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

Answer:

given =  He gas at 20 atm and CH4 gas at 'P' atm.

           

to find =  pressure

solution =   He = 80% of initial mole

                  rate =  pressure / $\sqrt{molecular mass }$

                   rate = P1 /P2 $\sqrt{M1} /\sqrt{M2}$

                   80/20 = 20/P$\sqrt{16/4}$

                   P =   5 X  $\sqrt{4}$

                   P= 5 X 2

                  P = 10atm

       ∴     The helium gas by mole the value of P is 10 atm.

Answer 2

Answer:

Concept:

Thomas Graham, a Scottish physical chemist, developed Graham's law of diffusion in 1848. Graham discovered through experimentation that a gas's rate of effusion is inversely equal to the square root of its particle's molar mass. Graham's law states that a gas's rate of effusion or diffusion is inversely proportional to the square of its molecular weight. As a result, if one gas has a molecular weight that is four times greater than another, it will diffuse through a porous plug or escape through a tiny hole in a vessel at a rate that is half as fast as the other gas (heavier gases diffuse more slowly). Years later, the kinetic theory of gases offered a thorough theoretical justification of Graham's law.

Given:

He gas at 20 atm and CH4 gas at "P" atm are both present in a vessel. The gases initially contain 80% He gas by mole when a pinhole is formed in them, hence the value of "P" is

Find:

find the answer for the given question

Answer:

given that:

He=80%

$rate=\frac{pressure}{\sqrt{molecular.mass} }$

$rate=\frac{P1}{P2} *\sqrt{\frac{M1}{M2} }$

sub the values in given equation

$\frac{80}{20} =\frac{20}{P2} *\sqrt{\frac{16}{4} }$

In order to find the value of P, we must solve this equation.

$4=\frac{20}{P2} *\sqrt{4}$

$P2=\frac{20}{4} *2$

$P2=5*2$

$P2=10atm$

According to Graham's law, a gas's rate of diffusion or effusion is inversely related to its molecular weight squared. Graham's law of gas effusion or diffusion states that rate 1 / rate 2 = (mass 2 / mass 1), where rate 1 and rate 2 are the rates of gas effusion or diffusion, measured in moles per unit time, of Gases 1 and 2, respectively. According to Graham's law of diffusion, the ratio of the square root of the molar masses of two gases to their diffusion rates is the same.

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