Gas at a pressure P0 in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to

To find the resulting pressure of the gas after the changes, we can use the kinetic theory of gases.

The pressure $P_0$ exerted by an ideal gas contained in a vessel of volume $V$ is given by the formula:
$P_0=\frac{1}{3}\frac{Nm}{V}{v}^2$
where:
- $N$ is the total number of gas molecules,
- $m$ is the mass of each molecule,
- $v$ is the speed (root-mean-square speed) of the molecules, and
- $V$ is the volume of the container.

Now, according to the question, the following changes are made:
1. The mass of each molecule is halved:
$m^{\prime }=\frac{m}{2}$
2. The speed of each molecule is doubled:
$v^{\prime }=2v$

Substituting these new values into the pressure equation, the new pressure $P$ is:
$P=\frac{1}{3}\frac{N{m}^{\prime }}{V}{\left(v^{\prime }\right)}^2$
$P=\frac{1}{3}\frac{N\cdot \left(\frac{m}{2}\right)}{V}{\left(2v\right)}^2$

Simplifying the expression:
$P=\frac{1}{3}\frac{Nm}{2V}\left(4v^2\right)$
$P=2\cdot \left(\frac{1}{3}\frac{Nm}{V}{v}^2\right)$

Since the term inside the brackets is the initial pressure $P_0$, we have:
$P=2P_0$

Therefore, the resulting pressure is 2P₀.