Question

An ideal gas at pressure ${P_o}$ in a vessel. If the masses of all the molecules are halved and their RMS speeds doubled, then the resultant pressure $P$ will be:
A. $4{P_o}$
B. $2{P_o}$
C. ${P_o}$
D. $\dfrac{{{P_o}}}{2}$

Answer 1

RMS speed means Root Mean Square Speed. It is defined as the average speed of gas particles. Root Mean Square Speed is inversely proportional to the molecular mass of gas. By applying this formula according to the instruction as given in the question we can get the resultant pressure.

Complete step by step answer:
According to the question, we have to calculate the root mean square speed of a gas at pressure $P$ when the masses of all the gas molecules are halved and the root mean square speed at pressure ${P_o}$ is doubled.
As we know the formula of RMS speed is given as:
${V_{RMS}} = \sqrt {\dfrac{{3RT}}{M}}$
Where, ${V_{RMS}}$ = RMS speed of gas molecules.
$R$ = Gas constant
$T$ = Temperature
$M$ = Molecular mass of gas
But we have to relate it with pressure so that the pressure can be determined. So as we know in the case of ideal gas:
$PV = RT$
Where, $P$ = Pressure of gas
$V$ = Volume of gas
By substituting these values in the above formula of root mean square speed we will get:
${V_{RMS}} = \sqrt {\dfrac{{3PV}}{M}}$
Let’s suppose the volume of ideal gas be $V$
And molecular mass at pressure ${P_o}$ be $M$.
So, RMS speed at pressure ${P_o}$ will be:
${V_{RMS}} = \sqrt {\dfrac{{3PV}}{M}}$
According to the question, the molecular mass of gas molecules get doubled at pressure $P$
So, molecular mass at pressure $P$ be $2M$
And also the root mean square speed of gas molecules get halved at pressure $P$
So, RMS speed at pressure $P$ be $\dfrac{1}{2}{V_{RMS}}$
So, RMS speed will be:
$\dfrac{1}{2}{V_{RMS}} = \sqrt {\dfrac{{3PV}}{{2M}}}$
By dividing both the root mean square speeds we will get:
$\dfrac{{{V_{RMS}}}}{{\dfrac{1}{2}{V_{RMS}}}} = \dfrac{{\sqrt {\dfrac{{3{P_o}V}}{M}} }}{{\sqrt {\dfrac{{3PV}}{{2M}}} }}$
$\ \Rightarrow 2 = \sqrt {\dfrac{{2{P_o}}}{P}} \\\ \Rightarrow P = 2{P_o} \\\ \$

Hence, option B is correct.
Note:
According to Kinetic Molecular Theory, an increase in temperature results in an increase in the average kinetic energy of the molecules. As the particles move faster, they will collide to the edge of the container more often. If the reaction is kept at constant pressure, the particles must stay farther apart, and an increase in volume will compensate for the increase in particle collision with the surface of the container.