91Ó°ÊÓ

1. Initial volume,$\left({\mathrm{V}}_{\mathrm{o}}\right)=1.0{\mathrm{m}}^3$
2. Final volume,$\left({\mathrm{V}}_1\right)=4{\mathrm{V}}_{\mathrm{o}}=4.0{\mathrm{m}}^3$
3. Initial pressure,$\left({\mathrm{P}}_{\mathrm{o}}\right)=40\mathrm{Pa}$
4. Final pressure,$\left({\mathrm{P}}_1\right)=\frac{{\mathrm{P}}_{\mathrm{o}}}{4.0}=10\mathrm{Pa}$

Work is any activity that includes exerting force on an object to displace (move) it. Work is defined as the application of a force to an item that causes it to be displaced (moved) across a certain distance in the force's direction. Energy transfer is another way to describe work. It is given that the volume changes from 1.0 m³to 4.0 m³and pressure changes from 40 Pa to 10 Pa.

Formulae:

When pressure is constant and volume over path changes, then we can calculate work done as: $\mathrm{W}=\mathrm{p}∆\mathrm{V}$ …(¾±).

When there is a change in pressure but the volume is constant, and then work done is given as: $\mathrm{W}={∫}_{{\mathrm{V}}_{\mathrm{o}}}^{{\mathrm{V}}_1}\mathrm{p}d\mathrm{V}$ …(¾±¾±)

As shown in fig., one part of path A represents the constant pressure, which is 40 Pa, and there is change in volume from 1 m³ to 4 m³.

Then the work done via path A can be calculated using equation (i) as:

$\begin{array}{rcl}{\mathrm{W}}_{\mathrm{A}}& =& \mathrm{p}\left({\mathrm{V}}_1-{\mathrm{V}}_{\mathrm{o}}\right)\\ & =& 40\mathrm{Pa}\left(4{\mathrm{m}}^3-1{\mathrm{m}}^3\right)\\ & =& 1.2×{10}^2\mathrm{J}\end{array}$

During the process, as shown inthe figure, for the other part of path A, the volume is constant, so we can say that no work is done during this process.

So the total work done over the entire path A is 120 J.

We know that the pressure is a function of volume, so take the volume integral as

${\mathrm{W}}_{\mathrm{B}}=∫\mathrm{pdV}$

As pressure is a linear function of volume, we write

$\mathrm{p}=\mathrm{a}+\mathrm{bV}$ …(¾±¾±¾±)

So we have to find first the values of constants a and b

We can write, $40\mathrm{Pa}=\mathrm{a}+\mathrm{b}×1{\mathrm{m}}^3$and $10\mathrm{Pa}=\mathrm{a}+\mathrm{b}×4{\mathrm{m}}^3$

From these, we get the values a = 50 Pa and b = -10 Pa/m³

We write equation (iii) as$\mathrm{p}=50\mathrm{Pa}-\left(10\mathrm{Pa}/{\mathrm{m}}^3\right)\mathrm{V}$

Using equation (ii), the work done in the path B is given as:

$\begin{array}{rcl}\mathrm{W}& =& {∫}_1^4\left(50-10\mathrm{V}\right)d\mathrm{V}\\ & =& {\left(50\mathrm{V}-10\frac{{\mathrm{V}}^2}{2}\right)}_1^4\\ & =& 200\mathrm{J}-50\mathrm{J}-80\mathrm{J}+5\mathrm{J}\\ & =& 75\mathrm{J}\end{array}$

Hence, the total work done by the body in the path B is 75 J

Now, along path C, one of the parts represents the constant pressure, which is 10 Pa and along the path, volume changes from 1 m³ to 4 m³.

So, the work done in the path C using equation (i) is given as:

$\begin{array}{rcl}{\mathrm{W}}_{\mathrm{C}}& =& 10\mathrm{Pa}\left(4{\mathrm{m}}^3-1{\mathrm{m}}^3\right)\\ & =& 30\mathrm{J}.\end{array}$

And other part of the path C is at constant volume, so no work is done. So, the total work done over the entire path C is 30 J