A 2.30 L sample of O2(g) was collected over a liquid at a total pressure of 785 torr and 25°C. When the O2(g) was dried (vapor of that liquid removed), the gas had a volume of 1.93 L at 25°C and 785 torr. Calculate the vapor pressure of this liquid at 25°C.
Use PV = nRT
Solve for n=PV/RT for the first set of conditions which will give you total n (moles vapor + moles O2).
Then calculate n = PV/RT for the second set of conditions which gives just moles O2.
Subtract to find moles liquid vapor, then use PV = nRT and solve for P of the liquid (vapor of the liquid).
To calculate the vapor pressure of the liquid at 25°C, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Let's first convert the given temperatures from Celsius to Kelvin:
25°C + 273.15 = 298.15 K
Now we can use the ideal gas law equation to find the number of moles of O2 before drying:
P1V1 = nRT1
We are given:
P1 = 785 torr
V1 = 2.30 L
T1 = 298.15 K
Let's plug in the values and solve for n:
n = (P1V1) / (RT1)
= (785 torr * 2.30 L) / (0.0821 L·atm/mol·K * 298.15 K)
= 10.52 mol O2
Now, we can use the ideal gas law equation again to find the number of moles of O2 after drying:
P2V2 = nRT2
We are given:
P2 = 785 torr
V2 = 1.93 L
T2 = 298.15 K
Let's plug in the values and solve for n:
n = (P2V2) / (RT2)
= (785 torr * 1.93 L) / (0.0821 L·atm/mol·K * 298.15 K)
= 8.39 mol O2
Since the difference in the number of moles is due to the removed vapor of the liquid, we can calculate the number of moles of the vapor:
n_vapor = n_before - n_after
= 10.52 mol - 8.39 mol
= 2.13 mol
Now we can calculate the vapor pressure using the ideal gas law equation once again:
P_vapor = (n_vapor * R * T2) / V2
= (2.13 mol * 0.0821 L·atm/mol·K * 298.15 K) / 1.93 L
= 85.4 torr
Therefore, the vapor pressure of the liquid at 25°C is approximately 85.4 torr.
To calculate the vapor pressure of the liquid at 25°C, we need to find the pressure difference between the total pressure and the pressure of the dried gas.
Let's denote the vapor pressure of the liquid as P_vapor.
Given:
Total pressure (P_total) = 785 torr
Volume of the collected sample (V_sample) = 2.30 L
Volume of the dried gas (V_dried) = 1.93 L
First, we need to find the partial pressure of the other gas component present in the collected sample. Since we are dealing with a gas mixture, we can assume that the pressure of the other gas component is related to its volume ratio.
Using the ideal gas law, we can write the equation as:
(P_total - P_vapor) * V_dried = P_total * V_sample
Rearranging the equation and solving for P_vapor, we have:
P_vapor = P_total - (P_total * V_sample) / V_dried
Substituting the given values:
P_vapor = 785 torr - (785 torr * 2.30 L) / 1.93 L
Now let's calculate it:
P_vapor = 785 torr - (785 torr * 2.30 L) / 1.93 L
P_vapor = 785 torr - 934.19 torr
P_vapor = -149.19 torr
However, vapor pressure cannot be negative. This indicates that there might be an error in the calculation or the assumption we made about the gas mixture is incorrect. I recommend double-checking the given values and calculations.