Question

A $20.0-\mathrm{L}$ cylinder containing $11.34\mathrm{k}\mathrm{g}$of butane, ${\mathrm{C}}_4{\mathrm{H}}_{10}$, is opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if the gas escapes until the pressure in the cylinder is equal to the atmospheric pressure, $0.983$atm,ata temperature of $27^{{}^{\circ }}\mathrm{C}.$

Short answer

The mass of the remaining gas in the cylinder is $46.4$ grams.

Step-by-step solution

Definition of ideal gas

The general gas equation is known to be an ideal gas law. Despite its flaws, the ideal gas law provides a goodapproximation of the behaviour of various gases under a variety of situations.

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,

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

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

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

Finding number of moles

Here, we have to find the amount (or) number of moles of gas present after the cylinder is opened.So, we just consider the given conditions when the cylinder is opened.

$\begin{array}{r}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}(\mathrm{V})=20\mathrm{L}.\\ \mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}(\mathrm{P})=0.983\mathrm{a}\mathrm{t}\mathrm{m}\\ \mathrm{T}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}(\mathrm{T})=27^{{}^{\circ }}\\ \mathrm{A}\mathrm{m}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{g}\mathrm{a}\mathrm{s}(\mathrm{n})=?\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\end{array}$

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

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

So,

$\begin{array}{r}27^{{}^{\circ }}=(27)+273.\\ 15\mathrm{K}=300.\\ 15\mathrm{K}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\end{array}$

Number of moles of gas

$(n)=\frac{PV}{RT}$

$\begin{array}{r}=\frac{\mathrm{R}\mathrm{T}}{(0.08206)\times (300.15)}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}.\\ =\frac{19.66}{24.63}\\ =0.8\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}.\end{array}$

Calculating final massof remaining gas in cylinder

We need to convert $0.8$ moles of${\mathrm{C}}_4{\mathrm{H}}_{10}$ into the equivalent mass figure. Atomic weight of carbon $\left(C\right)=12$g.

Atomic weight of hydrogen $(\mathrm{H})=1$ g.

So, molecular weight of ${\mathrm{C}}_4{\mathrm{H}}_{10}=(4)\cdot (12)+(10)\cdot (1)$ g

$\begin{array}{r}=48+10\\ =58\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}\end{array}$

i.e. 1 mol${\mathrm{C}}_4{\mathrm{H}}_{10}=58$ g${\mathrm{C}}_4{\mathrm{H}}_{10}$.

$0.8$mol$\mathrm{C}{\mathrm{O}}_2=(0.8)\cdot (58)$ g${\mathrm{C}}_4{\mathrm{H}}_{10}$.

$=46.4$g${\mathrm{C}}_4{\mathrm{H}}_{10}$.

Hence, the mass of the remaining gas in the cylinder is $46.4$ grams.