Question

Equal amounts of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is

• A. 0.55 atm
• B. 0.11 atm
• C. 1 atm
• D. 0.12 atm

Answer 1

Answer: (C)

1 atm

Answer 2

No. of moles of lighter gas $= \frac { m } { 4 }$

No. of moles of heavier gas $= \frac { m } { 40 }$

Total no. of moles $= \frac { m } { 4 } + \frac { m } { 40 } = \frac { 11 m } { 40 }$

Mole fraction of lighter gas $= \frac { \frac { m } { 4 } } { \frac { 11 m } { 40 } } = \frac { 10 } { 11 }$

Partial pressure due to lighter gas $= P _ { o } \times \frac { 10 } { 11 }$

$= 1.1 \times \frac { 10 } { 11 } = 1 \mathrm {~atm}$