A box of 1-litre capacity is divided into two equal compartments by a thin partition which is filled with 6g of $H_{2}$ and 16 g $C H_{4}$ respectively.The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be:

P

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2 P

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P / 2

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P / 4

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Solution

The correct option is C $P$
Number of moles, $n_{H_{2}} = 6 / 2 = 3$
and $n_{C H_{4}} = 16 / 16 = 1$

$P_{H_{2}} = \frac{3}{3 + 1} \times P = \frac{3}{4} P$
$P_{C H_{4}} = \frac{1}{3 + 1} \times P = P / 4$
$p_{total} = P_{H_{2}} + P_{C H_{4}} = \frac{3}{4} P + \frac{p}{4} = P$

Hence, option $A$ is correct.

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A box of 1-litre capacity is divided into two equal compartments by a thin partition which is filled with 6g of H2 and 16 g CH4 respectively.The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be:

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A box of 1-litre capacity is divided into two equal compartments by a thin partition which is filled with 6g of $H_{2}$ and 16 g $C H_{4}$ respectively.The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be: