Using the Clicker If you have a clicker

Using the “Clicker” If you have a clicker now, and did not do this last time, please enter your ID in your clicker. First, turn on your clicker by sliding the power switch, on the left, up. Next, store your student number in the clicker. You only have to do this once. Press the * button to enter the setup menu. Press the up arrow button to get to ID Press the big green arrow key Press the T button, then the up arrow to get a U Enter the rest of your BU ID. Press the big green arrow key.

Ice water 100 grams of ice, with a temperature of -10°C, is added to a styrofoam cup of water. The water is initially at +10°C, and has an unknown mass m. If the final temperature of the mixture is 0°C, what is the unknown mass m? Assume that no heat is exchanged with the cup or with the surroundings. Use these approximate values to determine your answer: Specific heat of liquid water is about 4000 J/(kg °C) Specific heat of ice is about 2000 J/(kg °C) Latent heat of fusion of water is about 3 x 105 J/kg

Ice water One possible starting point is to determine what happens if nothing changes phase. How much water at +10°C does it take to bring 100 g of ice at -10°C to 0°C? (The water also ends up at 0°C. ) You can do heat lost = heat gained or the equivalent method: Plugging in numbers gives: Lot's of things cancel and we're left with: 100 g = 2 m, so m = 50 g. So, that's one possible answer.

Ice water Challenge : find the range of possible answers for m, the mass of the water.

Ice water Another possibility is that all the ice melts. You can do heat lost = heat gained or the equivalent method: Plugging in numbers gives: Lot's of things cancel and we're left with: 100 g × 32 = 4 m, so m = 800 g. So, there's another possible answer.

Ice water If nothing changes phase, m = 50 g. If all the ice melts, m = 800 g. Are we done? In other words, is the answer that as long as m is greater than or equal to 50 g and less than or equal to 800 g the final temperature of the mixture will be 0°C, and for all values of m outside of this range the final temperature will be something other than 0°C? 1. Yes, that is correct 2. No, at some values of m between 50 and 800 g the final temperature will be something other than 0°C. 3. No, some values below 50 g would also work. 4. No, some values above 800 g would also work.

Ice water If 50 g of water is enough to bring the ice to 0°C while 800 g of water is enough to bring the ice to 0°C and completely melt it, what happens for any mass of water between 50 g and 800 g?

Ice water If 50 g of water is enough to bring the ice to 0°C while 800 g of water is enough to bring the ice to 0°C and completely melt it, what happens for any mass of water between 50 g and 800 g? Some of the ice melts, but not all, and the final temperature of the mixture is still 0°C. Are there any other possibilities that would result in a final temperature of 0°C?

Ice water If 50 g of water is enough to bring the ice to 0°C while 800 g of water is enough to bring the ice to 0°C and completely melt it, what happens for any mass of water between 50 g and 800 g? Some of the ice melts, but not all, and the final temperature of the mixture is still 0°C. Are there any other possibilities that would result in a final temperature of 0°C? Yes – some or all of the water could turn to ice.

Ice water Let’s see what happens if all the water freezes. You can do heat lost = heat gained or the equivalent method: Plugging in numbers gives: Lot's of things cancel and we're left with: 100 g × 2 = 34 m, so m ≈ 6 g. So, there's another possible answer.

Ice water, summary Amazingly, we can add anywhere from 6 g to 800 g of water at +10°C to the 100 g of ice at -10°C and get a mixture with an equilibrium temperature of 0°C. This is because the equilibrium temperature we were trying to achieve is the temperature at which liquid and solid water can co-exist, and we get such a wide ranges of possible masses because of the large amount of energy associated with a phase change.

Thermodynamics: a microscopic view We’ll now turn to looking at thermodynamics at the level of individual molecules.

Avogadro’s number A mole is very similar to a dozen, in the sense that it stands for a certain number of things. A dozen means 12, while a mole means 6. 02 x 1023. This is also known as Avogadro's number, NA.

The ideal gas law An ideal gas satisfies these conditions: • It consists of a large number of identical molecules. • The volume occupied by the molecules themselves is negligible compared to the volume of the container they're in. • The molecules obey Newton's laws of motion, and they move in random motion. • The molecules experience forces only during collisions; any collisions are completely elastic, and take a negligible amount of time. Simulation

The ideal gas law states that: PV = n. RT where P is pressure, V is volume, n is the number of moles, T is the absolute temperature, and R = 8. 31 J/(mol K) is the universal gas constant. The ideal gas law can be written in terms of N, the number of molecules, instead. N = n NA, so: where k is the Boltzmann constant, 1. 38 x 10 -23 J/K. This is the universal gas constant divided by Avogadro's number.

