Question

What is the volume of the container that holds $25.0g$ of carbon gas at $STP$ ?

Answer 1

In the above mentioned question , we will find out the volume by using the ideal gas law equation, and we will also discuss at what condition we generally use the ideal gas law. And we will discuss more about it.

Complete step by step answer:
Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation ${\mathbf{PV}} = {\mathbf{nRT}}$
$P$ represents pressure (may have units of $atm$ , depending on the units of the universal gas constant)
$V$ represents volume (must have units of liters)
$n$ represents the number of moles
$R$ is the proportionality constant (has units of $\dfrac{{L \times \,atm}}{{mol \times \,k}})$
$T$ represents the temperature, which must be in $Kelvins$ .
Next, list your known and unknown variables. Our only unknown is the volume of $C(g$ ) . Our known variables are $P,n,R,{\text{ }}and{\text{ }}T$ .
At $STP$ , the temperature is $273K$ and the pressure is $1{\text{ }}atm$ . The proportionality constant, $R$ , is equal to $0.0821\dfrac{{L \times \,atm}}{{mol \times \,K}}$ The only issue is the mass of $C\left( g \right),$ we need to convert it into moles of $C\left( g \right)\;$ in order to use the $ideal{\text{ }}gas{\text{ }}law.$
$\;25.0g \times \dfrac{{1molC}}{{12.01g}}\, = \,2.08\,mol\,C$ Now all we have to do is rearrange the equation and solve for $V$ like so:
$V = \dfrac{{n \times R \times T}}{P}$
$\Rightarrow \,\,\,\,\,\,\,V = \dfrac{{2.08\,\,mol \times 0.0821\dfrac{{L \times atm}}{{mol \times K}} \times \,(273K)}}{{1atm}}$
$\therefore V = 46.6L$

Note:
The ideal gas law allows us to calculate the value of the fourth quantity $(P,{\text{ }}V,{\text{ }}T,{\text{ }}or{\text{ }}n)$ needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of $P,{\text{ }}V,{\text{ }}T,{\text{ }}and{\text{ }}n$ ) are known.