Suppose that after the final process in Fig. 12.22 (see...

Name: Problem 11, Chapter 12: College Physics
Uploaded: 2021-05-18T03:31:40-08:00
Duration: 5 min
Channel: Satpal Satpal
Description: Suppose that after the final process in Fig. 12.22 (see Exercise 10), the pressure of the gas is decreased isometrically from 1.0 ×10^5 Pa to 0.70 ×10^5 Pa, and then the gas is compressed isobarically from 1.0 m^3 to 0.80 m^3. What is the total work done in all of these processes, including 1 through 5 ?

Suppose that after the final process in Fig. 12.22 (see Exercise 10$)$, the pressure of the gas is decreased isometrically from $1.0 \times 10^{5} \mathrm{~Pa}$ to $0.70 \times 10^{5} \mathrm{~Pa}$, and then the gas is compressed isobarically from $1.0 \mathrm{~m}^{3}$ to $0.80 \mathrm{~m}^{3}$. What is the total work done in all of these processes, including 1 through $5 ?$

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Key Concepts

Isobaric (Constant Pressure) Process

In an isobaric process, the pressure remains constant throughout the change in state. The work done in such a process is calculated by multiplying the constant pressure by the change in volume (W = P ?V), which represents the energy transferred as work during expansion or compression.

Work in Thermodynamic Processes

Work in thermodynamics is defined as the energy transferred by a system to or from its surroundings as a result of volume change under pressure. It is typically computed by integrating the pressure with respect to volume change, and the nature of the process (e.g., isometric, isobaric) determines how this integral is evaluated.

Isometric (Constant Volume) Process

This process involves changing other state variables while keeping the volume constant. In an isometric process, the work done by or on the system is zero because there is no change in volume, even if the pressure or temperature of the system changes.

Transcript

00:01 So let us calculate work done in this process.

00:05 First process 1 to 2, 1 to 2.

00:09 Work done in process 1 to 2 is equal to pressure multiplied by change in volume.

00:14 Since process 1 to 2 is a isometric process, so it will be pressure multiplied by change in volume will be 0.

00:23 It will be 0 jol.

00:26 Since delta v1 2 is equal to 0.

00:31 Now during process 2 to 3 work done in process 2 to 3 is equals to pressure multiplied by v2 minus v1.

00:42 Let us put the value of pressure as 0 .5 multiplied by 10 to the power of 5.

00:49 The value of v2 is 0 .75 minus v1 is 0 .25.

00:57 So work done in process 2 to 3 will come out to be 2 .5 multiplied by 10 to the 2 .5 multiplied by 10.

01:05 To the power of 4 jule.

01:08 This is the process work done in process 2 to 3.

01:10 Now work done in process 3 to 4, w 3 2 will be p multiplied by del v t 2 4.

01:27 Now since volume is constant in process 3 to 4 hence the work done will be 0.

01:35 Since volume v delta v here is 0 since del v is equal to 0 here.

01:45 Okay, now, ongdon in process 425 is equals to p2 -2 -5 is equal to p -2 -dus minus v1 -dase.

01:59 So let's put the values here...

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