Won't the pressure be the vapor pressure of the methanol (assuming there is enough liquid methanol to establish equilibrium between liquid and vapor)?
You can calculate how much of the methanol will evaporate by
n = PV/RT. There will be some liquid left at equilibrium.
7.40 g of liquid methanol, CH3OH, are placed in an empty 20.0 L container at 45.0°C. What will be the pressure in the container when equilibrium is established? Assume the temperature remains constant and the volume of any liquid is negligible. The vapor pressure of methanol at 45.0°C is 205 mm Hg.
1 answer
1 answer