Final answer:
A unit mass of air at 298.15 K is the mass of one mole of air at that temperature, which is approximately 29 g. Using the ideal gas law, one mole of any ideal gas at STP occupies about 22.4 L. The mass density of air at STP is 1.25 kg/m³, making the air in a typical living room weigh around 96 kg.
Step-by-step explanation:
The student's question about the unit mass of air at 298.15 K refers to the mass of one mole of air under specific conditions. Assuming we are dealing with a temperature of 298.15 K, which translates to around 25°C, and using the ideal gas law, we can determine the volume occupied by this amount of gas. It's important to note that the average molecular mass of air, which is approximately a mixture of 80% nitrogen (N2) and 20% oxygen (O2), is about 29 g/mol.
Using the provided ideal gas law equation, P(0.955 atm)V = (1 mol) (0.0821 L·atm/mol·K) (298 K), we can solve for the volume V. As this equation is dependent on pressure P, and the pressure is given as 0.955 atm, using simple algebra, the volume V can be determined to be approximately 22.4 L/mol, which is the accepted volume of one mole of any ideal gas at STP (standard temperature and pressure).
When considering the mass density of air, we know that one cubic meter of air at STP has a mass of 1.25 kg, and the mass of one mole of air at 298.15 K with a molar mass of 29 g/mol is essentially 29 g or 0.029 kg. If we were to apply this to a typical living room with dimensions 5 m x 5 m x 3 m, the mass of air inside the room would amount to approximately 96 kg, which is remarkably similar to the average mass of a human being.