Question

A vessel of 5.0 L capacity contains 1.4 g nitrogen at 1800 K. Assuming that at this temperature, 40% of molecules are dissociated into atoms and the gas is ideal, what is the gas pressure?

• A. 2.07 atm
• B. 1.05 atm
• C. 1.476 atm
• D. 2.67 atm

Answer 1

Answer: (A)

2.07 atm

Answer 2

Initial moles of $N_2 = \frac{1.4 \, \text{g}}{28 \, \text{g/mol}} = \frac{1}{20}$ mol.

The dissociation reaction is $N_2 \rightarrow 2N$. If 40% of $N_2$ dissociates, the moles of $N_2$ dissociated are $0.40 \times \frac{1}{20} = \frac{1}{50}$ mol.

The total number of moles at equilibrium is the sum of remaining $N_2$ and formed $N$ atoms: Total moles $n = (\frac{1}{20} - \frac{1}{50}) + 2(\frac{1}{50}) = \frac{1}{20} + \frac{1}{50} = \frac{5+2}{100} = \frac{7}{100} = 0.07$ mol.

Using the ideal gas law, $PV = nRT$: $P = \frac{nRT}{V} = \frac{0.07 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 1800 \, \text{K}}{5.0 \, \text{L}} \approx 2.07$ atm.