(T)$is in Kelvin$(K)$, pressure$(P)$is in atm.

So, with constant number of moles of an ideal gas at two different conditions we have Combined Gas law

$PV=nRT$

$\begin{array}{rl}\frac{P_1{V}_1}{T_1}& =nR\\ \frac{P_2{V}_2}{T_2}& =nR\end{array}$

So,$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Therefore,

$V_1,V_2=$Initial and Final Volumes.

$P_1,P_2=$Initial and Final Pressures.

$T_1,T_2=$Initial and Final Absolute Temperatures.

Step 3: Convert the temperatures in degree Celsius to kelvins.

Let us solve the given problem.

We know that given temperatures are in degree Celsius but not in kelvins (absolute scale).

So, we have to convert the temperatures in degree Celsius to kelvins.

We have, $0^{\circ }C=(0)+273.15K$

So,

$\begin{array}{rl}{21}^{\circ }C& =(21)+273.15K\\ & =294.15K\\ {5.24}^{\circ }C& =(5.24)+273.15K\\ & =278.39K\end{array}$

Therefore,

$\begin{array}{rlrlr}\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\left(V_1\right)& =100.21\mathrm{L}.& & \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\left(V_2\right)& =144.53\mathrm{L}.\\ \mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{T}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\left(T_1\right)& =294.15\mathrm{K}.& \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{T}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\left(T_2\right)& =278.39\mathrm{K}.\\ \mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\left(P_1\right)& =0.981\mathrm{a}\mathrm{t}\mathrm{m}& & \mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\left(P_2\right)& =?\mathrm{a}\mathrm{t}\mathrm{m}.\end{array}$

 Step 4: Calculate pressure experienced by the balloon

Consider the given problem and solve.

$\begin{array}{rl}PV& =nRT\\ \frac{P_1{V}_1}{T_1}& =\frac{P_2{V}_2}{T_2}\\ \frac{(0.981)\cdot (100.21)}{294.15}& =\frac{\left(P_2\right)\cdot (144.53)}{278.39}\\ \frac{98.306}{294.15}& =\frac{\left(P_2\right)\cdot (144.53)}{278.39}\\ 0.334& =\frac{\left(P_2\right)\cdot (144.53)}{278.39}\end{array}$

Final pressure

$\begin{array}{rl}{P}_2& =\frac{(0.334)\cdot (278.39)}{144.53}\\ P_2& =\frac{92.982}{144.53}\\ & =0.643\mathrm{a}\mathrm{t}\mathrm{m}.\end{array}$

Therefore, calculated pressure is $0.643\mathrm{a}\mathrm{t}\mathrm{m}.$