Liquid nitrogen has a density of 0.807 g/mL at -195.8 ^$\circ$C. If 1.00 L of N₂(l) is allowed to warm to 25^$\circ$C at a pressure of 1.00 atm, what volume will the gas occupy (R = 0.08206 L*atm/K*mol) A) 59.1 L B) 182 L C) 705 L D) 1.41 * 10³ L E) 1.97 * 10⁴ L

Question

Quizplus · Accepted Answer

First, we need to find the number of moles of nitrogen gas present in 1 L of liquid nitrogen:
Density = mass/volume
0.807 g/mL = mass/1 L
mass = 0.807 kg/L
Molar mass of nitrogen gas (N2) = 28.02 g/mol
Number of moles = mass/molar mass
Number of moles = 0.807 kg/L / (28.02 g/mol)
Number of moles = 0.0288 mol/L

Using the ideal gas law, we can find the final volume of the gas at 25°C and 1 atm pressure:
PV = nRT
V = nRT/P
V = (0.0288 mol/L)(0.08206 L atm/mol K)(298 K)/(1 atm)
V = 0.705 L

Therefore, the gas will occupy 0.705 L at 25°C and 1 atm pressure, which corresponds to choice C.

Liquid Nitrogen Has a Density of 0 ∘\circ∘ C If 1 ∘\circ∘

Liquid Nitrogen Has a Density of 0 $\circ$ C If 1 $\circ$