Question

Gas at pressure ${P_0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure $P$ will be equal to:
a. $4{P_0}$
b. $2{P_0}$
c. ${P_0}$
d. $\dfrac{{{P_0}}}{2}$

Answer 1

Hint: To solve the given problem remember that the given pressure is directly proportional to the product of mass and the square of root mean square of the value.

Formula used:
To calculate the value of the pressure $P$:
$P = \dfrac{1}{3}\dfrac{{mNC}}{V}rm{s^2}$
Where $m$ is the mass,
$N$ is the number of molecules,
$C$ speed of the gas molecules.
$rms$ means the root means square value.

Complete step by step answer:
Let us consider the given terms in the question. We have the value of the pressure ${P_0}$ of a gas that is contained in a gas vessel.
We need to find the value of the resulting pressure when the masses of every molecule when it is halved and the speed of the molecules when it is doubled.
Consider the kinetic theory of gases to solve the given problem. According to the kinetic theory of gases, which takes into account the microscopic structure of a matter, a gas is enclosed in a vessel contains a very large number of atoms and molecules moving in all directions with all the possible values of the velocities, thereby colliding with one and another.
The formula for the pressure is given by,
$ \Rightarrow P = \dfrac{1}{3}\dfrac{{mNC}}{V}rm{s^2}$
Where $m$ is the mass,
$N$ is the number of molecules,
$C$ speed of the gas molecules.
$rms$ means the root means square value.
Remember that the given pressure is directly proportional to the product of mass and the square of root mean square of the value. That is,
$ \Rightarrow P \propto mV_{rms}^2$
Let us consider two different pressures that are ${P_1}$ and ${P_2}$.
Consider the pressure when the molecule's mass is halved. We have,
$ \Rightarrow {P_1} = \dfrac{1}{3}\dfrac{{mN{C^2}}}{V}$
Consider the pressure when the speeds of the molecules are double. We have,
$ \Rightarrow {P_2} = \dfrac{1}{3}\left( {\dfrac{{mN}}{{2V}}} \right) \times {\left( {2C} \right)^2}$
By dividing ${P_2}$ by ${P_1}$ ,
$ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\dfrac{1}{3}\left( {\dfrac{{mN}}{{2V}}} \right) \times \left( {4{C^2}} \right)}}{{\dfrac{1}{3}\dfrac{{mN{C^2}}}{V}}}$
$ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{4}{2}$
$ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = 2$
Take the denominator ${P_1}$ to the left-hand side to get the answer. That is,
$ \Rightarrow {P_2} = 2{P_0}$
$\therefore {P_2} = 2{P_0}$

Hence, the correct answer is option (B).

Note: The whole problem is solved with the help of the kinetic theory of the gases. According to the kinetic theory of gases, the time of impact is negligible in comparison to the time taken to traverse the free path.