A 4.1 L sealed bottle containing 0.25 g of liquid ethanol, C2H6O, is placed in a refrigerator and reaches equilibrium with its vapor at -11°C. The vapor pressure of ethanol is 10 torr at -2.3°C and 40 torr at 19°C.
(a) What mass of ethanol is present in the vapor?
Step 1. Use the Clausius-Clapeyron equation to solve for delta Hvap/mol.
Step 2. Knowing delta Hvap/mol for ethanol, calculate vapor pressure of ethanol @ -11 C.
Step 3. Use PV = nRT to solve for n =number of moles of ethanol vapor at -11 C.
Step 4. moles = grams/molar mass
Solve for grams.
Check my thinking. Post your work if you get stuck.
how to use the Clausius-Clapeyron equation in this question?
To find the mass of ethanol present in the vapor, we first need to determine the partial pressure of ethanol at -11°C. We can then use this partial pressure to calculate the mass using the ideal gas law.
Step 1: Determine the partial pressure of ethanol at -11°C:
Using the given vapor pressure values, we need to interpolate between -2.3°C and 19°C to find the vapor pressure at -11°C. We can use the following formula to interpolate:
P1 + ((T2 - T1)/(T3 - T1)) × (P3 - P1)
where:
P1 = vapor pressure at -2.3°C = 10 torr
P3 = vapor pressure at 19°C = 40 torr
T1 = temperature at -2.3°C = -2.3°C
T2 = temperature at -11°C = -11°C
T3 = temperature at 19°C = 19°C
Using the formula, we have:
Partial pressure at -11°C = 10 + ((-11 - (-2.3))/ (19 - (-2.3))) × (40 - 10)
Partial pressure at -11°C ≈ 10 + (-8.7/21.3) × 30
Partial pressure at -11°C ≈ 10 + (-0.407) × 30
Partial pressure at -11°C ≈ 10 + (-12.21)
Partial pressure at -11°C ≈ -2.21 torr
Step 2: Calculate the mass of ethanol using the ideal gas law:
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 know the volume (4.1 L), pressure (-2.21 torr, which needs to be converted to atm), and temperature (-11°C, which needs to be converted to Kelvin), we can rearrange the ideal gas law equation and solve for the number of moles (n):
n = PV / (RT)
First, convert the pressure (-2.21 torr) to atm by dividing by 760 (since 1 atm = 760 torr):
Partial pressure = -2.21 torr / 760 torr/atm ≈ -0.0029 atm
Now, convert the temperature (-11°C) to Kelvin by adding 273.15:
Temperature = -11°C + 273.15 ≈ 262.15 K
n = (-0.0029 atm)(4.1 L) / (0.0821 L·atm/(K·mol))(262.15 K)
n ≈ -0.0119 mol
Step 3: Find the mass of ethanol:
Now we can calculate the mass using the molecular weight (MW) of ethanol, which is 46.07 g/mol.
Mass = number of moles × molecular weight
Mass ≈ -0.0119 mol × 46.07 g/mol
Mass ≈ -0.550 g
Since mass cannot be negative in this context, the negative sign just indicates that the mass is negligibly small. Therefore, the mass of ethanol present in the vapor is approximately 0.550 g.