Many analytical instruments require liquid nitrogen. While filling an instrument with liquid nitrogen, an analytical chemist slips and spills the contents of the dewer, 29.0 L, onto the floor. Since the boiling point of liquid nitrogen at 1.00 atm is 77.4 K, all the nitrogen boils away to gas very quickly. The density of liquid nitrogen is 804 kg/m^3. What volume of nitrogen gas is produced from the liquid nitrogen spill if the temperature of the room is 25 degrees C at 1.00 atm?

Question

Many analytical instruments require liquid nitrogen. While filling an instrument with liquid nitrogen, an analytical chemist slips and spills the contents of the dewer, 29.0 L, onto the floor. Since the boiling point of liquid nitrogen at 1.00 atm is 77.4 K, all the nitrogen boils away to gas very quickly. The density of liquid nitrogen is 804 kg/m^3. What volume of nitrogen gas is produced from the liquid nitrogen spill if the temperature of the room is 25 degrees C at 1.00 atm?

Answer 1

Here is what I would do.

First, your prof may not like this but I prefer to work in g/cc for density; therefore, convert 804 kg/m^3 = 0.804 g/cc.
m = volume x density
mass = 29,000 cc x 0.804 g/cc = ?? grams.
moles N2 = grams/molar mass = n
Then use PV = nRT and solve for volume.

Answer 2

To find the volume of nitrogen gas produced from the liquid nitrogen spill, we need to calculate the amount of liquid nitrogen that evaporates.

Given:
Volume of spilled liquid nitrogen (V_liq) = 29.0 L
Density of liquid nitrogen (ρ_liq) = 804 kg/m^3
Boiling point of liquid nitrogen (T_boil) = 77.4 K
Temperature of the room (T_room) = 25 °C = 298 K

First, we need to convert the volume of liquid nitrogen (V_liq) from liters to cubic meters:
V_liq = 29.0 L = 0.029 m^3

Next, we can calculate the mass of the spilled liquid nitrogen using the density:
m_liq = ρ_liq * V_liq = 804 kg/m^3 * 0.029 m^3 = 23.316 kg

Since liquid nitrogen boils at its boiling point (77.4 K) and quickly evaporates to gas at room temperature (298 K), we can assume the gas has equal molar volume at both boiling and room temperatures.

The ideal gas law equation can be used to calculate the volume of gas produced:
PV = nRT

where:
P = pressure (1.00 atm)
V = volume of gas produced
n = amount of gas in moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (298 K)

We need to find n, the number of moles of nitrogen gas produced. We can use the molar mass of nitrogen to convert mass (m_liq) to moles (n):
molar mass of nitrogen (M_N2) = 28.02 g/mol

n = m_liq / M_N2 = 23.316 kg / (28.02 g/mol) = 831.422 mol

Now, we can use the ideal gas law equation to solve for the volume (V) of gas:
PV = nRT
V = (nRT) / P
V = (831.422 mol * 0.0821 L·atm/mol·K * 298 K) / (1.00 atm)
V ≈ 20438.14 L

Therefore, the volume of nitrogen gas produced from the liquid nitrogen spill is approximately 20438.14 liters.