Question

Liquid nitrogen, which is used in many physics research labs, can present a safety hazard if a large quantity evaporates in a confined space. The resulting nitrogen gas reduces the oxygen concentration, creating the risk of asphyxiation. Suppose \(1.00 \mathrm{~L}\) of liquid nitrogen \(\left(\rho=808 \mathrm{~kg} / \mathrm{m}^{3}\right)\) evaporates and comes into equilibrium with the air at \(21.0^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\). How much volume will it occupy?

Short answer

Answer: The volume occupied by the evaporated nitrogen gas is 0.677 m^3.

Step-by-step solution

Find the mass of liquid nitrogen that evaporates

To determine the mass of the liquid nitrogen that evaporates, we will use the volume and density values provided. The formula to calculate mass is: Mass (m) = Volume (V) * Density (ρ) The volume of the liquid nitrogen is \(1.00 L\). To use the density in \(\frac{kg}{m^3}\), we need to convert the volume from liters to cubic meters. There are 1000 liters in one cubic meter, so: \(V = 1.00 L \times \frac{1 m^3}{1000 L} = 0.001 m^3\) Now, using the density \(\rho=808 kg/m^3\), we can calculate the mass: \(m = V \times \rho = 0.001 m^3 \times 808 kg/m^3 = 0.808 kg\)

Convert the temperature to Kelvin

To use the ideal gas law, we need the temperature in Kelvin. To convert from Celsius to Kelvin, we add 273.15 to the Celsius value: \(T = 21.0^\circ C + 273.15 K = 294.15 K\)

Use the ideal gas law to calculate the volume of nitrogen gas

The ideal gas law is given by the equation: \(PV = nRT\) Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We are given the pressure (\(P = 101 kPa = 101,000 Pa\)) and temperature (T = 294.15 K). To find n, the number of moles, we need to use the mass (m) and the molar mass (M) of nitrogen gas: \(n = \frac{m}{M} \Rightarrow n = \frac{0.808 kg}{0.028 kg/mol (Molar ~mass~ of~ N_2)} = 28.857 ~moles\) Now using the ideal gas constant \(R = 8.314 J/(mol \cdot K)\), we can rewrite the ideal gas law formula as follows: \(V = \frac{nRT}{P}\) Plugging in the values, we get: \(V = \frac{28.857 moles \times 8.314 J/(mol \cdot K) \times 294.15 K}{101,000 Pa} = 0.677 m^3\) As a result, the volume occupied by the evaporated nitrogen gas is \(0.677 m^3\).