Heat Processes (Isobaric, Isothermic, Isochoric, Adiabatic)

We started the class with understanding from past lectures about a system. System includes Temperature, Pressure, and Volume. Recall from last lecture, we learned about isothermal where temperature is constant, and adiabatic where heat does not change. Today we learn 2 more, which are Isobaric where pressure is constant, and Isovolumetric or Isochoric where Volume is constant.

We are given 6 questions from a website to see the relationship between Pressure, Volume, and Temperature at the 3 different situations that we discussed earlier. When the system is isobaric, we found the Volume and Temperature to be directly proportional by using the gas law formula that we have learned before. When it's isochoric, the Presure and Temperature is also directly proportional. When it's isothermal, we found the Pressure and Volume to be inversely proportional; thus, it gives a curve on the graph. 

After finding these relationships, we are given some knowns and find the unknown by using the relationships. For isobaric, we plug in the knowns and find the final volume to be 25.1 dm(cubic). For isothoric, we found the final pressure to be 126 kPa, and for isothermal, we found the initial pressure to be 124 kPa when final volume is 20 dm(cubic), abd 248 kPa when the final volume is 40 dm(cubic).

We are then given 4 different graphs and we need to predict which graphs shows the situations that we talked about earlier. For the first graph where the line is vertical, we predict it to be isochoric, since the volume is constant as pressure goes up. For the horizontal line graph, we predict it to be isobaric since pressure is constant as volume increases. For the third and forth graph is kind of tricky since the difference between them is only the length of the line. We predict the longer graph to be isothermal since it only depends on molar mass and Ideal gas constant; thus, we believe isothermal to have faster rate, whereas the adiabatic depends on mass and specific heat as well as those two in isothermal. Thus, it has a slower rate.

Next, we discuss about rubber band. When rubber band is heated, at first we thought that they will stretch and elongate. After seeing the youtube experiment, we found that the rubber band actually shrinks. This happens because rubber band is polymer. When it's heated, it curls up and shrink.

After understanding that, we now can use it in real life application. Since rubber band curls up when it gets heated, we can use it to transfer an object from lower to higher location. We are going to apply it to moving a can from a lower conveyer belt to higher conveyer belt. The steps are as below. When the engine moved by the conveyer belt there is a positive work done, but no heat in or out. When the rubber band is shrinking, there is heat going in, as well as work done on it, but when the rubber band cools down. It is the opposite for the hear and work.

Now, we imagine a situation where the outside temperature is fairly cold. Is it possible for the heat to fully goes into the rubber and have the work done? The answer is no, because when the rubber band cools down, there is heat loss, which means the heat can't be efficiently become work.

For the reason above, we take the heat loss into an account. Thus, work is equal to heat gain - heat loss. To calculate energy efficiency, we also learned a new formula, which e=1-Qh/Qc.

We are given another problem. A 1 mole of air is put into an iced water. When we transform this into a graph, the cycle looks like a rectangle. From there, we decide where the work and the heat are. Feom point 1 to 2, there is positive work done and heat in because the volume increase, but pressure is constant. From point 2 to 3, there is heat out but no work done because the volume does not change. From point 3 to 4, there is negative work done and heat out as well because the volume decrease, but the pressure does not change. From point 4 back to 1, volume is constant but pressure increases; therefore, there is only heat in, but no work done.

We are then given that internal energy is 3/2PV. We plugged in all the pressure and volume from each point an find the internal energy. From there, we want to find the change in internal energy by finding the difference between one point to the next point. After finding that, we make a table with heat and work done. We know from before that W=P*delta V. We use that equation to find work where the work is done. Then we'll find the heat by adding the change in internal energy and work.

By adding all the work done in the system, we find the net work on that cycle. When we calculate the area of the enclosed graph, it turns out to be exactly the same as the net work. Thus, we can calcualate the work both ways.

For last experiment. We use the syringe again thats connected to the flask. After running this experiment, we got the data as shown in the table below. Then we calculate the internal energy the same way as we did before as shown below.

Using point 3 where the temperature is closest to atmosphere temperature, we then calculate the mole using the ideal gas law with the known at that point. Then we found the mass by dividing the mole found by the molar mass of ? After finding the differences in internal energy of each point to the next point, we calculate the heat using the formula we have known (mc*delta t). Then we subtract internal energy by the heat to find the work. Finally, we recorded the data in the table as we did below.