Question

A room has a volume of \[220{m^3}\].
a.Calculate the amount in moles, of air particles in the room at \[{\text{2}}{{\text{3}}^{\text{o}}}{\text{C}}\] and a pressure of \[100kPa\].
b.Assume that \[20\% \] of the molecules in the air are oxygen molecules and the remaining molecules are nitrogen. Calculate the mass of air in the room.

Answer 1

Hint: Here, we will be using the ideal gas equation. So, looking briefly into that we should know that in chemistry, the state equation for a hypothetical ideal gas is \[PV = nRT\]. The ideal gas law shows how the behaviour of an ideal gas sample is connected to the gas sample's pressure (P), temperature (T), volume (V), and molarity (n). In the equation \[PV = nRT\], the term \[{\text{R}}\] stands for the universal gas constant.

Complete step by step answer:
a.
We have, \[PV = nRT\]
Where \[{\text{P}}\]= pressure in atm
\[{\text{V}}\]=volume in L
\[R = 0.082057Latm/mol\]
\[{\text{T}}\]=temperature in K
From the question,
\[V = 220{m^3}\]
\[ = 220{m^3} \times {\left( {\dfrac{{10dm}}{{1m}}} \right)^3}\]
On simplification we get,
\[ = 220000d{m^3} = 220000L\]
The pressure of the room is given by,
\[P = 100kPa \times \dfrac{{{{10}^3}Pa}}{{1kPa}} \times \dfrac{{1atm}}{{101325Pa}}\]
On simplification we get,
\[ = 0.987atm\]
Thus, at \[100kPa\] and \[23^\circ C\], the mol of air in the room is given by
\[n = \dfrac{{PV}}{{RT}}\]
Now we can substitute the given values we get,
\[ = \dfrac{{0.987atm \times 220000L}}{{0.082057L.atm/mol.K.(23 + 273.15K)}}\]
On simplification we get,
\[ = 8934.67mol\]
We know that one mole of anything has \[6.022 \times {10^{23}}\]particles.
So, the number of air particles present
\[ = 8934.67 \times 6.022 \times {10^{23}} = 5.4 \times {10^{27}}particles\]
b.
We can now find the mass of air in the room from the moles of air we found. According to the question, we assume that \[20\%\] is oxygen and \[80\%\] is nitrogen.
Then, molar mass of air is,
\[8934.67molair \times \dfrac{{28.812gair}}{{mol}}\]
On simplification we get,
\[ = 2.6 \times {10^5} {\text{gair or 260kgair}}\]

Note:
In the above calculation, we have taken some assumptions and it must be noted. It is given in the question that the room has a volume of \[220{m^3}\]. From this, we can thus assume that the air takes up the entire room which means the air also has a volume of \[220{m^3}\] . The room is also assumed to be empty for this calculation. The last assumption we have made is that the air contains only ideal gases, which is highly probable.