Skip to main content
Ch 19: The First Law of Thermodynamics

Chapter 19, Problem 19

Work Done in a Cyclic Process. (a) In Fig. 19.7a, consider the closed loop 1 → 3 → 2 → 4 → 1. This is a cyclic process in which the initial and final states are the same. Find the total work done by the system in this cyclic process, and show that it is equal to the area enclosed by the loop.

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
1636
views
Was this helpful?

Video transcript

Welcome back everybody. We are looking at a cyclical process on a PV diagram and we are tasked with finding two different things. First. We need to figure out the network done during the process and we then need to verify that the network is equal to the area of the loop. So let's go ahead and start with part one here. The way we're going to do this is in order to get the network, we have to add up the work of all the different parts of the loop. So you can see here, we kind of have the process from A. To B. B, two C, C to D and then D to A. So we're going to need to find four separate values and then add them together. So our network will be the work of A. B plus the work of Bc plus the work from C to D. Plus the work from D to A. We'll just take this one at a time here. Um Just as a reminder, the formula for work is going to be pressure times our change in volume, which is going to be the pressure times our final volume minus our initial volume. Let's go ahead and go through these values. So the work done from A to B. Well let's see here we have a two B. This is at a pressure P not in our final vol is be not minus our initial of b. one. Right, so that is our work for A to B. Now. What about our work from B to C. So at this point and this point I want to make an observation here, this is all at a volume of the not. So our delta v here is just going to be zero, meaning that our work from B to Z is going to be equal to zero. Now let's look at the work done from C to D, C to D. We are looking at P one as our pressure, our final volume is going to be V one minus our initial volume of the not. And I know that because I'm just kind of following the arrows of this cyclic process and finally work from D all the way back to A. We're dealing with kind of the same problem here as the work from B to C, R. Delta v is zero because our volume is just V one meaning that our work done from D to A is going to be zero. So we have found our four values. Now we are ready to go ahead and calculate our network. So this is going to be E zero times V zero minus V one plus zero plus P one times v 1 - plus zero. Of course the zero terms don't matter. But here's what I'm gonna do, I'm gonna rearrange this a little bit. I'm gonna take a negative sign out of this term and you'll see why in just a second this is going to give us E one, B one minus zero zero times be -30. You can see that these terms are the same. So we can go ahead and combine these terms into one. So our network is going to be, let me scroll down just a little bit here. Our network is going to be equal to E 1 - IMEs the -70. That is the answer to part A now, part B. They want us to confirm that our network is just equal to the area of the loop. So let me scroll back up here so we can look at our loop here. So our loop is just a rectangle. So so the area of a rectangle is just going to be the width times the height. More importantly it is going to be the difference in these two values times the difference in these two values. So on our X axis here we have V one minus V not. And then on our y axis we have P one minus P not. And if you flip these terms around you will be able to see that our area is P one minus P not Times V 1 -3 Not which is exactly equal to our net work. So we found our network, we compared it to the area and this gives us an final answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one
Related Practice
Textbook Question
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
383
views
Textbook Question
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
723
views
Textbook Question
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
468
views
Textbook Question
Five moles of an ideal monatomic gas with an initial temperature of 127°C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?
872
views
Textbook Question
A gas in a cylinder is held at a constant pressure of 1.80 * 10^5 Pa and is cooled and compressed from 1.70 m^3 to 1.20 m^3. The internal energy of the gas decreases by 1.40 * 10^5 J. (c) Does it matter whether the gas is ideal? Why or why not?
397
views
Textbook Question
Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 * 10^6 J/kg and the boiling point is 120°C. At this pressure, 1.00 kg of water has a volume of 1.00 * 10^-3 m^3 , and 1.00 kg of steam has a volume of 0.824 m^3. (b) Compute the increase in internal energy of the water.
455
views