A cylinder with initial volume V contains a sample of gas at pressure p. On one end of the cylinder, a piston is let free to move so that the gas slowly expands in such a way that its pressure is directly proportional to its volume. After the gas reaches the volume 3V and pressure 3p, the piston is pushed in so that the gas is compressed isobarically to its original volume V. The gas is then cooled isochorically until it returns to the original volume and pressure. Find the work W done on the gas during the entire process. Find the amount of work W done on the gas during the entire process. Express your answer in terms of some or all of the variables p and V.

For this question, we need to draw a PV diagram (a pressure-volume diagram), which I have made and attached in the attachment. So please refer to that attachment. I will be using this diagram to solve for the work done on the gas.

So, Please refer to the attachment number 1. where x -axis is of volume and y-axis is of pressure.

As we know that, the work done on the gas is equal to the area under the curve.

W = Area of the triangle

W = 0.5 x ( base) x ( height)

W = 0.5 x (BC) x (AC)

W = 0.5 x (3V-V) x (3P-P)

W = 2PV

Hence, the work done on the gas = 2PV

bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-02-04T13:08:31+01:00

Work done is defined as the area under the curve.The work done on the gas is [tex]2pV[/tex].

To find the work done, we need to know about the work done.

What is work done?

Work done is defined as the area under the curve.

Given information-

The initial volume of the cylinder is [tex]V[/tex].

The initial pressure of the gas inside the cylinder is [tex]p[/tex]

Let the gas is the ideal gas.

When the gas reaches the volume of [tex]3V[/tex] and pressure [tex]3p[/tex], the piston is pushed in so that the gas is compressed isobarically to its original volume [tex]V[/tex].

The isobaric process in which the pressure remain constant.

As the work done is equal to the area under the curve. To find the work done draw the pv curve for the given problem and find its area as shown in the figure attached below.

As the isobaric process is horizontal line on the pv graph.Therefore the work done can be calculate by the area under the triangle. thus,

[tex]W=\int {pv} \, dx[/tex]

As the area of the triangle is half of the product of the base and height of the triangle. Thus,

[tex]W=\dfrac{1}{2} \times MN\times LM[/tex]

Here in the triangle the height is,

[tex]LM=3p-p\\LM=2p[/tex]

The base of the triangle is,

[tex]MN=3V-V\\MN=2V[/tex]

Put the values and find the work done as,

[tex]W=\dfrac{1}{2} \times 2V\times 2p\\W=2pV[/tex]

Hence the work done on the gas is [tex]2pV[/tex].

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