Question

A container is filled with a gas at pressure ${P_0}$. Find the pressure of the gas if the mass of the molecules is halved and their rms speed is doubled.
A) $2{P_0}$
B) $4{P_0}$
C) $\dfrac{{{P_0}}}{4}$
D) $\dfrac{{{P_0}}}{2}$

Answer 1

Hint: The molecules in the container collide with each other and with the walls of the container thus exerting pressure on the walls of the container. Hence, the molecules move with different velocities. Thus the average speed of the molecules is considered. The rms speed ${v_{rms}}$ refers to the square root of the mean of the squared speed $\overline {{v^2}} $.

Formula used:
The pressure of the gas enclosed in a container is given by, $P = \dfrac{1}{3}Nm\overline {{v^2}} $ where $N$ is the number of molecules present in the container, $m$ is the mass of the molecules and $\overline {{v^2}} $ is the mean of the squared speed.

Complete step by step answer:
Step 1: Express the relation for the pressure of gas enclosed in a container.
Given, the pressure of the gas is ${P_0}$.
Then, the pressure of the gas molecules is given by, ${P_0} = \dfrac{1}{3}Nm\overline {{v^2}} $, where $N$ is the number of molecules present in the container, $m$ is the mass of the molecules and $\overline {{v^2}} $ is the mean of the squared speed.
Let ${v_{rms}}$ be the rms speed of the molecule. Since, the square root of the mean of the squared speed $\overline {{v^2}} $ is called the root mean square speed ${v_{rms}}$ we have, $v_{rms}^2 = \overline {{v^2}} $ .
Now the expression for pressure becomes ${P_0} = \dfrac{1}{3}Nmv_{rms}^2$ -------- (1).
Thus the pressure of the gas is directly proportional to the product of the mass of the molecules and the square of their rms speed.
Step 2: Find the pressure when mass is halved and rms speed is doubled using equation (1).
Let ${P_0}'$ be the new pressure of the gas when the mass of the molecules is halved and their rms speed double.
From equation (1), we get ${P_0}' = \dfrac{1}{3}nVmv_{rms}^2$ --------- (2).
Now mass is halved i.e., $m$ will be $\dfrac{m}{2}$ and rms speed is doubled i.e., ${v_{rms}}$ will be $2{v_{rms}}$ .
Substituting the above changes in mass and rms speed in equation (2) we get, ${P_0}' = \dfrac{1}{3}nV\left( {\dfrac{m}{2}} \right){\left( {2{v_{rms}}} \right)^2}$
or, ${P_0}' = \dfrac{1}{3}nV\left( {\dfrac{m}{2}} \right)4v_{rms}^2$
Reducing the above equation we get ${P_0}' = 2 \times \dfrac{1}{3}nVmv_{rms}^2 = 2{P_0}$

Therefore, when the mass of the molecules is halved and their rms speed doubled, the pressure will be $2{P_0}$.

Additional information:
According to the kinetic theory of gases, the molecules in the container have the same average kinetic energy. As a result, lighter molecules will move faster than heavier molecules.

Note:
The number of molecules present in the container is given by, $N = nV$ , where $n$ is the number of molecules per unit volume and $V$ is the volume of gas. A change in the mass of the molecules will not cause a change in the volume of gas. Thus $N$ remains constant.