Question

When 2 g of gas A are introduced into an evacuated flask kept at ${25^ \circ }C$, the pressure is found to be one atmosphere, if 3 g of another gas B are then added to the same flask, the total pressure becomes 1.5 atm. Assuming the ideal gas behaviour, calculate the ratio of molecular weights ${M_A}:{M_B}$.

Answer 1

Hint: Gaseous state is the less ordered state. The molecules in gases are separated from each other by large distances. The intermolecular forces are the weakest in them. The gases have all three types of motion i.e. the translatory, rotatory, and vibratory and possess high kinetic energy. Various laws are there to explain the properties of gases (Boyle’s law, Charlie’s law, Gay-Lussac’s law, Avogadro’s law).

Complete step by step answer:
In this question, the weight and the pressure of the two gases A and B is given and we have to calculate the ratio of the molecular weights of the two gases.

For an ideal gas, we know that,

$PV = nRT$

$ \Rightarrow PV = \dfrac{w}{M}RT$

$ \Rightarrow M = \dfrac{{wRT}}{{PV}}$

(We have modified the ideal gas equation and make the above formula to calculate the molecular weight of the gases)

Now,For gas A

Pressure $ = 1$ atm,Weight of the gas $ = 2$ g

$ \Rightarrow {M_A} = \dfrac{{wRT}}{{PV}}$

$ \Rightarrow {M_A} = \dfrac{{2 \times RT}}{{1 \times V}} \to (1)$

Similarly,

For gas B

Weight of the gas $ = 3$ gAfter adding 3 g of gas B the total pressure is found to be 1.5 atm .So Pressure of the gas B $ = $ Total pressure – Pressure of gas B

$ = 1.5 - 1$

$ = 0.5$ atm

${M_B} = \dfrac{{wRT}}{{PV}}$

$ \Rightarrow {M_B} = \dfrac{{3 \times RT}}{{0.5 \times V}} \to (2)$

The ratio of the molecular weights of gas A and gas B is

$\dfrac{{{M_A}}}{{{M_B}}} = \dfrac{{2 \times RT}}{{1 \times V}} \times \dfrac{{0.5 \times V}}{{3 \times RT}}$ (Here V and RT will be cancelled as they are same in both cases)

$ = \dfrac{{2 \times 0.5}}{3}$

$ = \dfrac{{2 \times 5}}{{3 \times 10}}$

$ = \dfrac{1}{3}$

${M_A}:{M_B} = 1:3$

(We have put the given equations (1) and (2) and after calculation, we got the value for the ratio of the molecular weights of the gas A and gas B).

So, the correct answer is Option A .

Note: Properties of gases.
1,Shape and Volume: - Since particles of gases are not held in fixed positions and move freely gases neither have definite shapes nor volumes. They fill the whole space available to them and take up the shape and volume of the vessel in which they are enclosed.

2.Homogeneous nature: - All the parts of a gas or a gaseous mixture composition throughout.

3.Compressibility: - On increasing pressure, gases can be readily compressed due to the presence of large empty spaces.

4.Random motion: - The molecules or atoms of a gas are in a state of continuous zig-zag motion in all directions.

5.Pressure: - Due to their random motion, the molecules of the gas collide on the walls of the container and thus exert a certain force on the walls of the container.
Diffusion: - Gases mix (diffuse) with each other freely due to the free movement of their molecules.
Density: - Due to the large separation of molecules, gases have a large volume and thus low densities.