Question

How much work is done on the gas in the process shown in the figure?

Answer 1

Concepts and reason

The concept used to solve this problem is work done.

First, by splitting the space under the curve into a rectangle and a triangle, find the area of the rectangular space. Then find the area of the triangular space. Finally, combine the areas to find the area under the curve and calculate the work done.

Fundamentals

Gasses can do work by either expansion or compression against an external pressure. This work done is called pressure-volume work. This work done is given by the expression,

$W = - PdV$

Here, $W$ is the work done against the external pressure, $P$ is the pressure and $dV$ is the change in volume.

The change in volume is,

$dV = {V_{final}} - {V_{initial}}$

Here, ${V_{final}}$ is the final volume and ${V_{initail}}$ is the initial volume.

Thus, the work done will be negative if the gas is expanding because the change in volume will be positive. Similarly, the work done is positive if the gas is compressed because the change in volume will be negative.

The area of a rectangle is given by the equation

$A = ab$

Here, $a$ is the length of the rectangle and $b$ is the breadth of the rectangle.

The area of a triangle is,

$A = \frac{1}{2}bh$

Here, $b$ is the base length of a triangle and $h$ is the height of the triangle from the base.

Find the area of the rectangular space.

The area of a rectangle is,

$A = ab$

Substitute $200\,{\rm{c}}{{\rm{m}}^3}$ for $a$ and $200\,{\rm{kPa}}$ for $b$ .

The area of the rectangular space is,

$\begin{array}{c}\\A = \left( {\left( {200\,{\rm{c}}{{\rm{m}}^3}} \right)\left( {\frac{{{{10}^{ - 6}}\,{{\rm{m}}^3}}}{{1\,{\rm{c}}{{\rm{m}}^3}}}} \right)} \right)\left( {\left( {200\,{\rm{kPa}}} \right)\left( {\frac{{{{10}^3}\,{\rm{Pa}}}}{{1\,{\rm{kPa}}}}} \right)} \right)\\\\ = 40\,{{\rm{m}}^3}.{\rm{Pa}}\\\\ = {\rm{40}}\,{\rm{J}}\\\end{array}$

Find the area of the triangular space.

The area of a triangle is,

$A = \frac{1}{2}bh$

Substitute $200\,{\rm{c}}{{\rm{m}}^3}$ for $b$ and $200\,{\rm{kPa}}$ for $h$ .

The area of the triangular space is,

$\begin{array}{c}\\A = \frac{1}{2}\left( {\left( {200\,{\rm{c}}{{\rm{m}}^3}} \right)\left( {\frac{{{{10}^{ - 6}}\,{{\rm{m}}^3}}}{{1\,{\rm{c}}{{\rm{m}}^3}}}} \right)} \right)\left( {\left( {200\,{\rm{kPa}}} \right)\left( {\frac{{{{10}^3}\,{\rm{Pa}}}}{{1\,{\rm{kPa}}}}} \right)} \right)\\\\ = 20\,{{\rm{m}}^3}.{\rm{Pa}}\\\\ = {\rm{20}}\,{\rm{J}}\\\end{array}$

Find the work done on the gas.

The total work done is the area under the given curve, which is the sum of the area of the rectangular space and the triangular space.

$\begin{array}{c}\\W = {A_{total}}\\\\ = {A_{{\rm{rectangle}}}} + {A_{{\rm{triangle}}}}\\\end{array}$

Substitute $40\,{\rm{J}}$ for ${A_{{\rm{rectangle}}}}$ and 20 J for ${A_{{\rm{triangle}}}}$

The work done is

$\begin{array}{c}\\W = 40\,{\rm{J + 20 J}}\\\\ = {\rm{60 J}}\\\end{array}$

The work done in expanding a gas is negative. In the graph shown, the volume of the gas is increasing. So, the gas is expanding and the work done is negative. Thus,

$W = - 60\,{\rm{J}}$

Ans:

The work done on the gas is $- 60\,{\rm{J}}$ .

Answer 2

is -60J.

Answer 3

Work done is given by the area under the PV diagram
Here Work done W = (1/2) base(height)
= (1/2) (300 cm^3- 100 cm^3)(400 kPa - 200 kPa)
= (1/2)(200 cm^3)(200 kPa)
= (1/2)(200*10^-6 m^3)(200*10^3 Pa)
= 20 J

Answer 4

area under the lines.
it is
(200+400)*(300-100)*2/2 (kPa*cm3)
1.2e5(kPa*cm3)=120(Pa*m3)=120(J)

Answer 5

The work done is -60J. You have to add the square box under the triangle as part of your area which adds another 40J to the 20J from the triangle.