Question

Pressure of ${\text{1g}}$ an ideal gas at ${\text{2}}{{\text{7}}^{\text{o}}}{\text{C}}$ is found to be${\text{4 bar}}$. When ${\text{2g}}$ of another ideal gas B is introduced in the same flask at the same temperature, the pressure becomes ${\text{6 bar}}$. Find the relationship between their molecular masses.

Answer 1

To answer this question, you should recall the concept of the ideal gas equation. Ideal gases are the gases which have elastic collisions between their molecules and there are no intermolecular attractive forces. In reality, there is no such thing as ideal gases. The gases just show ideal behaviour under certain conditions of temperature and pressure.
The formula used:
${\text{PV}} = {\text{nRT}}$ where P is pressure, V is volume, R is the universal gas constant, n is no. of moles and T is temperature

Complete step by step answer:
Here, it is assumed that:
The ideal gases are made up of molecules which are in constant motion in random directions.
The molecules of an ideal gas behave as rigid spheres.
All the collisions are elastic.
The temperature of the gas is directly proportional to the average kinetic energy of the molecules.
Pressure occurs due to the collision between the molecules with the walls of the container.
Let ${\text{M}}$ and ${\text{M'}}$ be the molar masses of ideal gases A and B respectively.
The number of moles of gas A and B is $\dfrac{1}{{\text{M}}}$ and $\dfrac{2}{{{\text{M'}}}}$ ​ respectively.
Let P and P’ be the pressures of gases A and B respectively:
${\text{P}} = 2\;{\text{bar}}$.
Total pressure ${\text{P}} + {\text{P}}\prime = 3\;{\text{bar}}$.
We can calculate the pressure P’ as: ${\text{P}}\prime = 3 - 2 = 1\;{\text{bar}}$
The ideal gas equations for two gases A and B are:
${\text{PV}} = {\text{nRT }}\left( i \right)$
${\text{P'V}} = {\text{n'RT}}\left( {ii} \right)$
Divide equation (i) by equation (ii):
$\dfrac{{\text{P}}}{{{\text{P'}}}}{\text{ }} = \dfrac{{\text{n}}}{{{\text{n'}}}} = \dfrac{{1 \times {\text{M}}\prime }}{{2 \times {\text{M}}}} = \dfrac{{{\text{M}}\prime }}{{2{\text{M}}}}$
Rearranging and solving
$\dfrac{{{\text{M}}\prime }}{{\text{M}}}{\text{ }} = 2\dfrac{{\text{P}}}{{{\text{P}}\prime }} = \dfrac{{2 \times 2}}{1} = 4$

The final value will be calculated as ${\text{M}}\prime = 4{\text{M}}$

Note:
Unless mentioned otherwise, we always assume the gas as an ideal gas.The five gas laws are:
Boyle’s Law establishes a relationship between the pressure and the volume of a gas.
Charles’s Law establishes a relationship between the volume occupied by a gas and the absolute temperature.
Gay-Lussac’s Law establishes a relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas.
Avogadro’s Law establishes a relationship between the volume occupied by a gas and the amount of gaseous substance.
After combining these four aforementioned laws we arrive at the Combined Gas Law