You build a heat engine that takes 100 mol of an ideal...

You build a heat engine that takes 1.00 mol of an ideal diatomic gas through the cycle shown in Fig. P20.41. (a) Show that segment ab is an isothermal compression. (b) During which segment(s) of the cycle is heat absorbed by the gas? During which segment(s) is heat rejected? How do you know? (c) Calculate the temperature at points $a, b,$ and $c .$ (d) Calculate the net heat exchanged with the surroundings and the net work done by the engine in one cycle. (e) Calculate the thermal efficiency of the engine.

Transcript

00:01 Shown in this figure is the pv diagram of the heat engine in the problem where the number of moles the gas is given with a value of 1 .0 mole and it was stated that the gas is diatomic.

00:14 As such, its heat capacity and constant pressure process is 7 over 2r, where r is the ideal gas constant with a value of 8 .314 joules per mole kelvin.

00:27 And its capacity of the gas at constant volume process is 5 over 2.

00:36 So let's start answering question a.

00:40 Question a wants us to show that process a b is isothermal compression.

00:58 So you can see from the graph the volume at point a is greater than the volume at point b.

01:06 As such, this process is indeed a compression.

01:09 Now, what we need to do is to prove that it is is an isothermal process.

01:15 For it to be in an isothermal process, the temperature throughout the process should be constant as such change in t is equals to zero, or temperature at a should be equal to the temperature at b, or the ratio of t b over t .a is equals to 1.

01:38 So to prove that this process is an isotermal compression, we have to prove that t b over t .a is equal to 1.

01:44 So from combined gas low, we have pressure at point a multiplied by the volume at point a over the temperature at point a is equal to pressure at point b multiplied by the volume at point b over the temperature at point b.

02:09 Rearranging, we have tb over t .a is equals to pb -b -v -b over p -a -v -a.

02:24 Inserting the values from the graph the pressure at point b is 4 .0 times 10 raised to 5 pascal the volume at point b is 0 .005 meter cube over the pressure at point a which is 2 .0 times 10 raised to 5 pascal and the volume at point a which is 0 .01 meter cube solving this equation the answer is 1.

03:06 As such, we proved that process a -b is indeed an isothermal compression.

03:16 Now, for b, it asks us which among the process the heat is absorbed.

03:26 So we have to find where q is positive.

03:33 Heat is absorbed when the q is positive.

03:40 Let's start with the first process, which is the.

03:44 A b.

03:47 And since this process, the temperature is constant, we know that it is called isothermal.

03:58 The formula for getting the heat for an isothermal process is this one, q, ab, equals to n, r, t, ln of vb over va.

04:17 From the graph, we can see that volume at point b is less than the volume.

04:27 Point a.

04:29 As such, the change in volume is negative.

04:33 When the change's volume is negative, then the heat on this process is also negative.

04:39 So in this process, the heat is not absorbed.

04:41 It is rejected.

04:47 Now for the next process, process b to c, from the graph, we can see that its pressure is constant.

04:56 So this process is called isobaric.

05:03 For an isobaric process, the heat is it.

05:07 Qdc is equal to n cp and change in temperature.

05:14 Using ideal gas law, we can derive that cp.

05:20 It is also equals to cp over r multiplied by the constant pressure and the change in volume.

05:28 From the graph again, we can see that the volume at point c is greater than the volume at point b...

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