A gas is compressed from an initial volume of 042 m3 to a...

A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pressure changes with volume according to the relation P = aV + b, where a = -1200 kPa/m3 and b = 600 kPa. Calculate the work done during this process (a) by plotting the process on a P-V diagram and finding the area under the process curve and (b) by performing the necessary integrations.

Transcript

00:01 In this problem, we have been given that the pressure varies with the volume in a thermodynamic process as per the relation p equals a, v plus b, provided that a and b are constants and we have been given their values.

00:15 So let's change the statement of p in terms of the given values of a and b.

00:21 So that would be minus 1200 into 10 raise to 3, v plus 600 into 10 raise to 3.

00:34 So that's the value of a &p.

00:36 So we have expressed this pressure in terms of the volume.

00:40 And we need to determine the work done when the gas is compressed from 0 .42 cubic meter, that's the initial volume, to a volume of 0 .1.

00:54 Two cubic meter.

00:56 So here first we can figure out the work using the pv graph.

01:01 So for that we observe that p equals a v plus b.

01:06 Here we compared this with the standard equation of the line y is equal to mx plus c and we observe that the slope here is negative.

01:16 That means the line of the straight line will be making an obtuse angle with respect to the volume axis.

01:27 So here we can put the volume v as zero and we can get p as 600 into 10 ways to 3.

01:35 So that is somewhere over this.

01:39 So that's 600.

01:41 Let's take this p, the pressure in terms of multiples of 10 ways to 3.

01:46 And when p is zero, at that instant, v here will be equal to 0 .5.

01:55 So take this v in terms of cubic meters.

01:58 So that will be somewhere over here.

02:01 So basically this will be a straight line joining.

02:04 And we need to determine the work done to compress the gas from 0 .42 that will be somewhere over here.

02:12 That's 0 .42 to 0 .12.

02:15 So we observe that since this is a pressure volume curve, so the area here will give us the required work done and the area should be below this.

02:26 So work done can be computed by taking the area.

02:28 Of the trapezium, that's half times the height and the height will be 0 .42 minus 0 .12 times the sum of parallel sides here.

02:42 So to compute the parallel sides, we can just put in the value of v once here as 0 .42 and get the value of b...

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