Phase+Changes

Phase Changes In order for a substance to move between the states of matter; for example, to turn from a solid into a liquid, which is called **fusion**, or from a gas to a liquid (**vaporization**), energy must be gained or lost. The **heat of fusion** (symbolized//H//fus) of a substance is the amount of energy that must be put into the substance for it to melt. For example, the heat of fusion of water is 6.01 kJ/mol, or in other terms, 80 cal/g. The **heat of vaporization**, not surprisingly, is the amount of energy needed to cause the transition from liquid to gas, and it is symbolized //H//vap. You will not be required to memorize heat of fusion or vaporization values.  media type="custom" key="25330690"   Changes in the states of matter are often shown on phase diagrams, and you will probably see at least one of two different types of phase diagrams on the SAT II Chemistry exam. Let’s start with the phase diagram for water. The **phase diagram** for water is a graph of pressure versus temperature. Each of the lines on the graph represents an equilibrium position, at which the substance is present in two states at once. For example, anywhere along the line that separates ice and water, melting and freezing are occurring simultaneously.

The intersection of all three lines is known as the **triple point** (represented by a dot and a //T// on the figure). At this point, all three phases of matter are in equilibrium with each other. Point //X// represents the **critical point**, and at the critical point and beyond, the substance is forever in the vapor phase. This diagram allows us to explain strange phenomena, such as why water boils at a lower temperature at higher altitudes, for example. At higher altitudes, the air pressure is lower, and this means that water can reach the boiling point at a lower temperature. Interestingly enough, water would boil at room temperature if the pressure was low enough! One final note: If we put a liquid into a closed container, the evaporation of the liquid will cause an initial increase in the total pressure of the system, and then the pressure of the system will become a constant. The value of this final pressure is unique to each liquid and is known as the liquid’s **vapor pressure**. Water has a relatively low vapor pressure because it takes a lot of energy to break the hydrogen bonds so that molecules enter the gas phase. Water and other liquids that have low vapor pressures are said to be **nonvolatile**. Substances like rubbing alcohol and gasoline, which have relatively high vapor pressures, are said to be **volatile**. Example  Explanation Looking at the phase change diagram for water and following the dashed line at 1 atm, you can see that water would begin as a solid (ice) and melt at 0ºC. All of the water would be in liquid form by the time the temperature reached 75ºC. The second type of phase change graph you might see is called a **heating curve**. This is a graph of the change in temperature of a substance as energy is added in the form of heat. The pressure of the system is assumed to be held constant, at normal pressure (1 atm). As you can see from the graph below, at normal pressure water freezes at 0ºC and boils at 100ºC.

The plateaus on this diagram represent the points where water is being converted from one phase to another; at these stages the temperature remains constant since all the heat energy added is being used to break the attractions between the water molecules. <span style="background-color: #ffffff; color: #666666; display: block; font-family: georgia,times,serif; font-size: 18px; text-align: left;">Specific Heat <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">You might see a diagram that looks something like this one, and you might come across a question that asks you to calculate the amount of energy needed to take a particular substance through a phase change. This would be one of the most difficult questions on the exam, but you might see something like it, or at least part of it. If you were asked to do this, you would need to use the following equation: <span style="background-color: #ffffff; color: #57585b; display: block; font-family: Helvetica,Verdana,Arial,sans-serif; font-size: 11px;"> <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">energy (in calories) = //mC////p// D//T//  <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">where //m// = the mass of the substance (in grams) <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">//C////p// = the specific heat of the substance (in cal/g ºC) <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">D//T//   =  the change in temperature of the substance (in either Kelvins or ºC, but make sure all your units are compatible!) <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">As you can see, this requires that you know the specific heat of the substance. A substance’s **specific heat** refers to the heat required to raise the temperature of 1 g of a substance by 1ºC. You will not be required to remember any specific heat values for the exam. <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">Work through the example below to get a feel for how to use this equation. <span style="background-color: #ffffff; color: #666666; display: block; font-family: georgia,times,serif; font-size: 18px; text-align: left;">Example <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">If you had a 10.0 g piece of ice at -10ºC, under constant pressure of 1 atm, how much energy would be needed to melt this ice and raise the temperature to 25.0ºC? <span style="background-color: #ffffff; color: #666666; display: block; font-family: georgia,times,serif; font-size: 18px; text-align: left;">Explanation <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">First, the temperature of the ice would need to be raised from -10ºC to 0ºC. This would require the following calculation. The specific heat for ice is 0.485 cal/g ºC. Substituting in the formula <span style="background-color: #ffffff; color: #57585b; display: block; font-family: Helvetica,Verdana,Arial,sans-serif; font-size: 11px;"> <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">energy = //mC////p// D//T//; energy = (10.0 g) (0.485 cal/g ºC) (10.0ºC) = 48.5 cal  <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">So 48.5 calories are needed to raise temperature. <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">Next, we must calculate the heat of fusion of this ice: we must determine how much energy is needed to completely melt the 10 g of it. <span style="background-color: #ffffff; color: #57585b; display: block; font-family: Helvetica,Verdana,Arial,sans-serif; font-size: 11px;"> <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">energy = //mH//fus  <span style="background-color: #ffffff; color: #57585b; display: block; font-family: Helvetica,Verdana,Arial,sans-serif; font-size: 11px;"> <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">energy = (10.0 g) (80 cal/g) = 800 cal   <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">So 800 cal of energy are needed to completely melt this sample of ice. <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">Next, we need to see how much energy would be needed to raise the temperature of water from 0ºC to 25ºC. The specific heat for liquid water is 1.00 cal/g ºC. So again use <span style="background-color: #ffffff; color: #57585b; display: block; font-family: Helvetica,Verdana,Arial,sans-serif; font-size: 11px;"> <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">energy = //mC////p// D//T// to get energy = (10.0 g) (1.00 cal/g ºC) (25.0ºC) = 250 cal  <span style="background-color: #ffffff; color: #333333; display: block; font-family: georgia,times,serif; font-size: 14px; text-align: left;">Finally, add together all of the energies to get the total: 48.5 + 800 + 250 = about 1100 calories are needed to convert the ice to water at these given temperatures.