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

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P/2

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P/4

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Solution

The correct option is A P
Volume of each portion $= 0.5 L$
$n_{1} = 2 / 2 = 1 m o l$
$n_{2} = 26 / 16 = 1 m o l$
Now,when portion is removed ,
let pressure $= P^{^{'}}$
$∴ \frac{P^{1} \times 1}{R \times T} = \frac{P \times 0.5}{R \times T} - \frac{P \times 0.5}{R T}$
$\Rightarrow P^{1} = P$

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A box of 1L capacity is divided into two equal compartments by a thin partition which are filled with 2g H2 and 16gm CH4 respectively. The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be:

The correct option is A PVolume of each portion =0.5L n1=2/2=1mol n2=26/16=1molNow,when portion is removed , let pressure =P′ ∴P1×1R×T=P×0.5R×T−P×0.5RT⇒P1=P

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

The correct option is A P
Volume of each portion $= 0.5 L$
$n_{1} = 2 / 2 = 1 m o l$
$n_{2} = 26 / 16 = 1 m o l$
Now,when portion is removed ,
let pressure $= P^{^{'}}$
$∴ \frac{P^{1} \times 1}{R \times T} = \frac{P \times 0.5}{R \times T} - \frac{P \times 0.5}{R T}$
$\Rightarrow P^{1} = P$