The vapor pressure of liquid bromine at room temp is 168 torr. If the bromine is introduced drop by drop into a closed system containing air at 775 torr, what would the total pressure (in torr) be when no more bromine vaporized and a few drops of liquid are present in the flask?
total pressure would be 775+168 torr
To calculate the total pressure in the flask when no more bromine vaporizes and a few drops of liquid are present, we need to consider the partial pressure of bromine vapor and the partial pressure of air.
1. Calculate the partial pressure of bromine vapor:
Since the vapor pressure of liquid bromine at room temperature is 168 torr, the partial pressure of bromine vapor will be equal to this value.
Partial pressure of bromine vapor = 168 torr
2. Calculate the partial pressure of air:
Initially, the flask contains air at a pressure of 775 torr. As drops of liquid bromine are introduced, some of the air molecules will be displaced and the partial pressure of air will decrease.
Partial pressure of air = 775 torr - (pressure of bromine vapor)
3. Calculate the total pressure in the flask:
The total pressure in the flask is the sum of the partial pressures of bromine vapor and air.
Total pressure = Partial pressure of bromine vapor + Partial pressure of air
Total pressure = 168 torr + (775 torr - partial pressure of bromine vapor)
Since we want to find the total pressure when no more bromine vaporizes and a few drops of liquid are present, we need to solve for the partial pressure of bromine vapor in the equation. We can set up an equation and solve it.
Total pressure = 168 torr + (775 torr - partial pressure of bromine vapor)
Keep in mind that the partial pressure of bromine vapor will change as the bromine evaporates.
To determine the total pressure when no more bromine vaporizes and a few drops of liquid are present in the flask, we need to consider the partial pressures of the air and bromine vapor.
First, let's calculate the initial partial pressure of the air in the closed system. The air pressure is given as 775 torr.
Next, we need to calculate the partial pressure of the bromine vapor. The vapor pressure of liquid bromine at room temperature is given as 168 torr.
As bromine droplets are introduced into the system, some of them will evaporate, increasing the partial pressure of bromine vapor. Once the partial pressure of the bromine vapor reaches 168 torr (equal to its vapor pressure), the bromine will stop evaporating, and the system will reach equilibrium.
At this point, the total pressure in the system will be the sum of the partial pressures of the air and bromine vapor. Therefore:
Total pressure = Air partial pressure + Bromine vapor partial pressure
Since the air partial pressure is 775 torr and the bromine vapor partial pressure is 168 torr, the total pressure would be:
Total pressure = 775 torr + 168 torr = 943 torr
So, when no more bromine vaporizes and a few drops of liquid are present in the flask, the total pressure in the system would be 943 torr.