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?

#tag_title#Calculations#tag_content# Step 1: Determine the mass of liquid nitrogen $$ m = \rho \times V_{liq} = 808 \mathrm{~kg/m^3} \times 1.00 \times 10^{-3} \mathrm{~m^3} = 0.808 \mathrm{~kg} $$ Step 2: Convert the temperature to Kelvin $$ T_K = T_C + 273.15 = 21.0^{\circ} \mathrm{C} + 273.15 = 294.15 \mathrm{K} $$ Step 3: Use the Ideal Gas Law to find the volume of nitrogen gas Find the number of moles: $$ n = \frac{m}{M_N} = \frac{0.808 \mathrm{~kg}}{28.02 \mathrm{~g/mol} \times 10^{-3} \mathrm{~kg/g}} = 28.84 \mathrm{mol} $$ Calculate the volume of gas: $$ V_{gas} = \frac{nRT}{P} = \frac{28.84 \mathrm{mol} \times 8.314 \mathrm{kPa \cdot L/(mol \cdot K)} \times 294.15 \mathrm{K}}{101 \mathrm{kPa}} = 722.9 \mathrm{L} $$ The volume occupied by 1.00 L of liquid nitrogen when it evaporates and comes into equilibrium with the air at 21.0°C and 101 kPa is approximately 722.9 L.