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
To determine the work done in each segment of the path, we need to calculate the area under the pressure-volume (P-V) curve for each segment.
(a) A to B:
Looking at the diagram, we see that the vertical position remains constant (3 up), while the horizontal position changes by 1 (2 to 3). To calculate the work, we need to find the area under the curve between these two points. Since the vertical divisions represent 4.0 x 10^5 Pa and the horizontal divisions represent 4.5 x 10^-3 m^3, the area under the curve can be calculated as:
Work (A to B) = (3 squares) * (4.0 x 10^5 Pa) * (1 square) * (4.5 x 10^-3 m^3) = 54,000 J (negative, as work is done BY the system)
(b) B to C:
In this segment, the vertical position remains constant (7 up), while the horizontal position changes by 3 (2 to 5). We can calculate the area as:
Work (B to C) = (7 squares) * (4.0 x 10^5 Pa) * (3 squares) * (4.5 x 10^-3 m^3) = 378,000 J (negative, as work is done BY the system)
(c) C to A:
To calculate the work for this segment, we need to find the area under the curve between points C and A. From the diagram, we can see that the vertical position changes by -4 (7 down to 3), while the horizontal position changes by -3 (5 to 2). The work can be calculated as:
Work (C to A) = (-4 squares) * (4.0 x 10^5 Pa) * (-3 squares) * (4.5 x 10^-3 m^3) = -216,000 J (positive, as work is done ON the system)
Therefore, the work done in each segment of the path is:
(a) A to B: -54,000 J
(b) B to C: -378,000 J
(c) C to A: 216,000 J