Question

a vessel contain He gas at 20 atm and CH4 gas at 'P' atm. When a pinhole is made in the gases, the gases out initially contains 80% He gas by mole the value of 'P' is​

Answer 1

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Answer 2

Given:

Given that, one vessel consists of He gas which is at 20 atm and CH4 gas at 'P' atm.

To find:

We need to find the He gas by mole the value of 'P'.

Solution:

Before solving the question, we should know about Graham's Law.

Graham's Law may be a relation which states that the speed of the effusion of a gas is inversely proportional to the square root of its density or molecular mass.

$\[\frac{Rat{{e}_{1}}}{Rat{{e}_{2}}}=\sqrt{\frac{{{M}_{1}}}{{{M}_{2}}}}\]$

Where,

Rate1 is that the rate of effusion of 1 gas, expressed as volume or as moles per unit time.

Rate2 is that the rate of effusion of the second gas.

M1 is that the mass of gas 1.

M2 is that the mass of gas 2.

He = 80% of initial mole

Formula of rate is,

$\[\text{rate = pressure }/~\sqrt{molecular\text{ }mass}\]$

$\[rate=\frac{{{P}_{1}}}{{{P}_{2}}}\times \sqrt{\frac{{{M}_{1}}}{{{M}_{2}}}}\]$

Substituting the required values in the above equation,

$\[\frac{80}{20}=\frac{20}{{{P}_{2}}}\times \sqrt{\frac{16}{4}}\]$

When we solve this equation, we are going get the value of P.

 $& 4=\frac{20}{{{P}_{2}}}\times \sqrt{4}$

$& {{P}_{2}}=\frac{20}{4}\times 2 \\ & {{P}_{2}}=5\times 2 \\ & {{P}_{2}}=10\text{ atm}$

Therefore, the value of P is 10 atm.