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An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas...

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.90 to 2.95 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36 L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.90 to 2.36 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

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Answer #1

When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.90 to 2.95L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36L .

Step 1 :

w1 = - p (V2 - V1)

    = 2.00 (2.95 - 5.90)

   = 5.9 L.atm

   = 597.8 J

w2 = 2.50 x (2.36 - 2.95)

     = 1.475 L.atm

     = 149.45 J

w = w1 + w2 = 747.25 J

step 2 :

w3 = 2.50 x (2.36 - 5.90 )

   = 8.85 L.atm

   = 896.7 J

change in work = w - w3

                         = 747.25 - 896.7

change in work = - 149.J

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