Two flasks I and II shown below are connected by a valve of negligible volume.

When the valve is opened, the final pressure of the system in bar is $\mathrm{x}\times {10}^{-2}$. The value of $\mathrm{x}$ is (Integer answer)
[Assume : Ideal gas; $1$ bar $={10}^5 \mathrm{Pa}$ : Molar mass of $\left.{\mathrm{N}}_2=28.0{ \mathrm{mol}}^{-1};\mathrm{R}=8.31 \mathrm{J}{ \mathrm{mol}}^{-1}{ \mathrm{K}}^{-1}\right]$

Applying ; ${\left(\mathrm{n}-\mathrm{I}+\mathrm{n}-\mathrm{II}\right)}_{\mathrm{initial}}$ $=$The final number of moles.

$\mathrm{n}-\mathrm{I} \mathrm{and} \mathrm{n}-\mathrm{II}$ are the number of moles of gas in each container.

$\Rightarrow$ Assuming the system attains a final temperature of $\mathrm{T}$ (such that $300<\mathrm{T}<60$)

$\Rightarrow \left\{\mid \right.$ Heat lost by ${\mathrm{N}}_2$ of container $\left(\mathrm{I}\right)|=|$ Heat gained by ${\mathrm{N}}_2$ of container $\mathrm{II}|\mathrm{\}}$

$\Rightarrow \left(\mathrm{n}-\mathrm{I}\right){\mathrm{C}}_{\mathrm{m}}\left(300-\mathrm{T})=\left(\mathrm{n}-\mathrm{II}\right){\mathrm{C}}_{\mathrm{m}}(\mathrm{T}-60)\right.$

$\Rightarrow \left(\frac{2.8}{28}\right)(300-\mathrm{T})=\frac{0.2}{28}( \mathrm{T}-60)$

$\Rightarrow 14(300-\mathrm{T})=\mathrm{T}-60$

$\Rightarrow \frac{(14\times 300+60)}{15}=\mathrm{T}$

$\Rightarrow \mathrm{T}=284 \mathrm{K}$ (final temperature)

$\Rightarrow$ If the final pressure $=\mathrm{P}$

$\Rightarrow {\mathrm{n}}_{\mathrm{final} }=\frac{3.0}{28}$

$\Rightarrow \frac{\mathrm{P}}{\mathrm{RT}}\left({\mathrm{V}}_{\mathrm{I}}+{\mathrm{V}}_{\mathrm{II}}\right)=\frac{3.0\mathrm{gm}}{28\mathrm{gm}/\mathrm{mol}}$

$\mathrm{P}=\frac{3}{28} \mathrm{mol}\times 8.31\frac{\mathrm{J}}{\mathrm{mol}-\mathrm{K}}\times \frac{284 \mathrm{K}}{3\times {10}^{-3}{ \mathrm{m}}^3}\times {10}^5\frac{\mathrm{bar}}{\mathrm{Pa}}$

$\Rightarrow 0.84287\mathrm{bar}$

$\Rightarrow 84.28\times {10}^{-2} \mathrm{bar}$

$\Rightarrow 84$