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Cell voltage example .txt | It gains those electrons that are released by this zinc solid metal. |
Cell voltage example .txt | So that means in step two, when we draw our electrochemical cell, the first half cell will contain our zinc solid. |
Cell voltage example .txt | And that's because zinc is oxidized, and the anode always contains the oxidation reaction. |
Cell voltage example .txt | So this is our zinc metal. |
Cell voltage example .txt | That means this must be our cadmium metal. |
Cell voltage example .txt | And so electrons will travel via the conductor, via this volt meter and into this cathode, into this electrode. |
Cell voltage example .txt | The volt meter, by the way, is the thing that reads zero point 36 volts. |
Cell voltage example .txt | So this sold bridge is placed here because it plays the role of closing the circuit. |
Cell voltage example .txt | Without the sole bridge, this guy would not function. |
Cell voltage example .txt | Electrons would not flow, and it's very important. |
Cell voltage example .txt | So, once again, our anode, our place where oxidation occurs, and our cathode, the place where reduction occurs. |
Cell voltage example .txt | So electrons travel from this way, from this electrode to this electrode. |
Cell voltage example .txt | So when they leave this cell, this solid zinc releases electrons, right? |
Cell voltage example .txt | Electrons begin to flow here. |
Cell voltage example .txt | It also releases a zinc ion into this solution. |
Cell voltage example .txt | And when the electrons travel this way, they combine. |
Cell voltage example .txt | When they reach this metal, they combine with the cadmium ions forming our cadmium solid. |
Cell voltage example .txt | So, in the third step, we basically want to use this cell voltage formula to find our cell voltage for the cathode. |
Cell voltage example .txt | Remember? |
Cell voltage example .txt | Now, we know that this guy, this value of negative zero point 76 represents the value for the anode. |
Cell voltage example .txt | So we use this formula to solve for our cathode. |
Cell voltage example .txt | So we know our 0.36 volts, which is told by the voltmeter. |
Cell voltage example .txt | Now, this we don't know. |
Cell voltage example .txt | So we let a DX minus now in parentheses we have a negative 0.76 volts, right? |
Cell voltage example .txt | So negative and negative becomes a positive. |
Cell voltage example .txt | Then we subtract both sides by this guy and we get x to be 0.36
-0.76 volts gives us negative 0.4 volts. |
Cell voltage example .txt | So this is our cell voltage for the cathode cell for the cadmium for that reduction reaction of cadmium. |
Charles Law .txt | Now we already spoke about a concept called the Kinetic Molecular Theory of Gases. |
Charles Law .txt | And what this theory did for us is it helped us gain more intuition about how individual gas molecules interact on a very small microscopic level. |
Charles Law .txt | Now we also spoke about a law called Boils Law. |
Charles Law .txt | And what Boils Law did for us is it helps help us gain more intuition about the macroscopic behavior of gases. |
Charles Law .txt | In other words, what happens at constant temperature when we take a balloon filled with air and squeeze it. |
Charles Law .txt | Well, we said that, and Boyle's Law explained this, that when we squeeze the balloon, we decrease volume, increase pressure, and eventually our balloon will pop. |
Charles Law .txt | Now we're going to look at another law called Charles Law which also helps us explain the macroscopic large scale behavior of many gas molecules and how they interact with one another and with our system. |
Charles Law .txt | So for Charles Law to work, two conditions must hold. |
Charles Law .txt | We must have constant pressure and we must have constant number of molecules. |
Charles Law .txt | So our N number of moles stays the same. |
Charles Law .txt | And what Charles Law does is it relates volume and temperature. |
Charles Law .txt | Remember Boyle's Law related volume and pressure? |
Charles Law .txt | Well, Charles Law relates temperature and volume when pressure is constant. |
Charles Law .txt | And what it says is that volume is directly proportional to temperature. |
Charles Law .txt | And this means if we bring the T over or if we multiply a constant by T and we bring T over, what we get is the following v divided by T is equal to a constant. |
Charles Law .txt | Remember, in Boyle's Law we saw that P times V gave us a constant and that if we increase pressure, our volume must decrease while keeping the constant the same. |
Charles Law .txt | Well, in this case, we have the same kind of situation, except now we have volume divided by temperature. |
Charles Law .txt | In other words. |
Charles Law .txt | Now for this constant to remain a constant and not change if we increase volume, say by two, our temperature also must increase by the same amount by two. |
Charles Law .txt | So whenever we increase volume, we increase temperature. |
Charles Law .txt | Or if we decrease volume, we must decrease the temperature for this guy to remain constant. |
