An open container of volume 50 cc holds air (mean molecular weight = 28.8 g ) and is exposed to the atmosphere where the pressure is 100 kPa . The container is placed in a bath containing boiling water ( 100^ C ). Determine the mass of air in the container once thermal equilibrium is reached.

Step-by-step solution

Using the ideal gas equation, PV = nRT Since n = m M , we have PV = m M RT Rearranging for mass m : m = PVM RT Given values: P = 100 kPa = 10^5 Pa V = 50 cc = 50 10⁻⁶ m ^3 M = 28.8 g/mol T = 100^ C = 373 K R = 8.314 J K ⁻¹ mol ⁻¹ Substituting the values: m = 10^5 50 10⁻⁶ 28.8 8.314 373 m = 5 28.8 3101.122 m = 144 3101.122 0.0464 g