Liquid nitrogen has a density of 0.807 g/mL at -195.8 °C.If 1.00 L of N₂(l)is allowed to warm to 25°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

Using the ideal gas law, PV=nRT, we can solve for the final volume V. We first need to find the number of moles of gas, which can be found using the density and the molar mass of nitrogen (28.02 g/mol):

density = mass/volume
volume = mass/density = (1000 mL) * (0.807 g/mL) = 807 g = 0.807 kg
n = mass/molar mass = 0.807 kg / 28.02 g/mol = 0.0288 mol

Next, we can use the initial and final temperatures in Kelvin (77.35 K and 298 K, respectively), and the given pressure of 1.00 atm:

V = nRT/P = (0.0288 mol)*(0.08206 L*atm/mol*K)*(298 K)/(1.00 atm) = 0.705 L

Therefore, the gas will occupy a volume of 0.705 L at 25°C and 1.00 atm.

Liquid Nitrogen Has a Density of 0