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The volume of connecting tube is negligible. Now, the stopcock is opened. The temperature of both vessels are maintained at 27° C.
[Take : $R = \frac{1}{12} L atm / mol K$ ]
Codes Description for final condition
01 2.5 moles of gas will transfer from vesse-I to vessel-II
02 2.5 moles of gas will transfer from vessel-II to vessel-I
03 Final moles of gases in both vessels will be same.
04 Final pressures of gases in both vessels will be same.
Fill OMR as sum of correct codes,

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answer is 5.

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Detailed Solution

We can calculate the pressure as;

$15 = \frac{P \times 60}{\frac{1}{12} \times 300} \Rightarrow P = 6.25$ atm (Same in both vessels)

$\frac{10}{n} = \frac{50}{(15 - n)} \Rightarrow n = \frac{15}{6} = 2.5 moles.$
Here, first and last description is correct.

Thus, the sum of their code is;

1+4=5

So the correct answer is 5.

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