The pressure and volume of a gas are changed along a path ABCA in the figure. The vertical divisions on the graph represent 4.0 x 10^5 Pa, and the horizontal divisions represent 4.5 x 10^-3 m3.

The diagram is a upright rectangle divided into small squares (8 up and 6 across). The horizontal 6 squares represents volume and the vertical 8 squares respresents pressure. Point A is 3 up and 2 across. Point B is 7 up and 2 across. Point C is 7 up and 5 across.

Determine the work done (including algebraic sign) in each segment of the path.

(a) A to B

(b) B to C

(c) C to A

For Further Reading

Physics HELP!!!!!!!! - drwls, Sunday, April 22, 2007 at 11:47pm
The work done BY the fluid is the integral under the PdV curve in each case. The dependence of P upon V will depend upon whether the volume change is isothermal, adiabatic, or something in between. You should be able to figure that relationship out from the location of the points

A to B = 0

B to C = 37800
W=[(7squares)(4.0 x 10^5Pa)][(3 squares)(4.5 x 10^-3 m^3)]= 37800

Am I right so far and if so how do I attempt #3.

Thanks

Your answers to (a) and (b) are correct. I am guessing that the line connecting C to A is a straight one, although in an actual engine pricess it may not be. For (c), the integral is negative since you are advancing backwards along the V axis. The work integral in the straight-line case is
-(1/2)(Pc-Pa)*(Vc-Va)
since (1/2)(Pc-Pa)
is the mean value of the pressure when going from C to A. You are basically caluculating the area of a trapezoid.

To determine the work done in each segment of the path, we need to calculate the area under the PdV curve. Let's go through each segment:

(a) A to B:
In this segment, the volume changes from 2 squares to 7 squares, while the pressure remains constant at 3 squares. Since the volume is changing at constant pressure, this implies an isobaric process.

To calculate the work done (W), we multiply the pressure (P) by the change in volume (ΔV):
W = P * ΔV

ΔV = 5 squares * 4.5 x 10^-3 m^3/square = 22.5 x 10^-3 m^3

P = 3 squares * 4.0 x 10^5 Pa/square = 12.0 x 10^5 Pa

W = 12.0 x 10^5 Pa * 22.5 x 10^-3 m^3 = 27000 J

Therefore, the work done in segment A to B is 27000 J.

(b) B to C:
In this segment, both the pressure and volume change. The volume increases from 7 squares to 7 squares, while the pressure remains constant at 7 squares. This indicates an isochoric (constant volume) process.

To calculate the work done, we again use the formula W = P * ΔV. However, in an isochoric process, the change in volume is zero, so the work done is also zero.

Therefore, the work done in segment B to C is 0 J.

(c) C to A:
In this segment, the volume decreases from 7 squares to 2 squares, while the pressure remains constant at 7 squares. This suggests an isobaric process.

To calculate the work done, we use the formula W = P * ΔV. However, in this case, the change in volume is negative (decreasing volume), so the work done will have a negative sign.

ΔV = -5 squares * 4.5 x 10^-3 m^3/square = -22.5 x 10^-3 m^3

P = 7 squares * 4.0 x 10^5 Pa/square = 28.0 x 10^5 Pa

W = 28.0 x 10^5 Pa * (-22.5 x 10^-3 m^3) = -63000 J

Therefore, the work done in segment C to A is -63000 J.

In summary, the work done in each segment is as follows:
(a) A to B: 27000 J
(b) B to C: 0 J
(c) C to A: -63000 J