Answer
Given,
Mass of Methane = $320 \mathrm{mg}=0.32 \mathrm{g}$
Mass of Argon $=175 \mathrm{mg}=0.175 \mathrm{g}$
Mass of Neon $=225 \mathrm{mg}=0.225 \mathrm{g}$
Computing the number of moles,
Moles of Methane $=\frac{\text { Mass of Methane }}{\text { Molar mass of Methane }}=\frac{0.32}{16}=0.02$ moles
Moles of Argon $=\frac{\text { Mass of Argon }}{\text { Molar mass of Argon }}=\frac{0.175}{40}=0.004$ moles
Moles of Neon $=\frac{\text { Mass of Neon }}{\text { Molar mass of Neon }}=\frac{0.225}{20}=0.01125$ moles
Mole fraction of Methane $=\frac{\text { Moles of Methane }}{\text { Total number of moles }}=\frac{0.02}{0.03525}=0.567$
Mole fraction of Argon $=\frac{\text { Moles of Argon }}{\text { Total number of moles }}=\frac{0.004}{0.03525}=0.113$
Mole fraction of Neon $=\frac{\text { Moles of Neon }}{\text { Total number of moles }}=\frac{0.01125}{0.03525}=0.32$
We know,
Partial pressure of Neon $=8.87 \mathrm{kPa}$
$\begin{aligned} \text { Partial pressure of Neon } &=\text { Mole Fraction of Neon } \times \text { Total Pressure } \\ \text { Total Pressure } &=\frac{\text { Partial pressure of Neon }}{\text { Mole fraction of Neon }} \\ &=0.32 \times 8.87=2.839 \mathrm{kPa} \\ &=0.028 \mathrm{atm} \end{aligned}$
Hence the pressure exerted by the system is 0.028 atm.
Substituting the values of Pressure, Temperature, and Moles in the ideal gas equation we get,
$\begin{aligned} p \cdot V &=n \cdot R \cdot T \\ V &=\frac{n \cdot R \cdot T}{p} \\ &=\frac{0.03525 \times 0.0821 \times 300}{0.028} \mathrm{litres} \\ &=31 \text { litres } \end{aligned}$
Hence the pressure of the system is 0.028 atm and volume is 31 liters
Work Step by Step
--
