Question

When $2g$ of a gaseous substance A is introduced into an initially evacuated flask at $25{}^\circ C$, the pressure is found to be $1$ atmosphere. $3g$ of another gaseous substance B is then added to it at the same temperature and pressure. The final pressure is found to be $1.5$ atmosphere. Calculate the ratio of molecular masses of A and B assuming ideal gas behaviour.

Answer 1

The Ideal Gas Law is a simple equation demonstrating the relationship between temperature, pressure, and volume for gases. These specific relationships stem from Charles’s Law, Boyle’s Law, and Gay-Lussac’s Law.Charles’s Law identifies the direct proportionality between volume and temperature at constant pressure, Boyle’s Law identifies the inverse proportionality of pressure and volume at a constant temperature, and Gay-Lussac’s Law identifies the direct proportionality of pressure and temperature at constant volume. Combining, these form the Ideal Gas Law equation: $$PV{ }={ }nRT$$$P$is the pressure,$V$is the volume,$n$is the number of moles of gas,$R$is the universal gas constant, and$T$ is the absolute temperature.

Complete step by step answer:
Let the molecular masses of A and B be ${{M}_{A}}~$ and ${{M}_{B}}$ ​ respectively.
Pressure exerted by the gas B$=\left( 1.5-1.0 \right)=0.5~atm$
Volume and temperature are the same in both the gases.
Now we will write all the values which are given to us
For gas A: $P=1~atm$, which is the initial pressure of the system.
$w=2~g$, which is the weight of the gaseous substance provided to us,
$M={{M}_{A}}$, which is the molecular mass of the substance which is unknown,
We know that , PV=$$$\dfrac{w}{M}RT$ Where $R$ is the gas constant whose value of known to us, since it is a constant, and the $T$ denotes the temperature of the system, which is also provided to us, w$$ is the given mass, and $M$ is the molar mass, $P$ is the pressure as usual and the $V$ denotes the volume of the gas.
So, on substituting the values , we get

$$1\times V=$$$\dfrac{2}{MA}RT$
Or

$${{M}_{A}}=$$$\dfrac{2RT}{V}$ ………(i)

Now we will do the same for gas B, and take the ratio of molecular masses of both the gasses.

For gas B: $P=0.5{ }atm$, the pressure of gas B was also given in the question

$W=3g$ is the given weight of the gas,

$M={{M}_{B}}$ is the molecular mass of the gas B,

$$0.5\times V=$$$\dfrac{3}{MB}RT$

Or

$${{M}_{B}}=$$$\dfrac{3RT}{0.5\times V}$ ……..(ii)

Now we will take the ratio of both the molecular gasses by dividing both of them,

Dividing equation(i) by equation(ii) we get,

$\dfrac{MA}{MB}$ =$\dfrac{2RT}{V}\times \dfrac{0.5\times V}{3RT}$

=$\dfrac{2\times 0.5}{3}$ =$\dfrac{1}{3}$

So, the ratio of molecular masses of both the gases came out to be, ${{M}_{A}}:{{M}_{B}}=1:3$.

Note: For a gas to be “ideal” there are following assumptions :
1.The gas particles have negligible volume.
2.The gas particles are equally sized and do not have intermolecular forces (attraction or repulsion) with other gas particles.
3.The gas particles move randomly in agreement with Newton’s Laws of Motion.
4.The gas particles have perfect elastic collisions with

ideal gases are strictly a theoretical conception, real gases can behave ideally under certain conditions. Systems that have either very low pressures or high temperatures enable real gases to be estimated as “ideal.” The low pressure of a system allows the gas particles to experience less intermolecular forces with other gas particles. Similarly, high-temperature systems allow for the gas particles to move quickly within the system and exhibit less intermolecular forces with each other. Therefore, for calculation purposes, real gases can be considered “ideal” in either low pressure or high-temperature systems.