Many analytical instruments require liquid nitrogen. While filling an instrument with liquid nitrogen, an analytical chemist slips and spills the contents of the dewer, 25L, onto the floor. Since the boiling point of liquid nitrogen at 1atm is 77.4 K, all the nitrogen boils away to gas very quickly. The density of liquid nitrogen is 804kg/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 atm?
What is the mass of 25 L N2 gas?
mass = volume x density = 25 L x 804 g/L = ?? grams.
How many moles is that?
moles = ??grams/molar mass
What volume is that?
PV = nRT
You have P, n, R, and T, solve for V.
Note: I might fuss a little about the problem. My opinion is that the gas will fill the room, whatever its dimensions."-)
To find the volume of nitrogen gas produced from the liquid nitrogen spill, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's convert the temperature from Celsius to Kelvin. Kelvin temperature is simply Celsius temperature + 273.15. So, the temperature of the room in Kelvin is 25 + 273.15 = 298.15 K.
Since the spill happens at 1 atmosphere of pressure and room temperature, we can assume the pressure and temperature remain constant throughout the process.
Now, let's calculate the number of moles of nitrogen gas produced from the spill. To do this, we need the initial number of moles of liquid nitrogen in the dewar.
The density of liquid nitrogen is given as 804 kg/m^3. We can use this information to find the mass of the liquid nitrogen spilled.
Mass = density × volume
The volume of liquid nitrogen spilled is given as 25 liters, which we can convert to cubic meters (1 m^3 = 1000 liters).
Volume = 25 × (1/1000) = 0.025 m^3
Mass = 804 kg/m^3 × 0.025 m^3 = 20.1 kg
Now, we can calculate the number of moles of liquid nitrogen.
Number of moles = mass / molar mass
The molar mass of nitrogen is approximately 28 g/mol.
Molar mass = 28 g/mol = 0.028 kg/mol
Number of moles = 20.1 kg / 0.028 kg/mol ≈ 718.57 mol
Since the boiling process happens quickly, we can assume the liquid nitrogen completely turns into gaseous nitrogen. So, the number of moles of nitrogen gas produced will be the same as the number of moles of liquid nitrogen.
Number of moles of nitrogen gas produced = 718.57 mol
With the given pressure, temperature, and the number of moles, we can rearrange the ideal gas law equation to solve for V, the volume of the gas.
V = (nRT) / P
Now, let's plug in the values:
V = (718.57 mol × 0.0821 L·atm/mol·K × 298.15 K) / 1 atm
V ≈ 17942.37 L
Therefore, approximately 17942.37 liters of nitrogen gas is produced from the liquid nitrogen spill.