Final answer:
To estimate the temperature at which 1.00 kg of steam at a pressure of 1.50×10⁶ Pa occupies a volume of 0.220 m³, we can use the ideal gas equation PV = nRT. By substituting the given values into the equation, we find that the approximate temperature is 1671 K.
Step-by-step explanation:
To estimate the temperature at which 1.00 kg of steam at a pressure of 1.50×10⁶ Pa occupies a volume of 0.220 m³, we can use the ideal gas equation, PV = nRT. Rearranging the equation, we have T = (PV) / (nR), where P is the pressure, V is the volume, n is the number of moles, and R is the gas constant. First, we need to convert the mass of steam to moles using the molar mass of steam. 1.00 kg of steam is equivalent to 1000 g, which is 1000 / 18.0 = 55.56 mol. Substituting the values into the equation, we get T = (1.50×10⁶ Pa * 0.220 m³) / (55.56 mol * 8.3145 J/(K•mol)). Evaluating this expression gives us the temperature, which is approximately 1671 K.