$N_{2}$ is found in a litre flask under $100 k P a$ pressure and $O_{2}$ is found in another $3$ litre flask under $320 k P a$ pressure. If the two flask are connected, the resultant pressure is:

265 k P a

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210 k P a

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420 k P a

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365 k P a

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Solution

The correct option is A $265 k P a$
At initial condition $P_{1} = 100 k P a$ $= 1 a t m V_{1}$
$= 1 L P_{2}$
$= 320 k P a = 3 a t m V_{2}$ = $3 L$
Applying ideal gas law, $\frac{P V}{R T} = n$
total mole at initial conditions,
$(1 + n_{2})$ = $n_{i} = \frac{2}{R T} + \frac{9.3}{R T} = \frac{11.3}{R T}$
After mixing the gases the number of moles remains constant gases do not undergo reaction.
$n_{i} = n_{f} = P_{f} \times \frac{V_{f}}{R T} W h e r e, P_{f} = f i n a l p r e s s u r e V_{f} = f i n a l v o l u m e V_{f} = V_{1} + V_{2} = 4 L t h u s, P_{f} \times \frac{4}{R T} = \frac{11.3}{R T} P_{f} = 2.85 a t m$
$P_{f} = 285 k P a \approx 265 k P a$

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N2 is found in a litre flask under 100 kPa pressure and O2 is found in another 3 litre flask under 320 kPa pressure. If the two flask are connected, the resultant pressure is:

$N_{2}$ is found in a litre flask under $100 k P a$ pressure and $O_{2}$ is found in another $3$ litre flask under $320 k P a$ pressure. If the two flask are connected, the resultant pressure is: