A sealed flask contains 0.60 g of water at 28 degrees celcius. The vapor pressure of water at this temperature is 28.36 mmHg.
What is the minimum volume of the flask in order that no liquid water be present in the flask?
v=nrt/p
v= (0.03333)(0.0821)(301)/0.037316
v= 22.07
n- 0.6 divided by the molar mass of water (18) to get moles
r- constant (0.0821)
t- temperature in kelvin ( 28 + 273)
p- pressure in atm ( convert mm mercury to atm by dividing 28.36 by 760)
To calculate the minimum volume of the flask required in order for no liquid water to be present, we need to determine the vapor pressure of water at the given temperature and the volume at which the vapor pressure of water equals the external pressure.
Step 1: Convert the given temperature from Celsius to Kelvin.
The temperature in Kelvin (K) is equal to the temperature in degrees Celsius (°C) plus 273.15.
28°C + 273.15 = 301.15 K
Step 2: Calculate the vapor pressure of water at the given temperature.
To determine the vapor pressure of water at 301.15 K, we use the Clausius-Clapeyron equation:
ln(Pvap1/Pvap2) = (ΔHvap/R) * (1/T2 - 1/T1)
Pvap1 = vapor pressure at temperature T1 = 28.36 mmHg (given)
T1 = temperature in Kelvin = 301.15 K (from step 1)
T2 = temperature in Kelvin = 373.15 K (boiling point of water at atmospheric pressure)
R = gas constant = 8.314 J/(mol*K) (universal gas constant)
ΔHvap = enthalpy of vaporization of water = 40.7 kJ/mol (constant value)
Plugging in the values into the equation:
ln(Pvap1/1) = (ΔHvap/R) * (1/T2 - 1/T1)
ln(28.36/1) = (40.7 * 10^3 J/mol / 8.314 J/(mol*K)) * (1/373.15 K - 1/301.15 K)
Solving for ln(Pvap1/1):
ln(28.36) = (40.7 * 10^3 J/mol / 8.314 J/(mol*K)) * (301.15 K - 373.15 K) / (373.15 K * 301.15 K)
Using the natural logarithm (ln) function on both sides:
28.36 = (40.7 * 10^3 J/mol / 8.314 J/(mol*K)) * (301.15 K - 373.15 K) / (373.15 K * 301.15 K)
Simplifying the equation:
28.36 = (40.7 * 10^3 J/mol / 8.314 J/(mol*K)) * (-72.0 K^2) / (373.15 K * 301.15 K)
Solving for the vapor pressure of water (Pvap2) at 373.15 K:
28.36 = (40.7 * 10^3 J/mol / 8.314 J/(mol*K)) * (-72.0) / (373.15 * 301.15) * Pvap2
Pvap2 = 28.36 * (8.314 J/(mol*K) / (40.7 * 10^3 J/mol)) * (-72.0) / (373.15 * 301.15)
Pvap2 ≈ 373.63 mmHg
Step 3: Determine the minimum volume of the flask required.
When the vapor pressure of water (Pvap2) equals the external pressure, no liquid water will remain in the flask.
Given:
External pressure = Pvap2 = 373.63 mmHg (from step 2)
Using the Ideal Gas Law, PV = nRT, where
P = pressure (mmHg)
V = volume (L)
n = number of moles
R = gas constant = 0.0821 L*atm/(mol*K) (Rearranged to mmHg units)
T = temperature (K)
Plugging in the values:
(373.63 mmHg) * V = n * (0.0821 L*atm/(mol*K)) * (301.15 K)
Simplifying:
V = (n * (0.0821 L*atm/(mol*K)) * (301.15 K)) / (373.63 mmHg)
The number of moles (n) can be calculated using the molar mass of water (18.015 g/mol) and the mass of water present (0.60 g).
n = (0.60 g) / (18.015 g/mol)
Solving for the minimum volume (V):
V = [(0.60 g) / (18.015 g/mol)] * (0.0821 L*atm/(mol*K)) * (301.15 K) / (373.63 mmHg)
To determine the minimum volume of the flask in order to have no liquid water present, we need to consider the equilibrium conditions between the liquid and vapor phases of water.
At 28 degrees Celsius, the vapor pressure of water is given as 28.36 mmHg. This means that when the vapor pressure inside the flask reaches this value, the liquid water will start to evaporate and coexist with the vapor phase.
To calculate the minimum volume of the flask, we can approach this problem using the Ideal Gas Law and the concept of partial pressure.
Step 1: Convert the water mass into moles.
The molar mass of water (H2O) is approximately 18 g/mol. Thus, the number of moles of water present can be calculated as follows:
moles of water = mass of water / molar mass of water
moles of water = 0.60 g / 18 g/mol
moles of water ≈ 0.0333 mol
Step 2: Determine the partial pressure of the water vapor inside the flask.
The partial pressure of the water vapor is given as 28.36 mmHg, which is its equilibrium vapor pressure at 28 degrees Celsius.
Step 3: Apply the Ideal Gas Law to find the volume required.
The Ideal Gas Law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L•atm/K•mol)
T = temperature (in Kelvin)
Since we are given the pressure in mmHg and the molar mass of water is in grams, we need to convert these units into atm and Kelvin, respectively.
Convert pressure from mmHg to atm:
1 atm = 760 mmHg
28.36 mmHg ÷ 760 mmHg/atm ≈ 0.0373 atm (to four significant figures)
Convert temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 28°C + 273.15 K ≈ 301.15 K (to four significant figures)
Now we can substitute the values into the Ideal Gas Law equation to solve for the volume (V):
V = (nRT) / P
V = (0.0333 mol)(0.0821 L•atm/K•mol)(301.15 K) / 0.0373 atm
V ≈ 0.665 L (to three significant figures)
Therefore, the minimum volume of the flask to ensure no liquid water is present at 28 degrees Celsius is approximately 0.665 liters.