Question

1 L of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$and $\text{ }\dfrac{7}{8}\text{ L }$ of $\text{ }{{\text{O}}_{\text{2}}}\text{ }$at the same temperature and pressure were mixed together. What is the relation between the masses of the two gases in the mixture?
A) $\text{ }{{\text{M}}_{{{\text{N}}_{\text{2}}}}}\text{ = 3}{{\text{M}}_{{{\text{O}}_{\text{2}}}}}\text{ }$
B) $\text{ }{{\text{M}}_{{{\text{N}}_{\text{2}}}}}\text{ = 8}{{\text{M}}_{{{\text{O}}_{\text{2}}}}}\text{ }$
C) $\text{ }{{\text{M}}_{{{\text{N}}_{\text{2}}}}}\text{ = }{{\text{M}}_{{{\text{O}}_{\text{2}}}}}\text{ }$
D) $\text{ }{{\text{M}}_{{{\text{N}}_{\text{2}}}}}\text{ = 16}{{\text{M}}_{{{\text{O}}_{\text{2}}}}}\text{ }$

Answer 1

Hint: At constant pressure and temperature, the volume of the gas is directly related to the number of moles of the gas,
$\text{ V }\propto \text{ n }\propto \text{ }\dfrac{\text{m}}{\text{M}}\text{ }$
Where, m is the weight of the gas and M is the molar mass of the gas. For the gaseous mixture of oxygen and nitrogen, the ideal gas equation establishes a relation between volume and the number of moles of gas.

Complete step by step solution:
We are given the following data:
The volume of nitrogen gas $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ is $\text{ }{{\text{V}}_{{{\text{N}}_{\text{2}}}}}\text{ = 1 L }$
The volume of oxygen gas $\text{ }{{\text{O}}_{\text{2}}}\text{ }$ is $\text{ }{{\text{V}}_{{{\text{O}}_{\text{2}}}}}\text{ = }\dfrac{7}{8}\text{ L }$
We have to find the relation between the masses of two gasses in the mixture of oxygen and nitrogen.
The ideal gas equation relates the variable of the gaseous system the ideal gas equation is given as follows,
$\text{ PV = nRT }$
Where P is the pressure, V is the volume of gas, n is the number of moles of gases, T is the absolute temperature and R is the gas constant.
Here, n is the number of moles, and the mole is given as,
$\text{ n = }\dfrac{\text{Mass}}{\text{molar mass}}\text{ = }\dfrac{\text{m}}{\text{M}}\text{ }$
The ideal gas equation becomes,
$\text{ PV = }\left( \dfrac{\text{m}}{\text{M}}\text{ } \right)\text{RT }$
We are interested to find out the relation between the masse of nitrogen and oxygen .Thus write down the ideal gas equation for the nitrogen and oxygen. The equations are as follows,
a) For nitrogen gas:
$\text{ P}{{\text{V}}_{{{\text{N}}_{\text{2}}}}}\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{\text{M}}\text{ } \right)\text{RT }$
We have given the volume of the nitrogen $\text{ }{{\text{V}}_{{{\text{N}}_{\text{2}}}}}\text{ = 1 L }$and the molar mass of nitrogen is 28. Thus the equation becomes,
$\text{ P}\times \text{1 = }\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{28}\text{ } \right)\text{RT }$ (1)
b) For oxygen gas:
\[\text{ P}{{\text{V}}_{{{\text{O}}_{\text{2}}}}}\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}}{\text{M}}\text{ } \right)\text{RT }\]
We have given the volume of the oxygen $\text{ }{{\text{V}}_{{{\text{O}}_{\text{2}}}}}\text{ = }\dfrac{7}{8}\text{ L }$and the molar mass of oxygen gas is 32. Thus the equation becomes,
$\text{ P}\times \text{ }\dfrac{7}{8}\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}}{28}\text{ } \right)\text{RT }$ (2)
Divide the equation (1) by (2) .we get,
$\begin{align}
& \text{ }\dfrac{ \text{1}}{ \text{ }\dfrac{7}{8}\text{ }}\text{ = }\dfrac{\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{28} \right)\text{}}{\left( \dfrac{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}}{32} \right)\text{}}\text{ } \\
& \Rightarrow \dfrac{8}{7}\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{28} \right)\times \left( \dfrac{32}{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}} \right)\text{ } \\
\end{align}$
On further solving .we get,
$\begin{align}
& \text{ }\dfrac{8\times 28}{7\times 32}\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}} \right)\text{ } \\
& \Rightarrow 1\text{ = }\left( \dfrac{{{\text{m}}_{{{\text{N}}_{\text{2}}}}}}{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}} \right) \\
& \therefore \text{ }{{\text{m}}_{{{\text{N}}_{\text{2}}}}}\text{ = }{{\text{m}}_{{{\text{O}}_{\text{2}}}}}\text{ } \\
\end{align}$
Thus, the mass of oxygen gas and nitrogen gas is equal in the gas mixture. That is, \[{{\text{m}}_{{{\text{N}}_{\text{2}}}}}\text{ = }{{\text{m}}_{{{\text{O}}_{\text{2}}}}}\text{ }\]

Hence, (C) is the correct option.

Note: Note that, if the pressure, temperature of the gas mixture is constant then the volume of the gas in the gas mixture is directly related to the number of moles of the gases. According to Avogadro's law at constant temperature and pressure the gas mixture has the same number of molecules of the same mass of gas. Thus here, the amount of nitrogen and oxygen is equal.