Question

The volume V of an enclosure contains a mixture of three gases, $16$ g of oxygen, $28$ g of nitrogen and $44$ g of carbon dioxide at absolute temperature T. Consider R as a universal gas constant. The pressure of the mixture of gases is:

• A. $\frac{\left(4RT\right)}{V}$
• B. $\frac{\left(88RT\right)}{V}$
• C. $\frac{5}{2}\frac{{\displaystyle RT}}{V}$
• D. $3RTV$

Answer 1

The correct option is C

$\frac{5}{2}\frac{{\displaystyle RT}}{V}$

The explanation for the correct answer:-

Option (C)

Finding the number of moles of gases:

The number of moles of O₂ $=n1=\frac{16}{32}=0.5$ mole.

The number of moles of N₂ $=n2=\frac{28}{28}=1$ mole.

The number of moles of CO₂ $=n3=\frac{44}{44}=1$ mole.

Total number of moles in container $=n=n1+n2+n3$

$n=0.5+1+1$

$n=\frac{5}{2}$ moles

$P=\frac{\left(nRT\right)}{V}$

$P=\frac{5}{2}\frac{{\displaystyle RT}}{V}$

Hence, $P=\frac{5}{2}\frac{{\displaystyle RT}}{V}$is the correct answer.

Therefore, option (C) is the correct answer.