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

Question

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 6.40 to 3.20L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.20 to 2.56L .

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 6.40 to 2.56L 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?

Express your answer with the appropriate units.

Explanation / Answer

Solution :-

Lets calculate the amount of the work done in two step process

W= - P* delta T

    = - 2.00 atm * (3.20 L – 6.40 L)

   = 6.4 L atm

W= - 2.5 atm * (2.56 L – 3.20 L)

    = 1.6 L atm

Total work = 6.4 L atm + 1.6 L atm = 8 L atm

Lets convert it to Joules

8 L atm * 101.3 J / 1 L atm = 810.4 J

Now lets calculate the work in the one step process

W= - P* delta V

    = -2.50 atm * (2.56 L – 6.40 L)

    = 9.6 L atm

9.6 L atm *101.3 J / 1 L atm = 972.48 J

Now lets find the difference in the enrgy of the both process

q =972.48 J – 810.4 J

   = 162.08 J

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