A 4: 1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is

Question

A 4: 1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is (A) 8: 1 (B) 8: 3 (C) 4: 1 (D) 1: 1

Tardigrade Admin · Accepted Answer

Pressure of helium = 8 bar, CH4 = 2 bar According to Grahamâs law, (r1/r2)=(P1/P2) √(M2/M1) (rHe/rCH4) =(P1/P2) √(MCH4/MHe) =(8/2) √(16/4) =(8/1) =8: 1

Q. A 4 : 1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is

8 : 1

8 : 3

4 : 1

1 : 1

Pressure of helium = $8 ba r, C H_{4} = 2 ba r$
According to Graham’s law,
$\frac{r _{1}}{r _{2}} = \frac{P _{1}}{P _{2}} \frac{M _{2}}{M _{1}}$
$\frac{r _{He}}{r _{C H_{4}}} = \frac{P _{1}}{P _{2}} \frac{M _{C H_{4}}}{M _{He}} = \frac{8}{2} \frac{16}{4} = \frac{8}{1} = 8 : 1$

Q. A 4 : 1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is

8 : 1

8 : 3

4 : 1

1 : 1

Pressure of helium = 8bar,CH4​=2bar According to Graham’s law, r2​r1​​=P2​P1​​M1​M2​​​ rCH4​​rHe​​=P2​P1​​MHe​MCH4​​​​=28​416​​=18​=8:1

Pressure of helium = $8 ba r, C H_{4} = 2 ba r$
According to Graham’s law,
$\frac{r _{1}}{r _{2}} = \frac{P _{1}}{P _{2}} \frac{M _{2}}{M _{1}}$
$\frac{r _{He}}{r _{C H_{4}}} = \frac{P _{1}}{P _{2}} \frac{M _{C H_{4}}}{M _{He}} = \frac{8}{2} \frac{16}{4} = \frac{8}{1} = 8 : 1$