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Initially at a temperature of 90.0 ∘C, 0.280 m3 of air expands at a constant gauge...

Initially at a temperature of 90.0 ∘C, 0.280 m3 of air expands at a constant gauge pressure of 1.38×105 Pa to a volume of 1.50 m3 and then expands further adiabatically to a final volume of 2.40 m3 and a final gauge pressure of 2.24×104 Pa

Compute the total work done by the air. CV for air is 20.8 J/(mol⋅K) .

I got 551460J and it was wrong.

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Homework Answers

Answer #1

Given is:-

Initial pressure    volume   

FInal pressure   

first of all we have to draw the P-V graph for the given data. Then the area under the curve gives us the desired result.

At first the pressure is constant, only volume is increasing, therefore the area under the curve is

In the second part gas increases it's volume adiabatically

Thus we have to calculate the area in two parts, area of triangle + area of rectangle

Area of triangle is

Finally

Thus total work done by air is

Thus the total work done by the air in whole number of processes is 199140 J

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