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
Your answers seem accurate:
1. Viscosity: increase
Boiling point: increase
Vapor pressure: decrease
Surface tension: increase
Heat of Vaporization: increase
2. A) Compound X expands when it freezes
3. 1.708 g
1. To determine how the intermolecular forces affect each property, we need to understand the relationship between these forces and the properties.
- Viscosity: Viscosity is a measure of a fluid's resistance to flow. Intermolecular forces increase the attraction between molecules, making it harder for them to slide past each other. Therefore, as intermolecular forces increase, the viscosity of a substance also increases.
- Boiling point: The boiling point is the temperature at which a substance changes from a liquid to a gas phase. Stronger intermolecular forces require more energy to break the bonds between molecules, so substances with stronger intermolecular forces have higher boiling points.
- Vapor pressure: Vapor pressure is the pressure exerted by a substance's vapor when it is in equilibrium with its liquid phase. As intermolecular forces increase, it becomes more difficult for molecules to escape from the liquid phase, resulting in a decrease in vapor pressure.
- Surface tension: Surface tension is the force that acts at the surface of a liquid, causing it to behave as if it had a thin, elastic skin. Stronger intermolecular forces lead to stronger surface tension because it requires more energy to increase the surface area of a liquid.
- Heat of vaporization: The heat of vaporization is the amount of heat energy required to convert a substance from a liquid to a gas phase. As the intermolecular forces increase, it becomes more difficult to break the forces holding the liquid together, leading to an increase in the heat of vaporization.
Based on this understanding, your answers are correct.
2. The triple point of a substance is the temperature and pressure at which all three phases (solid, liquid, and gas) can coexist in equilibrium. The melting point of a substance is the temperature at which it changes from the solid phase to the liquid phase at a specific pressure (in this case, 1.00 atm).
In the given scenario, the melting point of compound X is lower than the temperature of its triple point. This means that at the temperature of its triple point (45 degrees Celsius), compound X has not yet reached its melting point and remains in the solid phase.
When a substance freezes, it typically contracts and becomes denser. Therefore, your answer of compound X contracting when it freezes (option B) is incorrect.
3. To determine the number of grams of water that can be cooled through evaporation, we need to consider the heat of vaporization of water and the specific heat of water.
The heat of vaporization of water is 43 kJ/mol, and the specific heat of water is 4.18 J/g*C.
First, we need to calculate the heat required to cool down the water from 35 degrees Celsius to 20 degrees Celsius.
Heat absorbed = mass x specific heat x temperature change
Heat absorbed = 65 g x 4.18 J/g*C x (35-20) C
Heat absorbed = 65 g x 4.18 J/g*C x 15 C
Next, we need to determine how much heat can be absorbed by the evaporation of water. One mole of water has a heat of vaporization of 43 kJ.
Moles of water for evaporation = (heat absorbed) / (heat of vaporization)
Moles of water for evaporation = ((65 g x 4.18 J/g*C x 15 C) / 1000 J/kJ) / (43 kJ/mol)
Finally, we can calculate the mass of water that can be cooled through evaporation using the molar mass of water (18.015 g/mol).
Mass of water cooled = (moles of water for evaporation) x (molar mass of water)
Mass of water cooled = ((65 g x 4.18 J/g*C x 15 C) / 1000 J/kJ) / (43 kJ/mol) x 18.015 g/mol
Simplifying this expression will give us the answer, which is approximately 1.708 g.
Therefore, your answer is correct.