One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V

1.	One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V diagram shown below, calculate the net work done in the process.1 atm = 106 dyne cm-2.
Answer» The work done in cyclic process is numerically equal to the area of the cycle. As the cycle ABCD is traced in clock- wise direction, the work done is positive. W = + area ABCD = DC x AD Now, DC = 4 - 3 = 3 litres = 3 x 10³ cm³ ( ∴1 lit = 10³ cm³) and AD = 5 - 2 = 3 atm = 3 x 10⁶dyne cm^-1 work done = 3 x 10³ x 3 x 10⁶ = 9 x 10⁹ ergs.

One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V diagram shown below, calculate the net work done in the process.1 atm = 106 dyne cm-2.

Answer»

The work done in cyclic process is numerically equal to the area of the cycle.

As the cycle ABCD is traced in clock- wise direction, the work done is positive.

W = + area ABCD

= DC x AD

Now, DC = 4 - 3 = 3 litres

= 3 x 10³ cm³

( ∴1 lit = 10³ cm³)

and AD = 5 - 2 = 3 atm

= 3 x 10⁶dyne cm^-1

work done = 3 x 10³ x 3 x 10⁶

= 9 x 10⁹ ergs.

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One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V diagram shown below, calculate the net work done in the process.1 atm = 106 dyne cm-2.

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