Question

equal molecules of n2 and o2 are kept in a closed container at pressure p if n2 is removed from the system the pressure of the container ( please ignore this tick )

Answer 1

it is given that equal number of molecules are kept in a container so there would be half the pressure of N2 and half the pressure of O2 so if N2 is removed then the pressure would be halved so the pressure would become P/2 .
HOPE THIS HELPS YOU
THANK YOU

Answer 2

The correct answer is option (2). $\frac{P}{2}.$

Given:

It is given that equal molecules of $N_{2}$ and $O_{2}$ are taken in the closed container.

Pressure inside the container = P.

To Find:

With the given data, we have to find the pressure inside the container if $N_{2}$ is removed from the system.

Solution:

It is given that the number of molecules of $N_{2}$ and $O_{2}$ are equal.

∴, $y_{N_{2} = 0.5$ and $y_{O_{2} = 0.5$,

where $y_{N_{2}$ and $y_{O_{2}$ are mole fractions of $N_{2}$ and $O_{2}$, respectively.

Then,

$P_{N_{2}} - y_{N_{2}} . P_{T} = \frac{P}{2}$

$P_{O_{2}} - y_{O_{2}} . P_{T} = \frac{P}{2}$,

where $P_{N_{2}$ and $P_{O_{2}$ are the partial pressures of $N_{2}$ and $O_{2}$, respectively, and $P_{T}$ is the total pressure of the system.

If $N_{2}$ is removed from the system, then only $O_{2}$ is present.

∴,  $P_{T} = P_{O_{2}} = \frac{P}{2}.$

Hence, the pressure inside the container if $N_{2}$ is removed from the system is option (2). $\frac{P}{2}$.

#SPJ3