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  • Find out the magnitude of work (in kJ) done by one mole of an ideal gas for expansion. <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAOYAAADHCAYAAAANmi8oAAAgAElEQVR4nO2de3RU1fXH

Find out the magnitude of work (in kJ) done by one mole of an ideal gas for expansion.

Option: 1

-1


Option: 2

2


Option: 3

6


Option: 4

60


Answers (1)

best_answer

Work is a path function and not state function and area under \mathrm{P-V} curve gives work.

So, work will be the total area of the Trapezium

\therefore \mathrm{|work| = \frac{1}{2} \times \textrm{(sum of parallel sides)} \times \textrm{(distance between the parallel sides)}}

\therefore \mathrm{|work| = \frac{1}{2} \times 6 \times 20=60 \times 10^{-5} \ bar-lit}

\therefore \mathrm{|work| = 60 \times 10^{5} \ bar-lit=60 \times 10^{5} \times 10^{-3} bar-m^3=6000 J}

Thus, the magnitude of work done is 6 kJ

 

Posted by

Kuldeep Maurya

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