Question

A 4.7-L sealed bottle containing 0.33 g of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) is placed in a refrigerator and reaches equilibrium with its vapor at \(-11^{\circ} \mathrm{C}\). (a) What mass of ethanol is present in the vapor? (b) When the container is removed and warmed to room temperature, \(20 .{ }^{\circ} \mathrm{C},\) will all the ethanol vaporize? (c) How much liquid ethanol would be present at \(0.0^{\circ} \mathrm{C}\) ? The vapor pressure of ethanol is \(10 .\) torr at \(-2.3^{\circ} \mathrm{C}\) and \(40 .\) torr at \(19^{\circ} \mathrm{C}\).

Short answer

No ethanol fully vaporizes at -11°C. At 20°C, all 0.33g vaporizes. For 0°C, less liquid ethanol persists.

Step-by-step solution

- Understanding the Problem

We need to determine the mass of ethanol in the vapor phase at -11°C, evaluate if all the ethanol will vaporize at 20°C, and find out the amount of liquid ethanol present at 0°C.

- Use Clausius-Clapeyron Equation

To find the vapor pressure at -11°C, use the Clausius-Clapeyron equation. Given the vapor pressures at two different temperatures, use the formula: \[ \ln \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right), \] where \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \).

- Merge Given Data

Convert the temperatures to Kelvin: \( T_1 = -2.3 + 273.15 = 270.85 \) K \( T_2 = 19 + 273.15 = 292.15 \) K \( P_1 = 10 \) torr and \( P_2 = 40 \) torr.

- Solve for ΔHvap

Using the Clausius-Clapeyron equation, \[ \ln \left( \frac{10}{40} \right) = \frac{\Delta H_{vap}}{8.314} \left( \frac{1}{292.15} - \frac{1}{270.85} \right), \] solve for \( \Delta H_{vap} \).

- Calculate vapor pressure at -11°C

Use Clausius-Clapeyron equation again with -11°C (262.15 K) to find the vapor pressure: \[ \ln \left( \frac{P_{-2.3}}{P_{-11}} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{1}{262.15} - \frac{1}{270.85} \right). \]

- Determine ethanol mass using Ideal Gas Law

After finding the vapor pressure at -11°C, use the ideal gas law, given by \( PV = nRT \), to find the number of moles of ethanol in vapor, then convert to mass: \[ m = n \times M_{molar}, \] where \( M_{molar} \) is the molar mass of ethanol.

- Evaluate total vaporization at 20°C

Using the vapor pressure at 20°C, determine if it can hold all 0.33g of ethanol as vapor.

- Repeat Steps for 0.0°C

Calculate the conditions at 0°C similar to previous steps.

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is essential for understanding changes in vapor pressure due to temperature variations. This equation relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization (\(\text{Δ}H_{vap}\)). Here's the formula for clarity:
\[ \text{ln} \bigg( \frac{P_1}{P_2} \bigg) = \frac{\text{Δ}H_{vap}}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) \]
The terms are as follows:

• P_1 and P_2 are the vapor pressures at temperatures T_1 and T_2 respectively
• is the universal gas constant, equal to 8.314 J/mol·K
• \(\text{Δ}H_{vap}\) is the enthalpy of vaporization

First, we identify the necessary data: Given vapor pressures at -2.3°C (10 torr) and 19°C (40 torr). Converting the temperatures into Kelvin:

• \(\text{T_1 = -2.3 + 273.15 = 270.85 K}\)
• \(\text{T_2 = 19 + 273.15 = 292.15 K}\)

Using this data, we solve for the enthalpy of vaporization (\(\text{Δ}H_{vap}\)). This enables us to calculate the vapor pressure at -11°C using the rearranged Clausius-Clapeyron equation and the known values.

Ideal Gas Law

The Ideal Gas Law is an equation of state for an ideal gas. It describes the relationship between pressure (P), volume (V), the number of moles (n), the gas constant (R), and temperature (T). The formula is:
\[\ \text{PV = nRT}\]
Here's how each variable is defined:

• P: Pressure of the gas
• V: Volume of the gas
• n: Number of moles of the gas
• R: Universal gas constant (8.314 J/mol·K)
• T: Temperature in Kelvin

Once we know the vapor pressure of ethanol at -11°C, we use this formula to determine the number of moles (\(\text{n}\)) of ethanol vapor. Given the volume of the container (4.7 L), we can solve for the moles of ethanol, and then convert it to mass. The molar mass of ethanol is:

• C = 12 g/mol
• H = 1 g/mol
• O = 16 g/mol

which sums up to 46 g/mol. By multiplying the number of moles by the molar mass, we find the mass of the ethanol vapor.

Phase Equilibrium

Phase equilibrium refers to a state where liquid and vapor phases of a substance coexist at equilibrium conditions. This is significantly governed by temperature and pressure. For example, at a certain temperature, the vapor pressure of ethanol determines how much ethanol will be present in the vapor phase.
At -11°C, the ethanol reaches phase equilibrium, meaning some ethanol remains as liquid while some vaporizes, creating a balance. By calculating the vapor pressure at specific temperatures, it becomes possible to determine how much of the ethanol will be present in each phase.
When the container is warmed to room temperature (20°C), the new equilibrium conditions might make all the ethanol vaporize, depending on how the vapor pressure changes with temperature. We utilize the previously calculated vapor pressures and apply phase equilibrium principles to predict if the entire 0.33g of ethanol will vaporize at this new temperature.
To check if all ethanol vaporizes or not, compare the calculated vapor amount at 20°C with the container’s capacity. If the calculated amount is less than the volume of the container could hold as vapor at room temperature, it indicates complete vaporization.