1. As the intermolecular forces between molecules increase in magnitude, do you expect each of the following to increase, decrease or remain the same?
My answers:
Viscosity: increase
Boiling point: increase
Vapor pressure: decrease
Surface tension: increase
Heat of Vapirization: increase
2. The melting point of compound X is 34 degrees Celsius at 1.00 atm pressure. The temperature of its triple point is 45 degrees Celsius. It may be helpful to sketch a phase diagram.
A) Compound X expands when it freezes
B) Compound X contracts when it freezes
C) Compound X's volume does not change when frozen
My answer: A
3. Growing up in a dry climate,we cooled the house using evaporative cooling. Like air, other things can be cooled using evaporation. How many grams of water can be cooled from 35 degrees C to 20 C by the evaporation of 65 g of water? The heat of vaporization of water in this temperature range is 43 kJ/mol. The specific heat of water is 4.18 J/g C.
My answer : 1.708g
To answer the questions and explain how to arrive at the answers:
1. The intermolecular forces between molecules can affect different properties. Here's how you can analyze the effect on each of the given properties:
a. Viscosity: Viscosity is a measure of a fluid's resistance to flow. As intermolecular forces increase, the molecules become more strongly attracted to each other, making it more difficult for them to flow past one another. Therefore, viscosity increases as intermolecular forces increase.
b. Boiling point: Boiling point is the temperature at which a substance changes from a liquid to a gas phase. Intermolecular forces affect the strength of the bonds holding the molecules together. As these forces increase, more heat energy is required to overcome them, resulting in an increase in boiling point.
c. Vapor pressure: Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. Stronger intermolecular forces lead to increased attraction between molecules, reducing the tendency of molecules to escape into the gas phase. Hence, vapor pressure decreases as intermolecular forces increase.
d. Surface tension: Surface tension is the force that acts on the surface of a liquid and tends to minimize its surface area. This force is caused by the cohesive forces between liquid molecules. With stronger intermolecular forces, the cohesive forces increase, resulting in higher surface tension.
e. Heat of vaporization: Heat of vaporization is the amount of heat energy required to vaporize a specific amount of a substance at its boiling point. As intermolecular forces increase, more energy is needed to overcome these forces during vaporization, so the heat of vaporization increases.
2. The phase diagram is a useful tool to analyze this question. Compound X has a melting point of 34 degrees Celsius at 1.00 atm pressure, which is below its triple point temperature of 45 degrees Celsius. Since the freezing point is lower than the triple point temperature, it means that compound X contracts when it freezes, which is option B.
3. To determine how many grams of water can be cooled from 35 degrees Celsius to 20 degrees Celsius through the evaporation of 65 grams of water, we need to calculate the heat lost during evaporation. Here's how to calculate it:
First, determine the amount of heat energy required to evaporate 65 grams of water:
Heat energy = moles of water × heat of vaporization
Moles of water = mass of water / molar mass of water
Molar mass of water = 18 g/mol
Moles of water = 65 g / 18 g/mol = 3.61 mol
Heat energy = 3.61 mol × 43 kJ/mol = 155.23 kJ
Next, calculate the heat lost by the water when it cools from 35 degrees Celsius to 20 degrees Celsius:
Heat lost = mass of water × specific heat capacity × change in temperature
Specific heat capacity of water = 4.18 J/g °C
Heat lost = mass of water × 4.18 J/g °C × (35°C - 20°C)
Heat lost = mass of water × 4.18 J/g °C × 15°C
Since the heat lost during cooling is equivalent to the heat gained during evaporation, we can set up an equation:
Heat lost = Heat energy gained from evaporation
mass of water × 4.18 J/g °C × 15°C = 155.23 kJ
Now, solve for the mass of water:
mass of water = 155.23 kJ / (4.18 J/g °C × 15°C) ≈ 1.708 g
Therefore, approximately 1.708 grams of water can be cooled from 35 degrees Celsius to 20 degrees Celsius through the evaporation of 65 grams of water.