Trichloroethane, C2H3Cl3, is the active ingredient in aerosols that claim to stain-proof men's ties. Trichloroethane has a vapor pressure of 100.0 mm Hg at 20.0 C and boils at 74.1 C. An uncovered cup (1/2 pint) of trichloroethane (d=1.325 g/mL) is kept in an 18-ft3 refrigerator at 39 F. What percentage (by mass) of trichloroethane is left as a liquid when equilibrium is established?

Use the Clausius-Claypeyron equation and solve for delta Hvap.(You know p at 20 and you know p is 760 mm at 74.1.) Then use CC equation a second time and one of the other values to solve for vapor pressure at 277 K (39F). Use PV = nRT and solve for n. Convert n to grams, convert the 1/2 pt to grams, and go from there to solve for percent. Note it is percent remaining, not percent evaporated.

I found the first 2 steps, but don't know what to do next. how do i find the volume to plug it into the equation: PV=nRT?

To solve this problem, we need to calculate the amount of trichloroethane that has evaporated from the cup and compare it to the initial amount of trichloroethane.

First, let's convert the given measurements to SI units for consistency:

1) Convert temperature from Fahrenheit to Celsius: 39°F = 3.9°C.

2) Convert volume from cubic feet to liters: 1 ft^3 = 28.3168 liters. Hence, 18 ft^3 = 18 * 28.3168 liters ≈ 509.7 liters.

Next, we can use the ideal gas law to calculate the number of moles of trichloroethane in the cup at equilibrium:

PV = nRT

Where:
P = vapor pressure = 100.0 mmHg = 100.0 torr = 100.0 * 133.3 Pa
V = volume = 509.7 liters
n = number of moles (unknown)
R = ideal gas constant = 0.0821 L*atm/(mol*K)
T = temperature in Kelvin = (3.9 + 273.15) K

Rearranging the equation and solving for n:

n = (PV) / (RT)

n = (100.0 * 133.3) / (0.0821 * (3.9 + 273.15))

n ≈ 857.67 moles

Next, we can calculate the mass of trichloroethane in the cup:

mass = volume * density

mass = 509.7 liters * 1.325 g/mL * 1000 mL/L

mass ≈ 675,206.25 grams

Now, we need to determine how much trichloroethane has evaporated. At equilibrium, the total mass of trichloroethane in the system (liquid + vapor) remains constant. Therefore, we need to calculate the mass of trichloroethane in the vapor phase using the vapor pressure and temperature.

We can use the Clausius-Clapeyron equation to calculate the vapor pressure at 39°F (3.9°C):

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where:
P1 = vapor pressure at boiling point = 100.0 torr
P2 = vapor pressure at 39°F (3.9°C) (unknown)
ΔHvap = enthalpy of vaporization (unknown)
R = ideal gas constant = 8.314 J/(mol*K)
T1 = boiling point temperature = 74.1°C
T2 = temperature at 39°F (3.9°C) = 3.9°C

Rearranging the equation and solving for P2:

P2 = P1 * exp(-(ΔHvap/R) * (1/T2 - 1/T1))

Since we are told that the trichloroethane is at equilibrium, P2 equals the vapor pressure of the system.

By looking up the vapor pressure of trichloroethane at its boiling point (74.1°C or 347.25 K) in a reference source or through experimentation, we can substitute the values into the equation and solve for ΔHvap.

Assuming the vapor pressure of trichloroethane at 74.1°C is 760 torr (since it boils at this temperature), we can calculate ΔHvap:

760 torr = 100 torr * exp(-(ΔHvap/R) * (1/347.25 - 1/273.15))

ΔHvap = (log(760/100)) * (8.314 * (1/347.25 - 1/273.15))

ΔHvap ≈ 29,830 J/mol

Now, we can use the Clausius-Clapeyron equation to calculate the vapor pressure at 3.9°C:

P2 = 1000 torr * exp(-(29,830 J/mol / 8.314 J/(mol*K)) * (1/276.05 K - 1/298.05 K))

P2 ≈ 98.25 torr

Next, we can calculate the mass of trichloroethane in the vapor phase:

mass_vapor = (P2 * V) / (R * T)

mass_vapor = (98.25 torr * 509.7 liters) / (0.0821 L*atm/(mol*K) * (3.9 + 273.15) K)

mass_vapor ≈ 1,724.07 grams

Finally, we can calculate the percentage (by mass) of trichloroethane that is left as a liquid when equilibrium is established:

percentage_left = (mass_liquid / mass_total) * 100

percentage_left = (675,206.25 grams / (675,206.25 grams + 1,724.07 grams)) * 100

percentage_left ≈ 99.746%

Therefore, approximately 99.746% of the trichloroethane remains as a liquid when equilibrium is established.