Question

Suppose you heat 1.00 kg of water and convert it into steam at the boiling temperature of water, 373 K, and at normal atmospheric pressure, 1.00 atm. What is the volume of the steam?

Answer 1

The volume of the steam can be calculated using the ideal gas law, 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. Since the mass of the water is given, we can calculate the number of moles using the molar mass of water, which is 18.02 g/mol.

n = (1.00 kg) / (18.02 g/mol) = 0.055 mol

Using the ideal gas law, we can calculate the volume of the steam:

V = (nRT) / P

V = (0.055 mol * 8.314 J/mol*K * 373 K) / (1.00 atm)

V = 14.3 m^3