Question

Equal weights 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. $0.15$ atm
• C. $1$ atm
• D. $0.11$ atm

Answer 1

Answer: (C)

Number of moles of lighter gas $=\frac{m}{4}$

Number of moles of heavier gas $=\frac{m}{40}$

Total Number of moles $=\frac{m}{4}+\frac{m}{40}=\frac{11m}{40}$

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

Partial pressure due to lighter gas $=P_o\times \frac{10}{11}$

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