FIGURE CP19.80 shows a thermodynamic process followed by 0.015 mol of hydrogen. How much heat energy is transferred to the gas?

Video transcript

Hi, everyone in this practice problem, we're being asked to calculate the amount of heat released by the gas where we'll have 0.042 moles of carbon dioxide gas undergoing a thermodynamic process described by a PV diagram. As shown in the figure, the figure is showing a PV diagram with P in ATM in the Y axis and V in leader in the X axis. We're in the initial position of 1.0 liter comma 1.0 ATM. It will go into the final position or final point of 0. liter comma 10 A PM. The options given for the amount of heat released by the gas are a zero B 74 Jules c 1.2 times 10 to the power of three Jules and D 4.1 times 10 to the power of three jos. So in order for us to be able to solve this problem, we will have to make a couple assumptions. So first that we have to make is that the CO2 will be acting like an ideal gas. And then second, the process is going to be reversible. And at the same time it will be quasi static next or lastly, the system is going to be isolated, so isolated system here in the assumption. So we will divide up the uh process of actually solving the amount of heat released by the gas into a couple of steps. The first step here is to determine the work done on the gas during compression or determining W. So in this case, we have to convert all the units of measurements into si and W will be represented by the PV diagram as the area under the curve. So area under the curve and that will be divided into a one plus A two. In this case, I will say that the A one is going to be this triangle right here and the A two later on will be the rectangle underneath this triangle right here. So in uh red here is going to be a one and A two is going to be represented by this blue here which will be a rectangle. So this is a two just like. So, so a one plus A two, we can calculate that by taking the area under the curve. So let's start with a one. So a one for W here will be equal to half multiplied by. First is going to be the base which is the volume and the volume is given in leader. And I'm going to immediately convert that into um meter cube. So that'll be 0.01 m cube for the one liter minus 0.0001 m Q for the 0.1 liter or the final position multiply that with the height which is going to be 1.013 multiplied by 10 to the power of six pascals for 10 A PM minus 1. times 10 to the power of five pascals for the 1. A PM. This is going to be a one when a plus or sum that with a two, which is going to be 0.01 m cube minus 0.0001 m cube. The base is going to be the same but the length or the height of the rectangle is going to be different by the height of the triangle. So this will be equal to 1.013 times 10 to the power of five pascals for one A PM minus zero pascals for zero A PM. Calculating this, the work done will then come out to be 4501 2 multi uh 0.92 Joes for the Tora ale plus 1002.87 Jews for the rectangle or 4512.92 Jews for the area one or area in red and 1000 and 2.87 for the area in blue or area two in total. That will give us the work done on the gas during the compression to be 5515. jules. Awesome. Next, we want to move on to step two, which is determining the initial and final temperature using ideal gas law. So tintf so in this case, the I using ideal gas law will follow the equation of piv I divided by an R. Let's substitute all of our values. The initial pressure is going to be 01.013 times 10 to the power of five pascals or one A PM. And the initial volume is going to be 0.01 m cube or one liter to fight that with N which is going to be given in the problem statement to be 0.042 moles and multiply that with the uh ideal gas constant which is 8.31 joules per mole. Kelvin calculating this, we will obtain T I to be 2902.41. Kelvin moving on to TFTF will follow the same formula where TF will be equal to PFVF divided by NRPF or the uh final pressure will be 1.1 0.1 0.013 times standard of power of six pascals for Patter A PM and 0.0001 m four. Given in a problem statement to be 0.1 liter, the NNR will be the same which is 0.042 moles multiplied by 8.31 SB mall Kelvin and calculating this, we will obtain TF or the final temperature to be 290.24. Kelvin. Next, you want to calculate the change in the thermal energy or for step three is the delta eth. And if we recall, delta eth will follow the formula of N multiplied by CV, multiplied by delta T or essentially N multiplied by CV, multiplied by, in parenthesis, the F minus the I. So let's substitute all of our values. Delta Eth will then be equal to the most is 0. moles. Give it a problem statement. CV is the constant of 12.5 Joles per mole. Kelvin and TFS given to be uh or we just calculated that to be 290.24 Kelvin and Tis 2902.41. Kelvin calculating this to delta et I will then come out to a value of negative 371.39 jos lastly, what we need to calculate is the amount of heat released using the first law of thermodynamics. So four is going to be calculating Q using the first law of thermal dynamics. So delta EPH will be equal to Q minus W according to the first law of thermodynamics. And then rearranging Q or the amount of heat released will be equal to delta EPH plus W. So let's substitute the values that we obtain. The delta eth is negative 1000 371.39 jules and W is obtained in step one to be 5515. 15.79 Jules calculating this, the Q or the heat released will be equal to 4144.4 jules or rounding it so that it will come up in our answer choices that will correspond to 4.1 times 10 to the power of three Jews, which will be the final answer to this particular practice problem and will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. And thank you.