Question

If masses of all molecules of a gas are halved and their speed is double then the ratio of initial and final pressure will be
A. 2:1
B. 1:2
C. 4:1
D. 1:4

Answer 1

Consider a gas inside a cubical container with side length ‘ a’. Find the change in the momentum of a molecule when the molecule collides with it. Calculate the time taken for the molecule to travel from one wall to the opposite and back by using $\text{time = }\dfrac{\text{distance}}{\text{speed}}\text{ }$. Then find the force applied by the wall on this molecule by using $F=\dfrac{\Delta p}{t}$. Divide the force by area to find the value of pressure. Then calculate the initial and final pressures and then find the ratio.

Formula used:
$\text{time = }\dfrac{\text{distance}}{\text{speed}}\text{ }$
$F=\dfrac{\Delta p}{t}$

Complete step-by-step solution:
Consider a cubical container whose each side is of length a. The container is filled with gas. Let the mass and speed of each molecule of the gas be m and v.
Let us analyze the pressure applied on a wall of the container by one molecule of the gas. Then we can generalize it for all the molecules.
Suppose a molecule of the gas travels from one wall to the opposite wall of the container. Since the molecule is moving with velocity v its momentum will be mv. The molecule will collide with the opposite wall and will trace its path. Due to the collision, the wall will apply a force on the molecule, and the momentum of the molecule changes to –mv.
Therefore, the change in momentum of the molecule is $\Delta p$= –mv – (mv)= -2mv.
The force applied on the molecule is equal to $F=\dfrac{\Delta p}{t}=\dfrac{-2mv}{t}$ … (i).
Here, t is the time taken for the molecule to travel from one wall to the opposite wall and back to the wall. In this process, it travels a distance of 2a.
From the formula $\text{time = }\dfrac{\text{distance}}{\text{speed}}\text{ }$, we get $t=\dfrac{2a}{v}$.
Substitute the value of t in (i).
$\Rightarrow F=\dfrac{-2mv}{\dfrac{2a}{v}}=-\dfrac{m{{v}^{2}}}{a}$.
From Newton’s third law, we know that the molecule will apply a force on the wall of same magnitude but of opposite direction, i.e. $-F=-\left( -\dfrac{m{{v}^{2}}}{a} \right)=\dfrac{m{{v}^{2}}}{a}$.
The pressure is defined as the force per unit area. The area of the wall is ${{a}^{2}}$.
Therefore, pressure exerted by the molecule on the wall is $\dfrac{\dfrac{m{{v}^{2}}}{a}}{{{a}^{2}}}=\dfrac{m{{v}^{2}}}{{{a}^{3}}}$.
Therefore, the initial pressure is ${{P}_{i}}=\dfrac{m{{v}^{2}}}{{{a}^{3}}}$ …. (ii).
It is given that the mass of the molecule is halved and its velocity is doubled. Therefore, the final pressure is ${{P}_{f}}=\dfrac{\left( \dfrac{1}{2}m \right){{\left( 2v \right)}^{2}}}{{{a}^{3}}}=\dfrac{\left( \dfrac{1}{2}m \right)4{{v}^{2}}}{{{a}^{3}}}=\dfrac{2m{{v}^{2}}}{{{a}^{3}}}$ …. (iii).
Now, to find the ratio of the initial to the final pressures divide (ii) by (iii). $\Rightarrow \dfrac{{{P}_{i}}}{{{P}_{f}}}=\dfrac{\dfrac{m{{v}^{2}}}{{{a}^{3}}}}{\dfrac{2m{{v}^{2}}}{{{a}^{3}}}}=\dfrac{1}{2}$.
This means that the ratio of the initial pressure to the final pressure is 1:2.
Hence, the correct option is B.

Note: We can also solve the given question by the kinetic gas equation. According to the kinetic gas equation pressure of the gas is $P=\dfrac{Nm{{v}^{2}}}{3V}$, where, N is are the numbers of molecules inside the container, V is the volume of the container.
Therefore, ${{P}_{i}}=\dfrac{Nm{{v}^{2}}}{3V}$
And ${{P}_{f}}=\dfrac{N\left( \dfrac{1}{2}m \right){{(2v)}^{2}}}{3V}=\dfrac{2Nm{{v}^{2}}}{3V}$
Then,
$\dfrac{{{P}_{i}}}{{{P}_{f}}}=\dfrac{\dfrac{Nm{{v}^{2}}}{3V}}{\dfrac{2Nm{{v}^{2}}}{3V}}=\dfrac{1}{2}$