Question

a container contains a mixture of He gas and ch4 gas in the mass ratio 1 :2 at a 10 ATM pressure .a very fine hole is made into the container, the ratio of masses of He gas to that of ch4 gas coming out through the hole initially is

*1:2
*2:1
*1:1
*4:1​

Answer 1

Answer:

CHEMISTRY

A 4:1 molar mixture of He and CH

4

is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition ratio of the mixture effusing out initially?

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ANSWER

Pressure = 20 bar

Molar ratio = 4:1 (He and CH

4

)

P

He

=

5

4

×20=16bar

P

CH

4

=

5

1

×20=4bar

r

2

r

1

=

P

2

P

1

×

M

1

M

2

r

CH

4

r

He

=

P

CH

4

P

He

×

M

He

M

CH

4

=

4

16

×

4

16

=8

Therefore, the ratio of moles of helium and methane coming out initially is 8:1.

Hence, the correct option is A

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Answer 2

The ratio of masses of He gas to that of CH4 gas coming out through the hole will be 1:1.

According to Graham's law of diffusion,

                              $\frac{mHe}{mCH4}$ = $\frac{1}{2}$  

                               $\frac{nHe}{nCH4}$ = $\frac{1/4}{2/16}$ = 2:1

                    where, nHe - moles of Helium

                                nCH4 - moles of methane

                    Given, Pt i.e. Total pressure = 10 atm

                                pHe = mole fraction of He × Pt

                                        = $\frac{2}{3}$×10

                                        = $\frac{20}{3}$

                                pCH4 = mole fraction of CH4 × Pt

                                pCH4 = $\frac{1}{3}$×10

                                           = $\frac{10}{3}$

     According to Graham's Law of effusion,

                            $\frac{rHe}{rCH4}$ =  $\frac{pHe}{pCH4}$$\sqrt \frac{MCH4}{MHe}$

     where, rHe and rCH₄ are rates of effusion of respective compounds

                  MCH₄ and MHe are Molecular weights of respective compounds

                           

                         $\frac{rHe}{rCH4}$ = $(\frac{20/3}{10/3})$ $(\sqrt\frac{16}{4} })$

                         $\frac{rHe}{rCH4}$ = 4:1

     Mass ratio = ratio of rate of effusion × Ratio of molar mass

     Mass ratio = $(\frac{4}{1} )$×$(\frac{4}{16} )$

     Mass ratio = 1:1

Hence, After effusion through the hole, the ratio of mass of He:CH4 comes out to be 1:1