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

Applying ; ${\left(n - I + n -\text{ II }\right)}$_{initial} =$The final number of moles.</p><p>$n - Iandn - II$are the number of moles of gas in each container.</p><p>$⇒$Assuming the system attains a final temperature of$T$(such that$300 < T < 60$)</p><p>$⇒ \left\{∣\right) Heat lost by ${N}_{2}$ of container $( I ) | = |$ Heat gained by ${N}_{2}$ of container II | }

$⇒ \left(n - I\right){C}_{m}\left(300 - T ) = \left(n - II\right){C}_{m} ( T - 60 )\right)</p><p>$⇒ \left(\dfrac{2 . 8}{28}\right) ( 300 - T ) = \dfrac{0 . 2}{28} ( T - 60 )$</p><p>$⇒ 14 ( 300 - T ) = T - 60$</p><p>$⇒ \dfrac{( 14 \times 300 + 60 )}{15} = T$</p><p>$⇒ T = 284K$(final temperature)</p><p>$⇒$If the final pressure$= P$</p><p>$⇒ {n}_{final} = \dfrac{3 . 0}{28}$</p><p>$⇒ \dfrac{P}{RT}\left({V}_{I} + {V}_{II}\right) = \dfrac{3 . 0gm}{28gm / mol}$</p><p>$P = \dfrac{3}{28}mol \times 8 . 31\dfrac{J}{mol - K} \times \dfrac{284K}{3 \times {10}^{- 3}{m}^{3}} \times {10}^{5}\dfrac{bar}{Pa}$</p><p>$⇒ 0 . 84287bar$</p><p>$⇒ 84 . 28 \times {10}^{- 2}bar$</p><p>$⇒ 84