Answer:
The volume of 17.0 grams of water vapor at 112.0°C and a pressure of 0.822 atm is 26.66 L.
Step-by-step explanation:
We can use the ideal gas law to calculate the volume of water vapor at the given temperature and pressure:
PV = nRT
Where P is the pressure (0.822 atm), V is the volume, n is the number of moles of water vapor, R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in kelvin (385.15 K, calculated from 112.0°C), and m is the mass of water vapor.
First, we need to find the number of moles of water vapor, n:
n = m / M
Where m is the mass of water vapor (17.0 g) and M is the molar mass of water vapor (18.015 g/mol).
Substituting the values in the ideal gas law, we get:
PV = (m / M) * R * T
V = (m / M) * R * T / P
V = (17.0 g / 18.015 g/mol) * 0.0821 L·atm/mol·K * 385.15 K / 0.822 atm
V = 26.66 L
So, the volume of 17.0 grams of water vapor at 112.0°C and a pressure of 0.822 atm is 26.66 L.