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 5.40 to 2.70 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 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 5.40 to 2.16 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

According to first law of thermodynamics,

delta U = q + w --------(1)

since temperature is kept constant, delta U = 0,

So, q = - w -----------(2)

and we know w = - P * delta V -------------(3)

so, q = + P * delta V

Case (1)

q = 2.00(2.70 - 5.40) + 2.50 (2.16 - 2.70) = - 4.05 L. atm = - 4.05 * 101.32 J = - 410.346 J

Case(2)

q = 2.50 (2.16 - 5.40) = - 0.95 L atm = - 0.95 * 101.32 J = - 96.254 J

Difference between q for two step process to single step process = - 96.254 - (-410.346) = 314.092 J

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