Raising the temperature The temperature of an ideal gas is raised from 10°C to 20°C while the volume is kept constant. If the pressure at 10°C is P, what is Pf, the pressure at 20°C? 1. 2. 3. 4. 5. Pf < P Pf = P P < Pf < 2 P Pf = 2 P Pf > 2 P

Raising the temperature This is a situation in which we’re using an equation (the ideal gas law) with a T, so the temperatures should be in Kelvin. The temperature of an ideal gas is raised from 283 K to 293 K while the volume is kept constant. If the pressure at 283 K is P, what is Pf, the pressure at 293 K?

Raising the temperature This is a situation in which we’re using an equation (the ideal gas law) with a T, so the temperatures should be in Kelvin. The temperature of an ideal gas is raised from 283 K to 293 K while the volume is kept constant. If the pressure at 283 K is P, what is Pf, the pressure at 293 K? Not too much more than P.

Kinetic theory is the application of Newton's Laws to ideal gases. Start with one atom of ideal gas in a box and determine the pressure associated with that atom. Then add more atoms and sum (or average) over all the atoms to see what you get.

Results from kinetic theory, 1 1. Pressure is associated with collisions of gas particles with the walls. Dividing the total average force from all the particles by the wall area gives the pressure. Increasing temperature increases pressure for two reasons. There are more collisions, and the collisions involve a larger average force. For a fixed volume and temperature, adding more particles increases pressure because there are more collisions.

Results from kinetic theory, 2 What is the average velocity of the ideal gas particles?

Results from kinetic theory, 2 What is the average velocity of the ideal gas particles? 2. The average velocity is zero, because, on average, the velocity of particles going in one direction is cancelled by the velocity of particles going in the opposite direction. When you do the analysis, you find that what really matters is the rmsspeed (rms stands for root mean square). Square the speeds, take the average, and take the square root of that result.

Results from kinetic theory, 3 3. The really fundamental result of kinetic theory is that temperature is a direct measure of the average kinetic energy of the particles of ideal gas. Kinetic theory: Ideal gas law: This tells us that the average translational kinetic energy of the molecules is: Here we have a fundamental connection between temperature and the average translational kinetic energy of the atoms - they are directly proportional to one another.

Light and heavy atoms A box of ideal gas consists of light particles and heavy particles (the heavy ones have 16 times the mass of the light ones). Initially all the particles have the same speed. When equilibrium is reached, what will be true? 1. All the particles will still have the same speed 2. The average speed of the heavy particles equals the average speed of the light particles 3. The average speed of the heavy particles is larger than that of the light particles 4. The average speed of the heavy particles is smaller than that of the light particles 5.

Light and heavy atoms Coming to equilibrium means coming to a particular temperature, which means the average kinetic energy of the particles is a particular value. The average kinetic energy of the light particles equals the average kinetic energy of the heavy particles - this can only happen if the average speed of the heavy particles is smaller than that of the light particles. Smaller by a factor of what?

Light and heavy atoms Simulation Coming to equilibrium means coming to a particular temperature, which means the average kinetic energy of the particles is a particular value. The average kinetic energy of the light particles equals the average kinetic energy of the heavy particles - this can only happen if the average speed of the heavy particles is smaller than that of the light particles. Smaller by a factor of what? 4, so that:

Three cylinders Three identical cylinders are sealed with identical pistons that are free to slide up and down the cylinder without friction. Each cylinder contains ideal gas, and the gas occupies the same volume in each case, but the temperatures differ. In each cylinder the piston is above the gas, and the top of each piston is exposed the atmosphere. In cylinders 1, 2, and 3 the temperatures are 0°C, 50°C, and 100°C, respectively. How do the pressures compare? 1. 1>2>3 2. 3>2>1 3. all equal 4. not enough information to say

Free-body diagram of a piston Sketch the free-body diagram of one of the pistons.

Free-body diagram of a piston Sketch the free-body diagram of one of the pistons. The internal pressure, in this case, is determined by the free-body diagram. All three pistons have the same pressure, so the number of moles of gas must decrease from 1 to 2 to 3.

Introducing the P-V diagram P-V (pressure versus volume) diagrams can be very useful. What are the units resulting from multiplying pressure in k. Pa by volume in liters? Rank the four states shown on the diagram based on their absolute temperature, from greatest to least.

Introducing the P-V diagram P-V (pressure versus volume) diagrams can be very useful. What are the units resulting from multiplying pressure in k. Pa by volume in liters? Rank the four states shown on the diagram based on their absolute temperature, from greatest to least. Temperature is proportional to PV, so rank by PV: 2 > 1=3 > 4.

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