A evaluate the mean value of the voltage v volts over an...

(a) Evaluate the mean value of the voltage, v (volts) over an interval t = 2 to t = 10 seconds, where v = 6(1 - e-2t) Keep the answer as an exact value. [6 marks] (b) A gas is expanded in a piston according to the thermodynamics relation: PV1.4 = 1963, where P is the pressure and V the volume. The work done W by the gas is given by: V2 W = P dv V1 where V₁ and V₂ are the initial volume and the final volume of the gas respectively. Calculate W, the work done by the gas, if the volume V changes from 0.016 m³ to 0.09 m³. [6 marks]

Transcript

00:01 So here, a gas expands from an initial volume to a final volume.

00:05 And as that happens, the pressure is increasing linearly.

00:08 So we have to find the work done by the gas.

00:10 We have to look at the heat energy of weather efficiency and to determine heat that is absorbed, heat given off, absorbed from the hot reservoir and also heat that is giving off to the cold reservoir.

00:30 So that's what basically we are finding here in this question so we have been given the pressure and volume with this relation from the question you can see the pressure is 1 p minus 140 equals v minus 0 .5 multiplied by 260 minus 140 over 0 .72 minus 0 .75 now you you look at the graph here where we have p where we have the p against the p against the v the p against the v the pv graph here we have 0 .5 corresponding to the 140 kylpascal and 0 .72 corresponding to the 260 kilo pascal and this area here where we have shaded is the work done by the gas to work more on that later on so we have this p can also be written as 545 .45 v minus 132 .7 3 that's another way of putting it and so the work done by the gas is that area under the region so carrying on from there it means that the work done is given by half one 140 half 140 kilo pascal plus 260 plus 260 kilo pascal multiplied by the 0 .7 to minus 0 .5 so that the work done becomes equal to um 4 44 kilo pascal 44 kilo pascal 44 kilo pascal okay let me just convert everything straight away can look at the conversion later so we have the work done to be w equals 4400 0 newtine meter which is similar to 4400 joe which also is equal to 44 kilo joe okay now knowing this we cannot get to the a so for our a for our a we have the efficiency for the a for the a we have the efficiency of the heat engine we have efficiency heat engine efficiency of heat engine to be given us where we have the 8 .0 .35 so we are going to get this is the work done the work done all over the heat absorbed all over the heat absorbed from the reservoir from the reservoir that's ita one or tita one so this eater eater eater eater yeah eater okay so let's continue so we have 0 .35 equals 1 500 all over eta 1 and so we have a eta 1 to be 1 5 00 divided by 0 .35 divided by 0 .35 divided by 0 .35.

06:05 Okay so this becomes equals to 4 to 8 .5 .71 joe...

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