Problem 54 - General Chemistry: Principles and Modern Applications

Question

A \(2.50 \mathrm{g}\) sample of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is sealed in a \(5.00 \mathrm{L}\) flask at \(120.0^{\circ} \mathrm{C}\) (a) Show that the sample exists completely as vapor. (b) Estimate the temperature to which the flask must be cooled before liquid water condenses.

Step-by-Step Solution

Answer

The sample exists completely as vapor in the given exercise scenario, and the flask must be cooled to approximately \(80^{\circ} C\) for liquid water to start condensing. This performance of calculations assumes that water behaves ideally and the ideal gas law applies.

Step 1: Converting Mass to Moles

The mass of water is given as 2.50g. To make it useful for the gas law equations, it needs to be converted into moles. This is done using the molecular weight of water, which is about 18.015 g/mol. So the number of moles (\(n\)) of water = \(2.50 \, g / 18.015 \, g /mol = 0.1387 \, mol\).

Step 2: Realizing Water as Ideal Gas

Given that the water is sealed in the flask, operating at a high temperature of \(120.0^{\circ} C = 393.15 \, K\), the water can be approximated to behave as an ideal gas. Therefore, the ideal gas law \(PV = nRT\) (where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature) can be applied.

Step 3: Estimating the Pressure

Using the ideal gas law, it is possible now to calculate the pressure (\(P\)) inside the flask. \(P = nRT/V = 0.1387 \, mol * 0.08206 \, L-atm/K-mol * 393.15 \, K / 5.00 \, L = 0.912 \, atm\). The pressure inside the flask is below the vapor pressure of water at \(120.0^{\circ} C\) (which is around 1.99 atm). This indicates that water exists completely as vapor in the flask. Therefore, (a) is validated.

Step 4: Clausius-Clapeyron Equation

Next, to estimate the temperature at which the liquid water condenses, the Clausius-Clapeyron equation which describes the phase transitions can be used. The equation is \(ln(P_2/P_1) = -(\Delta H_{vap} / R) * (1/T_2 - 1/T_1)\). Here, \(P_1\) and \(P_2\) are the initial and final pressures, \(T_1\) and \(T_2\) are the corresponding temperatures, \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(R\) is the gas constant.

Step 5: Solving for Final Temperature

At \(1 atm\), water starts to condense. So, the final pressure (\(P_2\)) is 1 atm. Using \(P_2 = 1 atm\), \(P_1 = 0.912 atm\), \(\Delta H_{vap} = 40.7 KJ/mol = 40700 J/mol\), \(R = 8.314 J/mol-K\), and \(T_1 = 393.15 K\) in the Clausius-Clapeyron equation, solving for \(T_2\) gives \(T_2 =353 K\) or \(80^{\circ} C\). This means the flask must be cooled down to approximately \(80^{\circ} C\) for the water vapor to start condensing. So, the answer to (b) is \(80^{\circ} C\).

Key Concepts

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is crucial for understanding phase transitions between liquid and gas. This equation quantifies how vapor pressure changes with temperature. Imagine heating water in a sealed flask. As it warms, some of the liquid water evaporates, increasing the vapor pressure until equilibrium is reached. How do we estimate the temperature at which water starts condensing? The Clausius-Clapeyron equation provides just that. This equation is expressed as \(\ln(\frac{P_2}{P_1}) = -\frac{\Delta H_{vap}}{R} \times (\frac{1}{T_2} - \frac{1}{T_1})\), where:

• \(P_1\) and \(P_2\) are initial and final pressures.
• \(T_1\) and \(T_2\) are the corresponding temperatures.
• \(\Delta H_{vap}\) is the enthalpy of vaporization.

Using this equation helps us find out how much to cool the flask for condensation. By inputting known values and solving for unknown temperatures, we can map out phase transitions efficiently. This method is vital in chemistry and physics, particularly when predicting states and behaviors of fluids.

enthalpy of vaporization

Enthalpy of vaporization, often symbolized as \(\Delta H_{vap}\), is the energy required to convert one mole of liquid into vapor at constant temperature and pressure. It's an essential concept that describes how much energy is needed for phase change. In water, this value is particularly significant due to hydrogen bonding, requiring relatively high energy compared to many other liquids. For water, \(\Delta H_{vap}\) is approximately 40.7 kJ/mol.This energy is not just about breaking bonds; it's also about increasing the molecules' kinetic energy so they can enter the gas phase. Think of it as the cost of enabling molecules from the liquid to move freely as gas. Consider a situation where the water in a flask must evaporate completely with no increase in temperature. The heat provided reflects on this phase transition, hinting at \(\Delta H_{vap}\)'s role. This concept often combines with pressure and temperature data in equations, like the Clausius-Clapeyron, to predict and explain phase behaviors and transitions.

vapor pressure

Vapor pressure is a measure of the pressure exerted by a vapor in equilibrium with its liquid (or solid) form in a closed system. Interestingly, it depends solely on temperature. As the temperature increases, so does the vapor pressure, due to the added energy which allows more molecules to escape the liquid phase, forming vapor. When a liquid is trapped in a closed container, such as our flask with water, it will vaporize until it reaches a specific equilibrium pressure—this is the vapor pressure. At this point, the rates of condensation and evaporation are equal, meaning the quantity of liquid and vapor remains constant. In our exercise, the vapor pressure of water at 120°C is key to determining whether it will remain completely as vapor. By comparing it to the calculated pressure inside the flask, we deduced that the water remains in vapor form since the internal pressure is lower than its vapor pressure. This informs us about the conditions where no liquid is present, a crucial insight for understanding gaseous behaviors of fluids.