sapling learning A pressure versus volume (pvy diagram is shown in the figure be
ID: 1485046 • Letter: S
Question
sapling learning A pressure versus volume (pvy diagram is shown in the figure below for a system. The arrows of the curve indicate the direction of the process and the points of interest are labelled. Calculate the amount of work done on the system from 0-2 and then for the entire curve 0 5. The values for the points in the diagram are Pressure (Pa) volume (m3) Vo 258 po 1.37 x 104 p1 1.37 x 104 174 p2 5.76 x 103 p3 5.76 x 103 2.92 x 103 V4 13.9 bps-100 x 103 V5 9.04 Number W. 02 V (cubic p (Pa) NumberExplanation / Answer
The work done is equal to the area of the curve of the PV graph.
The process from 0 to 1 form a rectangle having length and breadth given as
Length = p0 = p1
= 1.37 x 104 Pa
Breadth = V0 – V1
= 25.8 m3 – 21.1 m3
= 4.7 m3
The work done from 0 to 1 is the area of the rectangle.
Thus,
W01 = area
= Length x Breadth
= (1.37 x 104 Pa)( 4.7 m3)
= 6.439 x 104 J
The process from 1 to 2 forms a trapezium having as follows:
Perpendicular distance between the parallel sides
h = V1 - V2
= 21.1 m3 – 17.4 m3
= 3.7 m3
Parallel sides
a = p1
= 1.37 x 104 Pa
b = p2
= 5.76 x 103 Pa
The work done from 0 to 1 is the area of the trapezium.
Thus,
W12 = area
= h {( a + b ) /2}
= (3.7 m3) (1.37 x 104 Pa + 5.76 x 103 Pa) / 2
= 3.6001 x 104 J
Thus, the work done from 0 to 2 is
W02 = W01 + W12
= 6.439 x 104 J + 3.6001 x 104 J
= 10.0391 x 104 J
Rounding off to three significant figures, the amount of work done on the system from 0 - 2 is 10.0 x 104 J.
The process from 2 to 3 form a rectangle having length and breadth given as
Length = p2 = p3
= 5.76 x 103 Pa
Breadth = V2 – V3
= 17.4 m3 – 13.9 m3
= 3.5 m3
The work done from 2 to 3 is the area of the rectangle.
Thus,
W23 = area
= Length x Breadth
= (5.76 x 103 Pa)( 3.5 m3)
= 2.016 x 104 J
The work done from 3 to 4 is zero as there is no change in the volume.
Thus,
W34 = 0 J
The process from 4 to 5 forms a trapezium having as follows:
Perpendicular distance between the parallel sides
h = V4 – V5
= 13.9 m3 – 9.04 m3
= 4.86 m3
Parallel sides
a = p4
= 2.92 x 103 Pa
b = p5
= 1.00 x 103 Pa
The work done from 4 to 5 is the area of the trapezium.
Thus,
W45 = area
= h {( a + b ) /2}
= (4.86 m3) (2.92 x 103 Pa + 1.00 x 103 Pa) / 2
= 0.95256 x 104 J
Now, the work done on the system from 0 – 5 is
W05 = W01 + W12 + W23 + W34 + W45
= 6.439 x 104 J + 3.6001 x 104 J + 2.016 x 104 J + 0 J + 0.95256 x 104 J
=13.008 x 104 J
Rounding off to three significant figures, the amount of work done on the system from 0 – 5 is 13.0 x 104 J.
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