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A closed vessel contains $He$ and $Ne$ in the mole ratio $3 : 1$ at a total pressure of $4 atm$ . Which statements are correct?

APartial pressure of $He$ is $3 atm$
BPartial pressure of $Ne$ is $1 atm$
CMass of $He$ (4) is greater than mass of $Ne$ (20) in the vessel
DIf $Ne$ is removed at constant $T$ , the pressure falls to $3 atm$

Answer: A, B, D

x_{He} = 0.75

,

x_{Ne} = 0.25

, so partial pressures are

3

and

1 atm

(A, B). Masses:

3 \times 4 = 12

versus

1 \times 20 = 20

(per 4 mol), so Ne outweighs He (C wrong). Removing Ne leaves only

p_{He} = 3 atm

(D).

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