Courses for Kids
Free study material
Offline Centres
Store Icon

How many grams of \[C{{O}_{2}}\] are in \[520mL\] of carbon dioxide gas at $0$ degrees Celsius and a pressure of \[750.3\] torr?

Last updated date: 20th Jun 2024
Total views: 373.5k
Views today: 4.73k
373.5k+ views
Hint: We know that the ideal gas law states certain assumptions about gases and theory which are unnecessarily false. Therefore this results that ideal gas law has some sort of limitation within it. For example; ideal gas law creates an assumption that the gas particles have neither volume nor are attracted to one another.

Complete step-by-step answer:
Firstly we have to put mass of \[C{{O}_{2}}\] under the given conditions $1.01g$
After that we have to determine \[mol\text{ }C{{O}_{2}}\] by using Ideal gas laws and multiplying \[mol\text{ }C{{O}_{2}}\]by its molar mass $44.0098\dfrac{g}{mol}$ . Whereas the ideal gas law equation is given by: $PV=nRT$.

Here we have $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is the gas constant and $T$ is temperature in Kelvins and the gas constant includes volume of unit in liters and the volume milliliters would be converted into liters.
 Given :
\[\Rightarrow P\text{ }=\text{ }750.3\text{ }Torr\]
\[\Rightarrow V\text{ }=\text{ }520mL\text{ }\times \text{ }\dfrac{1L}{1000mL}~~=\text{ }0.520\text{ }L\]
\[\Rightarrow R\text{ }=~~62.364\text{ }L\text{ }\dfrac{Torr}{Kmol}~\]
\[\Rightarrow T~=~0{}^\circ C~+~273.15~=~273\text{ }K\]
Now that we know the formula or as we say ideal gas equation; $PV=nRT$
Thus from here we can determine the formula for $n$ we get formula for $n=\dfrac{R\cdot T}{P\cdot V}$ and now by substituting the values we get:
$\Rightarrow n=\dfrac{750.3Torr\times 0.520L}{62.364L\dfrac{Torr}{Kmol}\times 273K}=0.02298mol\cdot C{{O}_{2}}$
Where mass of \[C{{O}_{2}}\] can be given by, multiplying the value of $n$ with $\dfrac{44gC{{O}_{2}}}{1molC{{O}_{2}}}$ we get;
$\Rightarrow 0.02298mol\cdot C{{O}_{2}}\times \dfrac{44gC{{O}_{2}}}{1molC{{O}_{2}}}$
$\Rightarrow 1.01g\cdot C{{O}_{2}}$
Therefore, $1.01$ grams of \[C{{O}_{2}}\] are in \[520mL\] of carbon dioxide gas at $0$ degrees Celsius and a pressure of \[750.3\] torr.

Note: Note that since this particle of ideal gases have neither volume therefore gas should be able to get condensed to the volume of zero. Whereas real gaseous particle that occupy space. A gaseous state will be more and more condensed in order to form liquidity and has volume. The gaseous law have no longer application which is because of substance is no longer in a gaseous state.