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

Express your answer with the appropriate units.

Explanation / Answer

Ist process

step 1 : work done = -PdV = - 2 x (3.2 - 6.4) = 6.4 J

step 2 : work done = -PdV = - 2.5 x (2.56 - 3.2) = 1.6 J

total work done = q = 6.4 + 1.6 = 8.0 J

IInd proces

work done = q = -PdV = - 2.5 x (2.56 - 6.40) = 9.6 J

The difference in q for two step process to one step process = 9.6 - 8.0 = 1.6 J

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