Final answer:
The density of water vapor that creates a partial pressure equal to the given vapor pressure can be found using the ideal gas law. By rearranging the equation and substituting the given values, we can calculate the moles of water vapor and then the mass. Dividing the mass by the volume will give us the density. The density of water vapor is approximately 0.00116/5 kg/m³. Therefore, the correct option is a).
Step-by-step explanation:
To calculate the density of water vapor that creates a partial pressure equal to the given vapor pressure, we need to use the ideal gas law equation:
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 in Kelvin.
We know the vapor pressure and the temperature, and we can assume the volume to be 1 m³. Rearranging the equation, we get:
n/V = P/RT
Substituting the given values:
n/1 = (0.025/5 atm) / (0.0821 L·atm/(mol·K) * (37/5 + 273) K)
n = (0.025/5) / (0.0821 * (37/5 + 273)) mol
Using the molar mass of water (18.015 g/mol), we can calculate the mass of water vapor:
mass = n * molar mass = (0.025/5) / (0.0821 * (37/5 + 273)) * 18.015 g
And finally, we can calculate the density by dividing the mass by the volume:
density = mass / volume = (0.025/5) / (0.0821 * (37/5 + 273)) * 18.015 / 1 kg/m³
Calculating the above expression, we find the density of water vapor to be approximately 0.00116/5 kg/m³.