Question

A mixture of $N_2$ and $Ar$ gases in a cylinder contains $7\,g$ of $N_2$ and $8\,g$ of $Ar$. If the total pressure of the mixture of the gases in the cylinder is $27\, bar$, the partial pressure of $N_2$ is : [Use atomic masses (in g $mol^{-1}$) :$N = 14,Ar = 40$]

• A. 9 bar
• B. 12 bar
• C. 15 bar
• D. 18 bar

Answer 1

The Correct Option is C

To find the partial pressure of \(N_2\) in the mixture, we can use Dalton's Law of Partial Pressures. According to this law, the partial pressure of a gas in a mixture is proportional to its mole fraction in the mixture.

1. First, calculate the number of moles of each gas.
   • The molar mass of \(N_2\) is \(28 \, g \, mol^{-1}\) (since \(N = 14\) and there are two nitrogen atoms in \(N_2\)).
   • The number of moles of \(N_2\) is \(\frac{7}{28} = 0.25 \, mol\).
   • The molar mass of \(Ar\) is \(40 \, g \, mol^{-1}\).
   • The number of moles of \(Ar\) is \(\frac{8}{40} = 0.2 \, mol\).
2. Find the total number of moles in the mixture:
   • Total moles = \(0.25 + 0.2 = 0.45 \, mol\).
3. Calculate the mole fraction of \(N_2\):
   • Mole fraction of \(N_2\) = \(\frac{0.25}{0.45}\).
   • Mole fraction of \(N_2\) = \(0.5556\).
4. Apply Dalton's Law to find the partial pressure of \(N_2\):
   • Partial pressure of \(N_2\) = Total pressure x Mole fraction of \(N_2\).
   • Partial pressure of \(N_2\) = \(27 \, \text{bar} \times 0.5556 = 15 \, \text{bar}\).

Thus, the partial pressure of \(N_2\) in the cylinder is 15 bar.