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.
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