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. $1:2$
• B. $1:4$
• C. $2:1$
• D. $4:1$

Answer 1

The correct option is A

$1:2$

Explanation for correct option:-

Option (A) $1:2$

Step 1: Finding the force applied on molecules

Let us assume a cubical container with each side of lenght $\mathrm{a}$ filled with a gas.

Let the speed of the molecules be $\mathrm{v}$ and the mass of the molecules be $\mathrm{m}$.

When the molecule travels from one wall to the other, the momentum of the molecule will be $\mathrm{mv}$

But due to the collision, the wall will apply a similar force on the molecule which changes the momentum of the molecule to $-\mathrm{mv}$

$\therefore$Change in momentum is $∆\mathrm{p}=-\mathrm{mv}-\left(\mathrm{mv}\right)=-2\mathrm{mv}$

$\therefore$Now the force applied on the molecule will be

$$\begin{gathered} \mathrm{F}=\frac{∆\mathrm{p}}{\mathrm{t}} \\ \Rightarrow \mathrm{F}=\frac{-2\mathrm{mv}}{\mathrm{t}}............\left(\mathrm{i}\right) \end{gathered}$$

Here $\mathrm{t}$ is the time which is required for the molecule to travel from one wall to another and then back.

Step 2: Finding the force applied by the molecule

As the molecule travels, it will cover a distance of $2\mathrm{a}$

Now we know $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

$\therefore$Time taken by the molecule to travel is $\mathrm{t}=\frac{2\mathrm{a}}{\mathrm{v}}$

By substituting the value of $\mathrm{t}$ in equation (i) we get,

$$\begin{gathered} \mathrm{F}=\frac{-2\mathrm{mv}}{\mathrm{t}} \\ \Rightarrow \mathrm{F}=\frac{-2\mathrm{mv}}{\left({\displaystyle \frac{2\mathrm{a}}{\mathrm{v}}}\right)} \\ \Rightarrow \mathrm{F}=\frac{-{\mathrm{mv}}^2}{\mathrm{a}} \end{gathered}$$

But according to Newton's third law, the molecule will apply the same force on the wall but in the opposite direction.

$\therefore$$$\begin{gathered} -\mathrm{F}=-\left(\frac{-{\mathrm{mv}}^2}{\mathrm{a}}\right) \\ =\frac{{\mathrm{mv}}^2}{\mathrm{a}} \end{gathered}$$

Step 3: Finding the pressure exerted by the molecule

We know that pressure is force per unit area, and since its a cubical container, the area of the wall will be ${\mathrm{a}}^2$

$\therefore$The pressure exerted by the molecule is:

$$\begin{gathered} \mathrm{P}=\frac{\mathrm{Force}}{\mathrm{Area}} \\ \Rightarrow \mathrm{P}=\frac{\left({\displaystyle \frac{{\mathrm{mv}}^2}{\mathrm{a}}}\right)}{{\mathrm{a}}^2} \\ \Rightarrow \mathrm{P}=\frac{{\mathrm{mv}}^2}{{\mathrm{a}}^3} \end{gathered}$$

$\therefore$Initial pressure is ${\mathrm{P}}_{\mathrm{i}}=\frac{{\mathrm{mv}}^2}{{\mathrm{a}}^3}.............\left(\mathrm{ii}\right)$

Step 4: Finding the ratio between initial and final pressure

According to the question, the mass of the molecule is halved and the velocity of the molecule is doubled

$\therefore$Final pressure:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{f}}=\frac{{\mathrm{mv}}^2}{{\mathrm{a}}^3} \\ \Rightarrow {\mathrm{P}}_{\mathrm{f}}=\frac{\left({\displaystyle \frac{1}{2}}\mathrm{m}\right){\left(2\mathrm{v}\right)}^2}{{\mathrm{a}}^3} \\ \Rightarrow {\mathrm{P}}_{\mathrm{f}}=\frac{\left({\displaystyle \frac{1}{2}\mathrm{m}}\right)4{\mathrm{v}}^2}{{\mathrm{a}}^3} \\ \Rightarrow {\mathrm{P}}_{\mathrm{f}}=\frac{2{\mathrm{mv}}^2}{{\mathrm{a}}^3}..............\left(\mathrm{iii}\right) \end{gathered}$$

$\therefore$the ratio of initial pressure${\mathrm{P}}_{\mathrm{i}}$ to the final pressure${\mathrm{P}}_{\mathrm{f}}$ is:

$$\begin{gathered} \frac{{\mathrm{P}}_{\mathrm{i}}}{{\mathrm{P}}_{\mathrm{f}}} \\ \Rightarrow \frac{{\mathrm{P}}_{\mathrm{i}}}{{\mathrm{P}}_{\mathrm{f}}}=\frac{\left({\displaystyle \frac{m{v}^2}{a^3}}\right)}{\left({\displaystyle \frac{2m{v}^2}{a^3}}\right)} \\ \Rightarrow \frac{{\mathrm{P}}_{\mathrm{i}}}{{\mathrm{P}}_{\mathrm{f}}}=\frac{1}{2} \\ \Rightarrow {\mathrm{P}}_{\mathrm{i}}:{\mathrm{P}}_{\mathrm{f}}=1:2 \end{gathered}$$

Hence it is the correct option

Therefore, the ratio of initial and final pressure will be an option (A) $1:2$