One mole of an ideal gas undergoes a cyclic change ABCD shown in the diagram. Find the net work done during the process. Take 1 atmospheric pressure = 10⁵ N/m².

Solution

Work done in a cyclic change is equal to the area of the cycle.
i.e. Work done, W = Area ABCDA
= DC X AD
This implies,

space DC equals 3 space lt space equals space 3 cross times 10 to the power of negative 3 end exponent straight m cubed space AD space equals space 3 Atm space equals space 3 cross times 10 to the power of 5 straight N divided by straight m squared

Thus,
Net work done,

straight W equals 3 cross times 10 to the power of negative 3 end exponent cross times 3 cross times 10 to the power of 5 space equals space 900 straight J

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