Charles Law .txt | Now notice this constant remains or depends on two things on the pressure and on the number of molecules. |
Charles Law .txt | For example, if we have more molecules, our constant will be higher. |
Charles Law .txt | And we'll learn more about that when we talk about the ideal gas law. |
Charles Law .txt | For now, it's sufficient to say that our cost depends on both of these guys. |
Charles Law .txt | So the same way we did for Boyle's Law, for Charles Law, we can rewrite this relation or this relation in the following manner. |
Charles Law .txt | Now, suppose I have some system, some gas system and I have two sets of different conditions for this gas system. |
Charles Law .txt | Well, I can relate them in the following manner. |
Charles Law .txt | Assuming that these guys are both constant. |
Charles Law .txt | In other words, note that the constant always stays the same when our pressure number of moles stays the same. |
Charles Law .txt | That means if we have some conditions, v one and T one and a second condition of V two and T two. |
Charles Law .txt | And these two conditions are under the same pressure and number of moles, that means our constant will be the same. |
Charles Law .txt | And so I can say V one over T one equals V two over t two equals that same constant. |
Charles Law .txt | So once again, for a gas sample with two different sets of T's and Vs for two different conditions, this is our law. |
Charles Law .txt | This is our Charles Law. |
Charles Law .txt | Now, whenever we talk about gases, and we talk about these different laws that revolve around gases, our temperature is never in Celsius, it's only in Kelvin. |
Charles Law .txt | And that means we have to convert Celsius to Kelvin. |
Charles Law .txt | And this is how you do it. |
Charles Law .txt | To get the Kelvin temperature, you take the Celsius temperature and you add 273.15 to it. |
Charles Law .txt | For example, if our Celsius temperature is ten degrees Celsius, I simply add 273.15 and I get 283.15 Kelvin. |
Charles Law .txt | That's our temperature in Kelvin. |
Charles Law .txt | So let's think of an example where Charles Law is evident. |
Charles Law .txt | So suppose it's someone's birthday and it's winter. |
Charles Law .txt | So it's cold outside and you need to go and buy somebody a present. |
Charles Law .txt | So he decides to buy a balloon. |
Charles Law .txt | So you go inside the store, you fill up the balloon with some helium. |
Charles Law .txt | Now suppose once you fill the balloon up, there are no holes in the balloon. |
Charles Law .txt | So the number of moles or number of molecules inside our balloon remains constant. |
Charles Law .txt | Also, let's suppose that our pressure inside the store and outside is one ATM atmospheric pressure. |
Charles Law .txt | So let's suppose that our pressure is also constant. |
Charles Law .txt | Now, obviously, if it's winter, it's much colder outside than inside. |
Charles Law .txt | So suppose I fill up my balloon with N number of moles of helium, and suppose now I go outside with that balloon. |
Charles Law .txt | Well, on the inside, inside the store, we were at one condition. |
Charles Law .txt | We had some D two and t two, right? |
Charles Law .txt | Once we step outside, our temperature drops. |
Charles Law .txt | So what happens to volume? |
Charles Law .txt | Well, according to Charles Law, this guy states that if we go outside and my temperature drops, then my volume must drop as well to make sure that our constant stays the same. |
Charles Law .txt | That means if I go outside with my balloon, my balloon will shrivel, it will become smaller, and that's because of Charles Law. |
Charles Law .txt | So once again, we see how macro scale events are explained by Charles Law, just like they were explained by Boyle's Law. |
Charles Law .txt | So we have yet another law that helps us explain how gas molecules behave when we have a lot of them together, not simply individual gas molecules. |
Charles Law .txt | Now, we can, of course, explain how Charles Law functions on an individual level using the kinetic theory. |
Charles Law .txt | Now, kinetic theory explains Charles Law on a microscopic level at constant pressure. |
Charles Law .txt | So we have constant pressure, right? |
Charles Law .txt | Our pressure doesn't change. |
Charles Law .txt | So if our temperature increases, what happens to the molecules? |
Charles Law .txt | The molecules gain more kinetic energy. |
Charles Law .txt | And if they gain more kinetic energy, they gain more speed. |
Charles Law .txt | They become quicker. |
Charles Law .txt | And that means the only way that our pressure stays constant with higher kinetic energy is if the volume expands. |
Charles Law .txt | So that's exactly what happens. |
Charles Law .txt | In order to ensure that there is constant pressure. |
Charles Law .txt | An increase in temperature means there's an increase in kinetic energy. |
Charles Law .txt | And this increase in kinetic energy means that there are more molecules, or the molecules are pushing against the wall of the container with greater force. |