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An ideal gas undergoes two processes. In the first one, the volume remains constant at 0.20...

An ideal gas undergoes two processes. In the first one, the volume remains constant at 0.20 m3 and the pressure increases from 9.0x104 Pa to 7.0x105 Pa. In the second one, the gas is compressed from 0.20 m3 to 0.12 m3 at a constant pressure of 7.0x105 Pa. Calculate the total work done on the gas.

a. - 48.0 kJ

b. - 56.0 kJ

c. - 64.0 kJ

d. - 72.0 kJ

e. None of the above

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Answer #1

as we know

W= P x dV

W1 = 0 J ( dV = 0 )

W2 = 7 x 105 Pa x (-0.08 m3)

W2 = -56 KJ . Answer

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