∙∙A cylinder with a piston contains 0.250 mol of ideal oxygen at a pressure of 2.40 ×10^5 Pa and

step 1 This is chapter 15, problem number 74. We have a cylinder with a piston, and we have 0 .25 moles

∙∙A cylinder with a piston contains 0.250 mol of ideal...

Key Concepts

Work in Thermodynamic Processes

Work in thermodynamic processes is the energy transfer due to volume change under pressure. The calculation of work depends on the type of process: it is the area under the curve on a p-V diagram, computed directly for constant-pressure processes, using logarithmic formulas for isothermal processes, and is zero for isochoric processes.

Isochoric Process

An isochoric process is characterized by a constant volume. Since the volume does not change, no work is done on or by the system during this process. However, energy transfer in the form of heat can change the pressure and temperature of the gas, as dictated by the ideal gas law.

Isothermal Process

An isothermal process occurs at a constant temperature. For an ideal gas, this means that the product of pressure and volume remains constant (Boyle's Law). The work done during an isothermal process is typically calculated using a logarithmic relation based on the ratio of the final and initial volumes.

Isobaric Process

An isobaric process is one in which the pressure remains constant throughout the process. In such processes, changes in volume are directly accompanied by proportional changes in temperature, and the work done by the gas can be calculated using the product of the constant pressure and the change in volume.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is a fundamental relation that interconnects pressure (P), volume (V), and temperature (T) for a given amount of gas (n) with the universal gas constant (R). It is essential for calculating any unknown state variable when a gas undergoes a thermodynamic process.

p-V Diagram

A p-V (pressure-volume) diagram is a graphical representation of the changes in pressure and volume in a thermodynamic system. It is a vital tool for visualizing processes, determining work done during expansion or compression (as the area under the curve), and understanding the sequence of state changes.

Transcript

00:01 This is chapter 15, problem number 74.

00:03 We have a cylinder with a piston, and we have 0 .25 moles of gas, of ideal oxygen in it.

00:11 So we're given the initial pressure as 2 .4 times 10 to 4 .5 pascals and the temperature to be 355 kelvin.

00:21 And there are a series of processes that we need to keep track on.

00:26 So first, the information that is given to us in a problem.

00:30 First, the gas expands isoparically, which means the pressure is constant, right? at least in this portion.

00:48 As the volume is going from an initial volume of v1 to double the v1.

00:58 Let's call this v2.

00:59 Okay.

01:02 And the second process, compressed, the gases then compressed to its initial volume, v1.

01:09 So there's compression, compression.

01:14 So to its initial value v1.

01:18 So from v2, we're going back to v1.

01:24 All right.

01:25 So the third process is that then the gas is cooled at constant volume, right? cooled at constant volume to its original pressure, to its original pressure of obviously p1, right? that's what it was.

01:56 In part a, we're asked to draw a pv diagram, which is going to be great, because it's going to make things clear for us.

02:04 Now, for all these three processes, we are asked to draw a pv diagram.

02:10 So then our y -axis is pressure, x being the volume.

02:16 So let's start with the first process.

02:18 The gas is expanding isolatedly, which means the pressure is going to be constant.

02:23 As the volume increases from an arbitrary, we want, let's say, to v2, right? so the corresponding pressure has to be the same throughout this point.

02:34 Process.

02:35 So this is a straight line.

02:37 As we're going to let's label it as one and two, something like this, right? the pressure is constant.

02:43 All right.

02:45 Now, the second process is the compression part, right? so as we compress the gas, initially it's going to give it, well, volume is going to be, it's going to go back to its initial volume.

03:04 So we're going to end up being somewhere here, right? but as we're compressing the gas, you expect the pressure to increase.

03:12 So the way to go there is by drawing something like this, right? as we're going from 2 through, let's say, 3 here, we're going back to the initial volume, and this is a compression, right? so, and then it's cooled at a constant volume, so we're still at v1 to its original pressure, p1.

03:33 So if this was p1, then we're going to go back here, right? all right.

03:40 So this is going to be, this is our pb diet room.

03:43 So we have three different processes, right? part b.

03:51 What is the temperature, t, during isothermal compression? we're referring to the temperature during, as we're going from 2 to 3, right? so during as we go into three, the temperature is constant.

04:18 So the temperature at two is going to be the same thing as the temperature at three, right? so this is either to t2 or t3.

04:29 It really does not matter.

04:32 But one thing that we know is the initial temperature, right? initial temperature here at this point.

04:38 So try to use that information.

04:40 So if we write down the ideal gas law, pvnrt, let's see what's going on in process one.

04:55 The gas is expanding the only thing that changes is the volume of the gas, so the pressure is constant.

05:02 So at point two, we still have the same pressure.

05:05 Then we come here, and then let's write it for the first point.

05:12 P v1 is going to be cool to nr t1 right so then um if we divide both sides by nr and nr so our t1 and then let's divide both size by v1 2 b1 so let's get rid of these two what we have is t1 over b1 on the right hand side equals to p over nr, right? so p over nr is not changing for the second point either.

05:54 So we can as well, we might as well do t2 over v2, right, equals p over nr.

06:01 So this equation also holds, which means t1 over v1 equals t2 over v2, right? and we're after finding t2.

06:11 So then t2 is going to be equal to t1.

06:15 V2 over v1 and we're given the fact that we 2 is twice of v1 so then let's write that down t2 equals t1 v2 is 2v1 over v1 these are going to cancel up then twice the t1 is going to give us our t2 t1 remember is 355 kelvin and 2 times 300, 355 kelvin gives us 710 kelvin, right? this also equals to t3 because that's an isothermal process along this line, right? so, all right...

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