Final answer:
To calculate the gas pressure after 30% of nitrogen molecules are dissociated, the molar quantity is adjusted and the ideal gas law is applied, resulting in a pressure of 1.92 atm, which corresponds with option D.
Step-by-step explanation:
The student is asking how to calculate the pressure of a gas within a vessel when a certain percentage of its molecules are dissociated into atoms. To solve this, we first need to understand the dissociation of nitrogen (N2) molecules into individual nitrogen atoms and then apply the ideal gas law to find the pressure after dissociation.
Given that we have 1.4 g of nitrogen, we can find the number of moles of N2 using its molar mass:
Moles of N2 = 1.4 g / 28.01 g/mol = 0.05 mol
With 30% dissociation, 0.015 mol of N2 dissociate, giving us 0.015 mol * 2 = 0.03 mol of N atoms. Since 70% of the original N2 remains, we have 0.035 mol of N2. The total number of moles of particles (N2 and N atoms) is then 0.035 mol + 0.03 mol = 0.065 mol.
We can now use the ideal gas law PV = nRT to find the pressure. Let's take R as 0.0821 L.atm.mol-1K-1 and convert the volume to L:
Volume = 5 L
Temperature = 1800 K
Pressure (P) = nRT / V
P = (0.065 mol) * (0.0821 L.atm.mol-1K-1) * (1800 K) / 5 L
P = 1.92 atm
The correct answer to the student's question is option D, 1.92 atm.